MATH 242: Algebraic number theory Matthew Morrow (
[email protected])
Contents 1 A review of some algebra
2
2 Quadratic residues and quadratic reciprocity
4
3 Algebraic numbers and algebraic integers
12
4 Algebraic number fields 4.1 First example: Gaussian integers . . . . . . . . 2πi/3 4.2 Second example: Z[ω] . . . . √ where ω = e 4.3 Third example: Z[ −5] . . . . . . . . . . . . . 4.4 Introductory tools for studying number fields: Norm, Trace, Discriminant, and Integral bases 4.4.1 Norm and Trace . . . . . . . . . . . . . 4.4.2 Discriminant . . . . . . . . . . . . . . . 4.4.3 Integral bases . . . . . . . . . . . . . . . 4.4.4 Applications of integral bases to number
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5 Main theoretic properties of DF 27 5.1 The class group, its finiteness, and cancellation of ideals . . . . . . . . . . . . . . . . . . . . . 28 5.2 Dedekind domains and Unique factorisation of ideals . . . . . . . . . . . . . . . . . . . . . . . 31 5.3 Norms of ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 6 Explicitly constructing ideals in DF and generators of ClF 7 Calculations of class groups of quadratic extensions, and applications 7.1 d = −5 . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 d = −6 . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 d = −7 . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 d = −10 . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 d = −13 . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 d = −14 . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 d = −15 . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 d = −17, −19 . . . . . . . . . . . . . . . . . . . . . . 7.9 d = −23 . . . . . . . . . . . . . . . . . . . . . . . . . 7.10 d = −30 . . . . . . . . . . . . . . . . . . . . . . . . . 7.11 d = 2, 3, 5, 6, 7, 11, 13, 17, 21, 29, 33, 37, 41 . . . . .
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42 43 46 46 46 47 47 48 48 48 48 49
8 Cyclotomic extensions and Fermat’s Last Theorem 50 8.1 Q(ζ) and its ring of integers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8.2 Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 9 Ramification theory 56 9.1 Ramification in quadratic extensions and quadratic reciprocity . . . . . . . . . . . . . . . . . 58 9.2 Ramification in cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 10 A new proof of quadratic reciprocity
62 1
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MATH 242 Algebraic Number Theory
1
A review of some algebra
In this course all rings R are commutative with unity. Algebraic number theory historically began as a study of factorization, and so we begin by reviewing properties of factorization in general commutative rings. You are expected to be familiar with this material, which may be found in any standard algebra textbook, since it will gradually be needed in the course. Definition 1.1. Let R be a ring (commutative with unity!). (i) R is an integral domain if and only if whenever a, b ∈ R satisfy ab = 0, then a = 0 or b = 0. (ii) We write a|b to mean that a divides b, i.e. that there exists c ∈ R such that ac = b. (iii) An integral domain R is a principal ideal domain (PID) if and only if every ideal I of R is principal, i.e. I = hai for some a ∈ R, where hai = aR is the principal ideal generated by a; this notation will be used throughout. (iv) An element u of R is called a unit if and only if it has a multiplicative inverse. R× denotes the group of units. An element p of R is called irreducible if and only if it is not a unit and whenever p = ab for some a, b ∈ R then a or b is a unit. Two elements a, b ∈ R are called associates if and only if there is a unit u ∈ R such that a = bu. An integral domain R is a unique factorization domain (UFD) if and only if every non-unit of R is a finite product of irreducible elements and whenever two such products p1 . . . pr , q1 . . . qs are equal, then r = s and, up to reordering, pi and qi are associates. (v) An integral domain R is a Euclidean domain (ED) if and only if there is a map ν : R \ {0} → Z≥0 with the following two properties: • For all a, b ∈ R with b 6= 0, there exist q, r ∈ R such that a = bq + r, where either ν(r) < ν(b) or r = 0. • For all non-zero a, b ∈ R, ν(a) ≤ ν(ab). The map ν is then called a Euclidean norm on R. (vi) R is a ED =⇒ R is a PID =⇒ R is a UFD. Example 1.2. The following examples will frequently appear and should be familiar to you: (i) The ring of integers Z is a ED, with Euclidean norm ν(n) = |n|. Hence it is also a PID and a UFD. (ii) If F is a field then the polynomial algebra F [X] is a ED with Euclidean norm ν(f (X)) = deg f (X) = d, if f (X) = a0 + a1 X + · · · + ad X d with ad 6= 0. Hence it is also a PID and a UFD. Next we review the rings Z/nZ, which also occur frequently in algebraic number theory: Definition 1.3. Given n ∈ Z, the ring Z/nZ is the quotient of Z by the principal ideal nZ. Given x, y ∈ Z, we say x is congruent to y mod n, and write x ≡ y mod n if and only if n|x − y, which is equivalent to saying that x + nZ = y + nZ, or that x and y have the same class in Z/nZ. We may write “x mod n” to mean the class of x in Z/nZ, when it is not likely to cause confusion. Note: I will never use the notation Zn for Z/nZ. If x ∈ Z, then x mod n is a unit in the ring Z/nZ if and only if x is coprime to n. In particular, if p ∈ N is a prime number, then (Z/pZ)× is a group of order p − 1: any element is equal to exactly one of 1, 2, . . . , p − 1 mod p. The following results will be used: 2
MATH 242 Algebraic Number Theory
3
Theorem 1.4. Let p ∈ N be a prime number. (i) (“Fermat’s little theorem”) If a ∈ Z is coprime to p, then ap−1 ≡ 1 mod p. (ii) (Z/pZ)× is a cyclic group. Proof. (i): [20.1, FRA]. (ii): [23.6, FRA]; more generally, any finite subgroup of the multiplicative group of a field is cyclic. Corollary 1.5. Let p ∈ N be a prime number. Then there exists g ∈ Z with the following properties: (i) If a ∈ Z is coprime to p, then a ≡ g r mod p for some r ≥ 0. (ii) If r ∈ Z is such that g r ≡ 1 mod p, then p − 1|r. Proof. Let g ∈ Z be such that g mod p generates the cyclic group (Z/pZ)× ; then (i) and (ii) are restatements of that fact that (Z/pZ)× is cyclic of order p − 1. The classical name for an integer g with the properties of the following corollary is a primitive root modulo p, but we will use this notation very little.
3
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MATH 242 Algebraic Number Theory
2
Quadratic residues and quadratic reciprocity
In this section we study so-called quadratic residues modulo an odd prime number p, which is essentially an analysis of when the equation X 2 ≡ a mod p has an integer solution, for a fixed value of a ∈ Z. These results and ideas will frequently reappear during the course when we study explicit examples, while the key theorem, namely the Law of Quadratic Reciprocity, will play a key role in describing the arithmetic of quadratic number fields in section 9.1. Definition 2.1. Let p ≥ 3 be a prime number, and suppose that a ∈ Z is coprime to p. Then a is said to be a quadratic residue modulo p if and only if there exists x ∈ Z satisfying x2 ≡ a mod p. (Note: The condition that a is coprime to p is important; 0 is not a quadratic residue modulo p, even though 02 ≡ 0 mod p.) Example 2.2. Here are basic examples: (i) 3 is a quadratic residue modulo 13, since 42 = 16 ≡ 3 mod 13. (ii) 02 ≡ 0 mod 3, 12 ≡ 1 mod 3, and 22 ≡ 1 mod 3. So if x is an integer then x2 ≡ 0 or 1 mod 3. Therefore 2 is not a quadratic residue mod 3. The following characterisation of quadratic residues is needed for later proofs: Lemma 2.3. Let p ≥ 3 be a prime number and suppose that a ∈ Z is coprime to p. Then a is a quadratic p−1 residue modulo p if and only if a 2 ≡ 1 mod p. Proof. ⇒: Suppose first that a is a quadratic residue modulo p; then a ≡ x2 mod p for some x ∈ Z. Then x is also coprime to p, and so xp−1 ≡ 1 mod p by Fermat’s little theorem. Therefore a
p−1 2
≡ (x2 )
p−1 2
= xp−1 ≡ 1 mod p.
p−1
⇐: Conversely, assume that a 2 ≡ 1 mod p. Let g be a primitive root modulo p (see the end of section 1); then a ≡ g r for some r ≥ 0. Our assumption implies that (g r )
p−1 2
≡ 1 mod p.
As g mod p generates the cyclic group (Z/pZ)× of order p − 1, this implies that p − 1|r p−1 2 . That is, r/2 is an integer, and so a ≡ (g r/2 )2 mod p; i.e. a is a quadratic residue modulo p. In order to develop a way of manipulating quadratic residues, and to clarify their properties, the following piece of notation is absolutely fundamental: Definition 2.4. The Legendre symbol, for p ≥ 3 a prime number and a ∈ Z, is defined by if p - a and a is a quadratic residue modulo p, 1 a = −1 if p - a and a is not a quadratic residue modulo p, p 0 if p|a. Example 2.5. To clarify the definition we offer the following examples: (i) 3 is a quadratic residue modulo 13, by the previous example, so
3 13
= 1.
(ii) 5 is not a quadratic residue modulo 7 (since 12 ≡ 62 ≡ 1, 22 ≡ 52 ≡ 2, 32 ≡ 42 ≡ 4 mod 7), so 5 13 = −1. (iii) −6 = 0 since 3 divides −6. 3 4
MATH 242 Algebraic Number Theory
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The Legendre symbol is a convenient tool for discussing and manipulating quadratic residues; the following are some of its key properties: Lemma 2.6. Let p ≥ 3 be a prime number and let a, b ∈ Z. Then p−1 (i) ap ≡ a 2 mod p (“Euler’s lemma”); (ii)
ab p
=
a p
b p
;
(iii) If a ≡ b mod p then
a p
=
b p
.
Proof. If p divides a or b then these identities are all trivial; so assume that p does not divide a or b. (i) By Fermat’s little theorem, ap−1 ≡ 1 mod p. So p divides ap−1 − 1 = (a p−1
p−1
p−1 2
− 1)(a
p−1 2
+ 1),
p−1
whence p divides a 2 − 1 or a 2 + 1, i.e. a 2 ≡ ±1 mod p. But according to lemma 2.3, a is a p−1 quadratic residue modulo p if and only if a 2 ≡ 1 mod p. p−1 p−1 p−1 b (ii) Since (ab) 2 = a 2 b 2 , part (i) implies that ab ≡ ap mod p. As all these terms are ±1, p p b and since p is odd, it follows that ab = ap p p . (iii) This is quite clear: the definition of a quadratic residue only depends on a mod p.
From the proposition we obtain an important and useful corollary: Proposition 2.7 (Legendre symbol of −1). Let p ≥ 3 be a prime number. Then p−1 (i) −1 = (−1) 2 ; p (ii) −1 is a quadratic residue modulo p if and only if p is congruent to 1 modulo 4. Proof. For (i), apply part (i) of the previous proposition to a = −1, and then, as in the previous proposition, use the fact that a congruence mod p when both sides are either 1 or −1 is actually an equality. For p−1 (ii), note that if p ≡ 1 mod 4 then (−1) 2 = 1 ≡ 1, whereas if p 6≡ 1 mod 4 then p ≡ 3 mod 4 and so p−1 (−1) 2 = −1. Here is an interesting application of the results so far: Corollary 2.8. There are infinitely many positive prime numbers which are congruent to 1 mod 4. Proof. Suppose not, and let {p1 , . . . , pn } be the finite set of all positive prime numbers which are ≡ 1 mod 4. Put M = (2p1 . . . pn )2 + 1; then M > 1, so M is divisible by some positive prime number p. Since M is odd, p 6= 2. Now observe that −1 ≡ (2p1 . . . pn )2 mod p, so −1 = 1; the previous proposition therefore implies that p p ≡ 1 mod 4. So p = pi for some i, whence M ≡ 1 mod p; this contradicts p|M . Remark 2.9. A deep result, beyond this course, is the following theorem of Dirichlet: if a, n are coprime integers, then there are infintely many prime numbers which are congruent to a modulo n. As we develop more tools during the course, we will see other special cases of this result. 5
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MATH 242 Algebraic Number Theory
The aim of this section is to prove the next two theorems (due to Gauss, Legendre, and Eisenstein), which allow us to easily calculate any Legendre symbol and therefore determine exactly when a is a quadratic residue modulo p. Their more theoretic importance will become clear in section 9.1. The proofs will be postponed until the end of the section. Theorem 2.10 (Legendre symbol of 2). p ≥ 3 a prime number. Then ( 1 p ≡ ±1 mod 8 2 = . p −1 p ≡ ±3 mod 8 Exercise 2.1 (Restatement of the Leg. sym. of 2 theorem). Suppose that b is an odd integer; show that 2 b2 − 1 is divisible by 8, and that b 8−1 is even if b ≡ ±1 mod 8 and is odd if b ≡ ±3 mod 8. Deduce that the Legendre symbol of 2 theorem can be rewritten as the statement that 2 2 = (−1)(p −1)/8 . p This is sometimes a useful formulation. Example 2.11. 61 ≡ −3 mod 8 and so 2 is not a quadratic residue modulo 61; checking this any other way but by using the theorem would be extremely time consuming. 73 ≡ 1 mod 8 and so 2 is a quadratic residue mod 73; notice that we have proved this without actually finding any integer x satisfying x2 ≡ 2 mod 73. The following is the second main theorem. Despite its simple statement, it is an extraordinary result, telling usthat if p,q ≥ 3 are prime numbers then there is a mysterious relationship between the Legendre q symbols p and pq . A priori, there is absolutely no reason that these should be related: although both symbols encode the solubility of an equation, the first is really an equation in Z/pZ and the second is in Z/qZ. These two equations should have nothing to do with one another! Theorem 2.12 (Law of Quadratic Reciprocity). Let p, q ≥ 3 be distinct primes. Then p−1 q−1 p q = (−1) 2 2 . q p Exercise 2.2 (Restatement of the LQR). Show that the Law of Quadratic Reciprocity can be restated in the follow way: If p, q ≥ 3 are prime numbers (not necessarily distinct), then ( 1 if p or q is ≡ 1 mod 4 p q =ε , where ε = q p −1 if both p and q are ≡ 3 mod 4 It is almost always in this form that one uses the Law of Quadratic Reciprocity. Example 2.13. We now demonstrate the power of the previous two theorems by effortlessly calculating some Legendre symbols: (i)
61 89
33 59
=
89 61
3 59
=
28 61
=
4 61
7 61
59 11
=
61 7
5 7 2 = = = = −1 7 5 5
(ii)
=
11 59
=−
59 3
·−
6
=
−1 3
4 11
= (−1)(3−1)/2 = −1
7
MATH 242 Algebraic Number Theory
(iii)
67 89
=
89 67
=
22 67
=
2 67
11 67
= −1 · −
67 11
=
1 11
=1
Before proving the two main theorems, we provide some sample applications to demonstrate their usefullness in questions of number theory; this week’s homework includes modifications of all these results: Proposition 2.14. p an odd prime. Then −2 is a quadratic residue modulo p if and only if p is congruent to 1 or 3 modulo 8. Proof. We have
−2 p
=
−1 p
2 2 2 = (−1)(p−1)/2 (−1)(p −1)/8 = (−1)(p−1)/2+(p −1)/8 , p
and p2 + 4p − 5 (p − 1)(p + 5) p − 1 p2 − 1 + = = . 2 8 8 8 One now checks that 1 3 p≡ −3 −1 (e.g. p ≡ −3 mod 8 ⇒ p = 8m − 3 ⇒
0 (p − 1)(p + 5) 0 ≡ mod 8 ⇒ 8 1 1 (p−1)(p+5) 8
1 3 p≡ −3 −1
mod 2.
= 8m + 1 ≡ 1 mod 2). Therefore
1 1 −2 mod 8 ⇒ = p −1 −1
.
Proposition 2.15. There are infinitely many positive primes in Z which are congruent to −1 modulo 8. Proof. As usual, we prove this by contradiction, assuming that the set of all such primes is finite: {p1 , . . . , pm }. Let M = 8(p1 . . . pm )2 − 1, and let M = q1r1 . . . qtrt be the prime factorization of M . If qi ≡ 1 mod 8 for all i, then M = q1r1 . . . qtrt ≡ 1r1 . . . 1rt ≡ 1 mod 8, which is false. Therefore there is i such that qi 6≡ 1 mod 8. Notice that qi is odd since M is odd. Moreover, we have (4p1 . . . pm )2 = 2(M + 1) ≡ 2 mod qi , so 2 is a quadratic residue modulo qi . The “Legendre symbol of 2” theorem now implies qi ≡ ±1 mod 8, and so qi ≡ −1 mod 8. Therefore qi = pj for some j, whence we get the usual contradiction: q|p1 . . . pm and q|M implies q|2. Proposition 2.16. Let p > 3 be a Fermat prime (this means p = 2n + 1 for some n ≥ 1). Then 3 is a primitive root modulo p. 7
8
MATH 242 Algebraic Number Theory n−1
Proof. Since the group Z/pZ× has p − 1 = 2n elements, any element g of the group which satisfies g 2 6= 1 2n−1 is a generator. Therefore it is enough to prove that 3 ≡ 6 1 mod p. But 2n−1 = (p − 1)/2 and 3(p−1)/2 = p3 ; therefore we must show that p3 6= 1; i.e. that 3 is not a quadratic residue mod p. Since p ≡ 1 mod 4, the Law of Quadratic Reciprocity implies p3 = p3 ; since 2 ≡ −1 mod 3, we see that p ≡ (−1)n + 1 mod 3, which is 0 or 2 modulo 3. But p is prime, so not divisible by 3. Therefore p ≡ 2 mod 3; 2 is not a quadratic residue modulo 3, so now we know p3 = −1. Therefore p3 = −1, completing the proof. We now begin the proofs of the two main theorems: “Legendre symbol of 2” and “Law of Quadratic Reciprocity”. Although Ireland and Rosen follows Gauss’ methods, we will follow a proof due to Eisenstein (there are at least 233 different proofs in existence today1 ). The following technical lemma is essential: Lemma 2.17 (Eisenstein’s lemma). Let p be an odd prime, and a ∈ Z not divisible by p; then a = (−1)s , p where (p−1)/2
s=
X
k=1
2ka . p
Proof. For any integer k, we will write r(k) to denote the unique integer in the range 0, . . . , p − 1 which is congruent to k modulo p; in other words, r(k) is the remainder when k is divided by p. The proof is a little tricky so we will label the steps: (1). Firstly, if 1 ≤ l ≤ p − 1 then r((−1)l l) is even and non-zero. Indeed, if l is even then r((−1)l l) = r(l) = l (which is even); if l is odd, then r((−1)l l) = r(−l) = p − l (which is even); and it is non-zero since r((−1)l l) ≡ (−1)l l 6≡ 0 mod p. (2). So, if k ∈ {1, 2, . . . , (p − 1)/2} then 2ka is coprime to p and therefore r(2ka) ∈ {1, . . . , p − 1}; so paragraph (1), with l = r(2ka), shows r((−1)r(2ka) r(2ka)) ∈ {2, 4, . . . , p − 1}. Thus there is a well-defined map R : {1, . . . , (p − 1)/2} → {2, 4, . . . , p − 1}, k 7→ r((−1)r(2ka) r(2ka)), which we next prove is injective. So suppose that 1 ≤ k, k 0 ≤ (p − 1)/2 and R(k) = R(k 0 ); this implies 0
(−1)r(2ka) r(2ka) ≡ (−1)r(2k a) r(2k 0 a) mod p. 0
Therefore (−1)r(2ka) 2ka ≡ (−1)r(2k a) 2k 0 a mod p; since 2a is a unit modulo p, we may divide by it to deduce 0 0 that (−1)r(2ka) k ≡ (−1)r(2k a) k 0 mod p. So k ≡ εk 0 mod p, where ε = (−1)r(2k a)−r(2ka) ∈ {0, 1}. But k, k 0 are both in the range 1 ≤ k, k 0 ≤ (p − 1)/2, so k ≡ −k 0 is impossible, and k ≡ k 0 happens if and only if k = k 0 . This proves that R is injective. (3). Since the codomain and domain of R both have cardinality (p − 1)/2, we see that R is actually a bijection: in other words, as k runs over the integers 1, 2, . . . , (p − 1)/2, then r((−1)r(2ka) r(2ka)) runs over the integers 2, 4, . . . , (p − 1). Therefore (p−1)/2
Y
(p−1)/2
r((−1)r(2ka) r(2ka)) =
k=1 1 http://www.rzuser.uni-heidelberg.de/
Y l=1
~hb3/fchrono.html
8
2l,
MATH 242 Algebraic Number Theory
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and so (p−1)/2
Y
(p−1)/2
Y
2l ≡
l=1
(−1)r(2ka) r(2ka) mod p.
(†)
k=1
(4). With identity (†) established, we may now complete the proof: (p−1)/2
a(p−1)/2
Y
(p−1)/2
Y
2k =
k=1
2ka
k=1 (p−1)/2
Y
≡
r(2ka) mod p
k=1 (p−1)/2 P
≡ (−1)
k
Y
r(2ka)
(−1)r(2ka) r(2ka) mod p
k=1 (p−1)/2 P
≡ (−1)
k
r(2ka)
Y
2k mod p
k=1 P Q(p−1)/2 to p, it follows that a(p−1)/2 ≡ (−1) k r(2ka) mod p. But Euler’s lemma says k=1 2k iscoprime P a(p−1)/2 ≡ ap , so ap ≡ (−1) k r(2ka) mod p. Since two powers of (−1) are congruent modulo p if and P only if they are equal, we deduce ap = (−1) k r(2ka) . Finally, the definition of the floor function and r function mean that 2ka 2ka = p + r(2ka), p k j P 2ka a k b p c , as required. ≡ r(2ka) mod 2. Therefore which implies 2ka p p = (−1)
Since
Now we may prove the first of the main theorems: Theorem 2.18 (Proof of “Legendre symbol of 2”). p an odd prime. Then ( 1 p ≡ ±1 mod 8 2 = p −1 p ≡ ±3 mod 8 Proof. If k is an integer in the range 1j ≤kk ≤ bp/4c then 1 ≤ 4k ≤ p, so actually 1 ≤ 4k < p (since p is prime) and so 0 ≤ 4k/p < 1; therefore 4k = 0. If instead k is the in the range bp/4c < k < (p − 1)/2 then j k p 2 4k a similar argument shows p = 1. So p = (−1)s (by Eisenstein’s lemma) where (p−1)/2
s=
X
k=1
2k2 = p
X
1=
k s.t. bp/4c 1, and 9 = N (3) = N (α)N (β). √
So N (α) = N (β) = 3. Let α = a + b −5 for some a, b ∈ Z; then 3 = N (α) = a2 + 5b2 , which is clearly impossible. This contradication shows that 3 must be irreducible. √ (ii) 7. Suppose 7 is not irreducible; then 7 = αβ for some α, β ∈ Z[ −5] such that neither α or β are units. Therefore N (α), N (β) > 1, and 49 = N (7) = N (α)N (β). √
So N (α) = N (β) = 7. Let α = a + b −5 for some a, b ∈ Z; then 7 = N (α) = a2 + 5b2 , which is clearly impossible. This contradication shows that 7 must be irreducible. 20
MATH 242 Algebraic Number Theory
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√ √ √ √ (iii) 1 + 2 −5. Suppose 1 + 2 −5 is not irreducible; then 1 + 2 −5 = αβ for some α, β ∈ Z[ −5] such that neither α or β are units. Therefore N (α), N (β) > 1, and √ 21 = N (1 + 2 −5) = N (α)N (β), so N (α) = 3 or √ 7. But calculations (i) and (ii) showed that this is impossible. This contradication shows that 1 + 2 −5 must be irreducible. √ (iv) 1 − 2 −5. Repeat the previous argument verbatim. This completes the proof. √ Theorem 4.8. Z[ −5] is not a unique factorization domain. Proof. We have
√ √ (1 + 2 −5)(1 − 2 −5) = 1 + 4 · 5 = 21
and also 3 · 7 = 21. √ We have therefore produced two decompositions of 21 into √ irreducibles, so to √ prove that Z[ −5] is not a UFD it is enough to show of 1 + 2 −5 or of 1 − 2 −5, i.e. that there is no unit √ that 3 is not an associate √ u such that 3 = u(1 + 2 −5) or 3 = u(1 − 2 −5). As the only units are −1, 1, this is clear. √ Corollary 4.9. There is no Euclidean norm on Z[ −5]. √ Proof. If there were a Euclidean norm on Z[ −5] then it would be a Euclidean domain, hence a PID and a UFD; but this contradicts the previous theorem.
4.4
Introductory tools for studying number fields: Norm, Trace, Discriminant, and Integral bases
Here we first explain the notion of norm, trace, and discriminant for any finite field extension; we will then specialise to the case of number fields and introduce in addition the notion of an integral basis. These ideas, which use little more than some linear algebra, will be used throughout the rest of the course, both theoretically and in examples. Recall from Algebra that an extension of fields L/F is simply a pair of fields F ⊆ L, where F is a subfield of L. The usual algebraic operations make L into a vector space over F , and one defines the degree of L/F , denoted |L : F |, to be dimF L. One says that L/F is a finite field extension of finite extension of fields if and only if this dimension is finite. For example, if α is an algebraic number, then Q(α)/Q is an extension of fields of degree |Q(α) : Q| = deg α, by proposition 3.9, hence is a finite field extension. Remark 4.10 (Tower Law). If L/M and M/F are finite extension fields, and ω1 , . . . , ωr is a basis for L/M , then L = M ω1 ⊕ · · · ⊕ M ωr . Taking dimensions as F -vector spaces gives dimF L = r dimF M , i.e. |L : F | = |L : M ||M : F |. This is called the tower law. For example, if F is a number field and α ∈ F , then |F : Q| = |F : Q(α)||Q(α) : Q|. As |Q(α) : Q| = deg α we see in particular that deg α divides |F : Q|. √ √ Here is an example of how this is useful: Just above we saw that α = 2 + 3 has minimal polynomial X 4 − 10X 2 + 1, but here is quicker proof: 21
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MATH 242 Algebraic Number Theory
Put F = Q(α) and notice that F contains and the tower law implies
√
2 and
√
3. For example,
α3 −9α −4
=
√
√ 2. Therefore Q( 2) ⊆ F
√ √ √ |F : Q| = |F : Q( 2)||Q( 2) : Q| = |F : Q( 2)|2. √ √ √ √ It is easy to check that no element of Q( 2) is a square root of 3, i.e. 3 ∈ / Q( 2), and so |F : Q( 2)| > 1. It follows that |F : Q| is an even number which is > 2. But since |F : Q| = deg α ≤ 4 (as α does satisfy a degree 4 monic polynomial), we deduce that deg α = 4; so α has minimal polynomial of degree 4, and this minimal polynomial divides, and therefore equals X 4 − 10X 2 + 1. 4.4.1
Norm and Trace
Definition 4.11. Let L/F be a finite extension of fields, and let α ∈ L. The norm of α, denoted NL/F (α) is defined to be the determinant of the following F -linear map of vector spaces: L → L,
β 7→ αβ.
I.e., NL/F (α) = det φα ∈ F . Similarly, the trace of α, denote TrL/F (α), is defined to the trace of φα ; i.e., TrP L/F (α) = Tr φα ∈ F . n In other words, let β1 , . . . , βn be a basis of L as an F -vector space, write αβi = j=1 ci,j βj with ci,j ∈ F , and put C = (ci,j ); then NL/F (α) = det C,
TrL/F (α) = Tr C =
n X
ci,i .
i=1
Example 4.12. Here we compute norms and traces of typical elements in some extensions. (i) C/R has basis 1, i. If x + iy ∈C, with x, y ∈ R, then the matrix for multiplication by x + iy with x −y respect to this basis is whence y x NC/R (x + iy) = x2 + y 2 ,
TrC/R (x + iy) = 2x.
(ii) Q(i)/Q has basis 1, i, and exactly the same argument as in the previous example shows that NQ(i)/Q (x + iy) = x2 + y 2 ,
TrQ(i)/Q (x + iy) = 2x,
for x, y ∈ Q. (iii) Q(ω)/Q, where ω 2 + ω + 1 = 0. This has basis 1, ω, and multiplication by a + bω in this basis is given by the matrix a −b . b a−b So NQ(ω)/Q (a + bω) = a2 + b2 − ab,
TrQ(ω)/Q (a + bω) = 2a − b.
The next two properties summarise the main theoretical properties of the Norm and Trace for a finite extension of fields L/F : Proposition 4.13 (Main theoretic properties of the Norm). NL/F : L → F has the following properties: (i) NL/F (α1 α2 ) = NL/F (α1 )NL/F (α2 ) for α1 , α2 ∈ L. (ii) NL/F (a) = a|L:F | if a ∈ F . 22
MATH 242 Algebraic Number Theory
23
(iii) NL/F (α) = 0 if and only if α = 0. (iv) If M is an intermediate field between L and F , then NL/F = NM/F ◦ NL/M . Proof. For any α ∈ F , let φα : L → L,
β 7→ αβ
be the F -linear map ‘multiplication by α’, so that NL/F α = det φα . (i): For any β ∈ F , φα1 (φα2 (β) = α1 α2 β = φα1 α2 (β), and so φα1 ◦ φα2 = φα1 α2 . Therefore NL/F (α1 )NL/F (α2 ) = det φα1 · det φα2 = det(φα1 ◦ φα2 ) = det(φα1 α2 ) = NL/F (α1 α2 ). (ii): If a ∈ F then the matrix for φa , with respect to any basis α1 , . . . , αn , of L/F is the n × n diagonal matrix with a’s all along the diagonal; this has determinant an . (iii): If α = 0 then φα = 0 so NL/F (α) = det φα = 0; conversely, if α 6= 0 then 1 = NL/F (1) = NL/F (αα−1 ) = NL/F (α)NL/F (α−1 ), so NL/F (α) 6= 0. (iv): We only sketch the proof in the special case that α ∈ M . Let ω1 , . . . , ωr be a basis for L/M ; then (by the very definition of what a basis is), L = M ω1 ⊕ · · · ⊕ M ωr . Since φα (M ωi ) = αM ωi ⊆ M ωi , for each i, we see that φα restricts to each F -vector space M ωi and that φα = φα |M ω1 ⊕ · · · ⊕ φα |M ωn . By a ‘block diagonal’ argument, we see that det(φα ) = det(φα |M )r ,
i.e. NL/M (α) = NM/F (α)|L:M | .
Therefore NL/M (α) = NM/F (α|F :M | ) = NM/F (NL/M (α)).
Proposition 4.14 (Main theoretic properties of the Trace). TrL/F : L → F has the following properties: (i) TrL/F (α1 + α2 ) = TrL/F (α1 ) + TrL/F (α2 ) for α1 , α2 ∈ L. (ii) TrL/F (aα) = a TrL/F (α) for a ∈ F , α ∈ L. (iii) TrL/F (a) = |L : F |a if a ∈ F . (iv) If M is an intermediate field between L and F , then TrL/F = TrM/F ◦ TrL/M . Proof. Copy the previous proof, replacing addition by multiplication. Remark 4.15 (Relation to Galois theory). Readers who know Galois theory may find the following formulae (exercise!) useful, though we will not need them: If L/F is a finite Galois extension of fields and α ∈ L, then Y X NL/F (α) = σ(α), TrL/F (α) = σ(α) σ∈Gal(L/F )
σ∈Gal(L/F )
More generally, if L/F is not necessarily Galois, then these formulae remain true if we let σ run over all F -linear field embeddings of L into a fixed normal extension of F containing L, e.g. the normal closure of L/F . 23
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MATH 242 Algebraic Number Theory
In the special case of a number field, the norm and trace maps are compatible with the ring of integers: Proposition 4.16. F a number field, and α ∈ DF . Then TrF/Q (α) and NF/Q (α) belong to Z. Proof. Let M = Q(α); then the previous two propositions imply that NF/Q (α) = NM/Q (α)|F :M | ,
TrF/Q (α) = |F : M | TrM/Q (α)
Therefore it remains only to show that NM/Q (α) and TrM/Q (α) are in Z; this is left to the homework. 4.4.2
Discriminant
Using the trace maps we can construct an interesting invariant known as the discriminant: Definition 4.17. L/F a finite extension of fields, and α1 , . . . , αn ∈ L some elements. Then the associated discriminant is ∆(α1 , . . . , αn ) = det(TrL/F (αi αj )) Example 4.18. Here is a computation of a discriminants: Let d ∈ Z \ {0, 1} not be a cube, and set F = Q(d1/3 ). Then ∆(1, d1/3 , d2/3 ) is the determinant of the matrix TrF/Q 1 TrF/Q d1/3 TrF/Q d2/3 3 0 0 TrF/Q d1/3 TrF/Q d2/3 TrF/Q d = 0 0 3d 0 3d 0 TrF/Q d2/3 TrF/Q d TrF/Q d4/3 So ∆(1, d1/3 , d2/3 ) = −27d2 . The next two results are the main theoretic properties of the disciminant: Proposition 4.19 (Discriminant detects bases). L/F a finite extension of fields, and α1 , . . . , αn ∈ L. If ∆(α1 , . . . , αn ) 6= 0 then α1 , . . . , αn are linearly independent over F . Conversely, if α1 , . . . , αn form a basis for L/F , and F has characteristic zero, then ∆(α1 , . . . , αn ) 6= 0. Proof. Suppose that α1 , . . . , αn are linearly dependent; so there are a1 , . . . , an ∈ F , not all zero, such that Pn i=1 ai αi = 0. Therefore, for any j, 0 = TrL/F (αj
n X
ai αi ) =
ai Tr(αi αj ),
i=1
i=1
which can be rewritten as
n X
a1 (Tr(αi αj )) ... = 0, an
where the column vector is non-zero. So det(Tr(αi αj )) = 0, as required. Conversely, for a contradiction suppose that α1 , . . . , αn are a basis for L/F and that ∆(α1 , . . . , αn ) = 0. So the F -linear map L → i αj )) has a kernel, i.e. there are a1 , . . . , an ∈ F , PnL attached to the matrix (Tr(αP n not all zero, such that i=1 ai Tr(αi αj ) = 0. Set α := i=1 ai αi ∈ L; this is non-zero Pnsince α1 , . . . , αn form a basis and a1 , . . . , an are not all zero. However, if β ∈ L, then we may write β = j=1 bj αj for some b1 , . . . , bn ∈ F , and we deduce that TrL/F (βα) = TrL/F (
n X j=1
bj αj α) =
n X
bj TrL/F (ααj ) =
j=1
n X j=1
bj
n X
ai TrL/F (αi αj ) = 0.
i=1
In particular, taking β = α−1 , we have just shown that n = TrL/F (1) = 0 in the field F , which contradicts the assumption that char F = 0. 24
MATH 242 Algebraic Number Theory
25
Proposition 4.20 (Change of basis formula for discriminant). L/F a finite extension of fields, with bases α1 , . . . , αn and Pnβ1 , . . . , βn . Let C be the change of basis matrix from the β-basis to the α-basis, i.e. C = (ci,j ) where αi = j=1 ci,j βj with ci,j ∈ F . Then ∆(α1 , . . . , αn ) = det(C)2 ∆(β1 , . . . , βn ). Proof. We have αi αk =
n X n X
ci,j ck,l βj βl ,
j=1 l=1
and taking traces gives TrL/F (αi αk ) =
n X n X
ci,j TrL/F (βj βl )ck,l .
j=1 l=1
In other words, A = CBC t , where A = (TrL/F (αi αj ) B = (TrL/F (βi βj )) Take determinants to get the desired result. 4.4.3
Integral bases
Over the next few results we introduce the notion of an integral basis for a number field; this will be an important tool in proving all the main results in the next section. Definition 4.21. Let F be a number field, I a non-zero ideal of DF , and ω1 , . . . , ωn ∈ F a basis for F/Q. Then ω1 , . . . , ωn is called an I-integral basis if and only if I = Zω1 + · · · + Zωn . Pn In other words, if an element α ∈ F is expressed in the basis as α = i=1 ci ωi , where ci ∈ Q, then α ∈ I ⇐⇒ ci ∈ Z for all i. If I = DF (which is by far the most important case), simply say integral basis instead. So in this case, P ' DF = Zω1 + · · · + Zωn ; in other words, the isomorphism of Q-vector spaces Qn → F , (ci ) 7→ i ci ωi restricts ' to an isomorphism of abelian groups Zn → DF . Lemma 4.22. F a number field. (i) If I is a non-zero ideal of DF , then I contains a basis for F/Q. (ii) If ω1 , . . . , ωn ∈ DF are a basis for F/Q, then ∆(ω1 , . . . , ωn ) is a non-zero integer. Proof. (i): Let α1 , . . . , αn ∈ F be a basis for F/Q; by lemma 4.5 there is a non-zero integer m such that ' mαi ∈ DF for all i. Let β ∈ I be non-zero. Then multiplication by mβ is an F -linear isomorphism L → L, so mβα1 , . . . , mβαn is also a basis for F/Q, and all terms belong to I. (ii): By proposition 4.16, the matrix (TrF/Q (ωi ωj )) has entries in Z, so its determinant ∆(ω1 , . . . , ωn ) is in Z; it is non-zero by proposition 4.19. Proposition 4.23. F a number field and I a non-zero ideal of DF . Then an I-integral basis exists. 25
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MATH 242 Algebraic Number Theory
Proof. Among all possible bases ω1 , . . . , ωn of F/Q which are contained within I, pick one which minimises |∆(ω1 , . . . , ωn )|. This makes sense by the previous lemma! Pn Certainly Zω1 + · · · + Zωm ⊆ I, so we must prove the converse. Let α ∈ I and write α = i=1 ai ωi for some ai ∈ Q; we must prove that ai ∈ Z for all i. If not, then by rearranging (for simplicity of notation), we may assume that a1 ∈ / Z. So a1 = ba1 c + θ for some 0 < θ < 1. Then ω10 := ω10 − ba1 c ω1 = θω1 + a2 α2 + · · · + an ωn , ω20 := ω2 , . . . , ωn0 := ωn 0 is also a basis Pn for F/Q belonging to I. The change of basis matrix C = (ci,j ) from the ω-basis to the ω -basis 0 (i.e. ωi = j=1 ci,j ωj with ci,j ∈ Q) is θ a2 a3 . . . an 0 1 0 . . . 0 .. . 0 C = 0 0 1 , . . .. . . .. . . . . . . . 0 0 0 ... 1
which has determinant θ2 . According to proposition 4.20, |∆(ω10 , . . . , ωn0 )| = θ2 |∆(ω1 , . . . , ωn )|, contradicting the minimality assumption. Exercise 4.2. Let F be a number field, and let ω1 , . . . , ωn ∈ DF be an integral basis. The absolute discriminant of F is defined by ∆F := ∆(ω1 , . . . , ωn ). Using the change of basis formula for the discriminant, show that ∆F does not depend on the chosen integral basis. √ If d ∈ Z \ {0, 1} is a square-free integer and F = Q( d), show that ( 4d if d ≡ 2, 3 mod 4 ∆F = d if d ≡ 1 mod 4 4.4.4
Applications of integral bases to number fields
In the next section we will use integral bases in almost all our proofs; here are two important applications already: Definition 4.24. F a number field and I a non-zero ideal of DF . The norm of I, denoted N (I) is defined to be the size of DF /I: N (I) = #DF /I. This definition makes sense because of the next lemma: Proposition 4.25. F a number field, and I a non-zero ideal of DF . Then DF /I is a finite ring. Proof. Let m ∈ I ∩ Z be non-zero (for example, let α ∈ I be non-zero, with minimal polynomial f (X) = X n + · · · + a0 ∈ Z[X]; then a0 = −a1 α − · · · − αn ∈ I ∩ Z). Let ω1 , . . . , ωn ∈ DF be an integral basis for F/Q. Pn For any α ∈ DF /I, write α = i=1 ci ωi , with ci ∈ Z; then use Euclidean division to write ci = mqi + ri where qi ∈ Z and 0 ≤ ri < qi . So now α=
n n n n X X X X (mqi + ri )αi = m( qi αi ) + ri αi ≡ ri αi mod I. i=1
i=1
i=1
26
i=1
MATH 242 Algebraic Number Theory
27
Pn Therefore any element of DF /I can be represented as i=1 ri αi , with 0 ≤ ri < m; so DF /I is finite, with at most mn elements. (Quick proof: DF /hmi ∼ = (Z/mZ)n , and DF /I is a quotient of this ring since m ∈ I.) Our second applications of integral bases is to the cancellation of ideals: Lemma 4.26 (Very useful!). F a number field, I a non-zero ideal of DF and α ∈ F such that αI ⊆ I; then α ∈ DF . Proof. By the existence of an I-integral basis, I is a finitely-generated abelian group; since αI ⊆ I, the standard test (lemma 3.3) implies α is an algebraic integer. So α ∈ F ∩ Zalg = DF . Proposition 4.27. F a number field, I, J non-zero ideals of DF , and α ∈ DF such that αI = JI; then J = hαi. Proof. We first prove this in the case that α = 1: so we assume I = JI and we to prove that J = DF . Pwant n Let ω1 , . . . ωn ∈ I be an I-integral basis for F/Q. Then ωi ∈ I = JI, so ωi = j=1 βi,j ωj for some βi,j ∈ J. Put B = (βi,j ); then ω1 ω1 .. .. B . = . ωn
ωn
and so det(Id − B) = 0. Expanding this determinant shows that 1 is a sum of products of terms from J, so 1 ∈ J, i.e. J = DF . Now we prove the result for general α. Suppose β ∈ J; then β 1 1 I ⊆ JI = αI = I α α α and so the previous lemma implies that β/α ∈ DF . This is true for all β ∈ J, so α−1 J is a non-zero ideal of DF . Moreover, I = (α−1 J)I, so the special case α = 1 implies that α−1 J = DF , i.e. J = hαi. The previous proposition will be significantly improved in theorem 5.11.
5
Main theoretic properties of DF
We have seen that the ring of integers of a number field need not be a UFD. We will see later in the course that if it were always a UFD, then Fermat’s last theorem would be easy to prove (and Kummer is famous of having presented this erroneous proof...). Attempts to fix this lack of UFDness led to two discoveries: firstly, the ideals of DF do satisfy a certain unique factorization property; secondly, the failure of the UFDness isn’t arbitrarily bad – it can be measured by an abelian group which turns out to be finite. In this section we prove two main theorems about DF and establish some additional theoretic properties (which are needed to study examples): (i) Subsection 1 introduces the class group and proves it is finite. (ii) Subsection 2 explains the unique factorisation of ideals of DF . (iii) Subsection 3 investigates the norm of an ideal, which will be a useful tool in the rest of the course. Remark 5.1. Both the main theorems involve multiplying ideals together: In a commutative ring R, the product IJ of two ideals I, J is by definition the ideal generated by ab, for a ∈ I, b ∈ J; this product is associative, commutative, and the ideal R is the identity element; moreover, if I = hai and J = hbi then IJ = habi.). √ Example 5.2. Let F = Q( −5), so that DF is not a UFD. Then √ √ h3, 1 + 2 −5ih3, 1 − 2 −5i = h3i. 27
28
5.1
MATH 242 Algebraic Number Theory
The class group, its finiteness, and cancellation of ideals
The absolutely most important object associated a number field F is an object known as its class group, denoted ClF . This is a finite abelian group whose elements are equivalence classes of non-zero ideals of DF , and whose composition law is given by multiplication of these ideals. We will see later, in section 7, that the class group encodes some very subtle information about F which can be used to prove or disprove existence of Diophantine equations. In this section we will define ClF as a set of equivalence classes of ideals, prove that it is finite (which is usually considered to be the first major theorem of algebraic number theory), and then use this finiteness to prove important results about the multiplication of ideals of DF . Definition 5.3. Let F be a number field. Non-zero ideals I, J ⊆ DF are called principally equivalent, written I ∼ J, if and only if there are non-zero α, β ∈ DF such that αI = βJ. This is an equivalence relation on the set of non-zero ideals (check this!), and the class group of F (or of DF , depending on the preferred terminology) is defined to be the set of equivalence classes: ClF := {non-zero ideals of DF }/principal equivalence. The equivalence class of a non-zero ideal I ⊆ DF is denoted [I] ∈ ClF . The following lemma establishes the basic relation between principal equivalence and principal ideals, and shows that ClF is trivial precisely when DF is a PID: Lemma 5.4. Let F be a number field. (i) Let I ⊆ DF be a non-zero ideal; then I is principal if and only if it is principally equivalent to DF . In other words, the equivalence class [DF ] is the set of principal ideals. (ii) #ClF = 1 if and only if DF is a PID. Proof. (i): The implication ⇒ is easy, since a principal ideal I = hαi is seen to be principally equivalent to DF via the equality 1I = αDF . Conversely, suppose that I ⊆ DF is a non-zero ideal which is principally equivalent to DF . Then αI = βDF for some non-zero α, β ∈ DF ; setting ω := β/α ∈ F× , we see that ωI = βI ⊆ I and so lemma 4.26 implies that ω ∈ DF . Since I = ωDF , this shows that I is principal. (ii): ClF consists of a single equivalence class if and only if every non-zero ideal of DF belongs to the equivalence class [DF ]; but (i) implies that this is the same as all ideals being principal. Example 5.5. Here are some first examples of class groups: (i) The class groups of Z, Z[i] and Z[ω] are trivial, since we know that these rings of integers are PIDs. √ √ (ii) If F = Q( −5) then we have seen that DF = Z[ −5] is not a PID; so ClF consists of > 1 equivalence classes of ideals. In fact, we will see in section 7 that #ClF = 2, which means (check this!) that any two non-principal ideals of DF are principally equivalent. We now turn to the proof that ClF is a finite set, starting with a rather technical (and slightly tricky) lemma which analyses the extent to which α 7→ |NF/Q (α)| may not be a Euclidean norm on DF ; it should be compared with the manipulations used to prove that Z[i] and Z[ω] are Euclidean domains. Lemma 5.6. Let F be a number field. There exists an integer M ≥ 1 (depending only on F ) with the following property: for any γ ∈ F there exist ω ∈ DF and an integer t ∈ {1, . . . , M } such that |NF/Q (tγ − ω)| < 1. Proof. Let ω1 , . . . , ωn ∈ DF be an integral basis for F/Q. For each r, j = 1, . . . , n, we may write ωr ωj =
n X i=1
28
br,i,j ωi
29
MATH 242 Algebraic Number Theory
for some br,i,j ∈ Z. Put X
C=
n X n Y
|br,i,σ(i) |,
σ∈Sym(n) i=1 r=1
√ pick an integer m > n C and put M = mn . We will show that M has the desired property. To do this we use a geometric argument: by expanding any element of F as a Q-linear sum of ω1 , . . . , ωn , define a function n X f : F → [0, 1]n , ai ωi 7→ ({a1 }, . . . , {an }), i=1
Pn where {a} = a − bac denotes the fractional part of any a ∈ Q. Now fix γ = i=1 ai ωi ∈ F . If we split the cube [0, 1]n into mn subcubes of side length 1/m, then at least two of the mn + 1 elements f (γ), f (2γ), . . . , f ((mn + 1)γ) must lie in the same subcube; in other words, there are integers 1 ≤ h < h0 ≤ mn + 1 such that |{h0 ar } − {har }| ≤ 1/m. So (h0 − h)γ =
n X
(bh0 ar c − bhar c)ωi +
r=1
n X
(r = 1, . . . , n)
({h0 ar } − {har })ωi ,
r=1
which can be rewritten as tγ = ω + δ P Pn n 0 where 1 ≤ t = h0 − h ≤ mn = M , ω = r=1 (bh ar c − bhar c)ωr ∈ DF , and δ = r=1 δr ωr where δr = ({h0 ar } − {har }). We claim that this value of t and ω have the desired property; to prove this we must show that |NF/Q (δ)| < 1. Since we must calculate the norm of δ, we first calculate the matrix D for “multiplication by δ” with respect to the basis ω1 , . . . , ωn : well, for j = 1, . . . , n, ! n n X X δωj = δr br,i,j ωj , i=1
r=1
Pn and so D = ( r=1 ar br,i,j )i,j . Since |δr | ≤ 1/m for all r, we have |Di,j | ≤
1 m
|NF/Q (δ)| = | det D| n X Y =| Di,σ(i) | (−1)sign σ i=1
σ∈Sym(n)
≤
X
n Y
|Di,σ(i) |
σ∈Sym(n) i=1
≤
X
n Y
σ∈Sym(n) i=1
1 C mn 1 C = M < 1, =
by choice of M . This completes the proof. 29
n
1 X |br,i,σ(i) | m r=1
!
Pn
r=1
|br,i,j | for all i, j, and so
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Corollary 5.7. Let F be a number field and let M ≥ 1 satisfy the condition of the previous lemma. Then for any α, β ∈ DF , with β 6= 0, there exist ω, r ∈ DF and an integer t ∈ {1, . . . , M } such that • tα = ωβ + r, and • |NF/Q (r)| < |NF/Q (β)|. Proof. Apply the previous result with γ = α/β to obtain ω ∈ DF and t ∈ {1, . . . , M }; then set r := tα − ωβ and use multiplicativity of NF/Q . Exercise 5.1. Let F be a number field. Check the details of the previous proof. Then check that the following are equivalent: (i) The function ν : DF → N, α 7→ |NF/Q (α)| is a Euclidean norm on DF . (ii) For any γ ∈ F there exists ω ∈ DF such that |NF/Q (γ − ω)| < 1. (iii) M = 1 satisfies the property of the previous lemma and corollary. Thus the magnitude of M in general measures the extend to which |NF/Q | is not a Euclidean norm on DF . √ Prove that condition (ii) is satisfied for Q( d) when d = −1, −2, −3, −7, and −11, and so deduce that the rings of integers of these fields are PIDs. From the previous corollary we may prove the promised theorem that ClF is always finite: Theorem 5.8. Let F be a number field. Then ClF is a finite set; in other words, there are only finitely many equivalence classes of ideals up to principal equivalence. Proof. Let M ≥ 1 be the integer of the previous lemma and corollary. Then DF /hM !i is finite by proposition 4.25, so S := {ideals of DF containing M !} is a finite set (recall from algebra that ideals of DF /hM !i are in one-to-one correspondence with elements of S). We will show that any non-zero ideal I of DF is principally equivalent to an ideal in S, whence ClF has at most #S elements. For any non-zero β ∈ I, the value of |NF/Q (β)| is a non-zero integer, so we may pick and fix a non-zero β ∈ I which minimises this value. For any α ∈ I the previous corollary yields ω ∈ DF and an integer t ∈ {1, . . . , M } such that |NF/Q (tα − ωβ)| < |NF/Q (β)|. Since tα − ωβ ∈ I, our minimality choice of β means that tα − ωβ must be zero. Therefore βt α = ω ∈ DF and so Mβ ! α ∈ DF ; this is true for all α ∈ I, so we conclude M! J := I ⊆ DF . β Hence J is a non-zero ideal of DF which by construction satisfies M !I = βJ. So I is principally equivalent to J. Moreover, M ! = Mβ ! β ∈ J since β ∈ I, so J ∈ S, as required. Definition 5.9. For a number field F , the size of the class group ClF is denoted hF = #ClF and is called the class number of F . It is an extremely important invariant of the number field. Rephrasing lemma 5.4, we see that hF = 1 if and only if DF is a PID. The finiteness of ClF has an extremely useful corollary for manipulating ideals: Corollary 5.10. Let F be a number field and let I ⊆ DF be a non-zero ideal. Then I k is a principal ideal for some integer k ∈ {1, . . . , hF }. 30
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Proof. By the pigeon holeprinciple, at least two of the ideals I, I 2 , . . . , I hF +1 must be principally equivalent; that is, there are integers 1 ≤ i < j ≤ hF + 1 and non-zero α, β ∈ DF such that αI i = βI j . Then α i β I = Ij = Ij ⊆ Ii β β and so ω := β/α is in DF by Lemma 4.26. Moreover, ωI i = I j−i I i , so proposition 4.27 implies that I j−i = hωi, which completes the proof. Exercise 5.2. Let F be a number field. Define a product operation on ClF by [I][J] := [IJ]. Prove that this makes ClF into a finite, abelian group, with identity element [DF ] (you must check associativity, commutativity, and existence of inverses); this is why ClF is known as the class group. Exercise 5.3. Let F be a number field and I ⊆ DF a non-zero ideal. Using the previous exercise, prove the following: (i) I hF is a principal ideal. (ii) If I k is a principal principal ideal for some integer k ≥ 1 which is coprime to hF , then I is already principal. Back in proposition 4.27 we proved a result concerning the cancellation of ideals; here is a significant improvement: Theorem 5.11. Let F be a number field. (i) (“Cancellation of ideals”) Let I, J, K be non-zero ideals of DF . If IJ = IK then J = K. (ii) (“Containment equals division”) Let I, J be non-zero ideals of DF . Then I ⊆ J if and only if there is a non-zero ideal K ⊆ DF such that JK = I. Proof. (i): Assume IJ = IK. By corollary 5.10 there is k ≥ 1 such that I k = hαi for some non-zero α ∈ DF . Certainly I k−1 IJ = I k−1 IK, whence αJ = αK; now multiply both sides by α−1 to deduce J = K. (ii): Implication ⇐ is true in all rings since JK ⊆ J, so we only need to prove the ⇒ direction. Suppose I ⊆ J. By corollary 5.10, there is k ≥ 1 such that J k = hαi for some non-zero α ∈ DF . Then J k−1 I ⊆ J k−1 J = hαi, so K :=
1 k−1 I αJ
is a non-zero ideal of DF . Moreover, JK =
1 1 k J I = αI = I, α α
as required.
5.2
Dedekind domains and Unique factorisation of ideals
Although DF may not be a UFD, we will show in this section that the ideals of DF do satisfy unique factorisation into prime ideals: Theorem 5.12 (Unique factorization of ideals). Let F be a number field and I a non-zero ideal of DF . Then I may be written as a product of powers of prime ideals: I = pe11 . . . pekk ,
(pi primes ideals of DF , ei ≥ 1)
and this expression is unique up to reindexing. 31
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Remark 5.13. In fact, the previous theorem is true for any integral domain D satisfying the following three properties: (i) Every non-zero prime ideal of D is a maximal ideal; (ii) D is Noetherian (i.e., any ascending chain of ideals I1 ⊆ I2 ⊆ · · · of D is eventually stationary). (iii) D is integrally closed. This means that if an element α of the fraction field of D is a root of a monic polynomial in D[X] then α ∈ D. Rings with these three properties are called Dedekind domains; we will prove in the next lemma that the ring of integers of a number field is a Dedekind domain. For example, any PID R is a Dedekind domain. The prime ideal = maximal ideal and Noetherian properties are standard results from algebra, while it easily follows from Gauss lemma and lemma 3.11 that UFDs, hence PIDs, are integrally closed. Theorem 5.11 is true for any Dedekind domain, although our proof relied on finiteness of the class group. The proofs in this section use only properties (i)–(iii) and Theorem 5.11, hence they are “purely algebraic” and are true for any Dedekind domain. Moreover, principal equivalence and the class group may be defined in the same way for Dedekind domains, but it will no longer necessarily be true that it is finite. However, we will continue to state our results only for rings of integers of number fields. Exercise 5.4. Using the fact that a PID is a UFD, prove directly that the above Unique factorisation of ideals theorem is true for a PID. Lemma 5.14. If F is a number field then DF is a Dedekind domain. Proof. We begin by proving property (i), namely that every non-zero prime ideal of DF is a maximal. If p is a non-zero prime ideal of DF then R := DF /p is a finite integral domain. We claim that this is enough to imply it is a field, whence p is a maximal ideal. Well, if r ∈ R is non-zero, then R → R, a 7→ ra is an injective map; since the domain and codomain have the same finite size, it is a bijection, and so there is a ∈ R such that ra = 1. We secondly prove that DF is Noetherian, so suppose that I1 ⊆ I2 ⊆ . . . is an ascending chain of ideals of DF . If Ii = 0 for all i ≥ 1 then the chain is already stationary and there is nothing to prove. Else Ii is a non-zero ideal for some i, whence DF /Ii is a finite ring and hence only has finitely many ideals. Therefore DF has only finitely many ideals containing Ii , so the chain Ii ⊆ Ii+1 ⊆ · · · must clearly eventually be stationary. Finally we prove that DF is integrally closed, so suppose that α ∈ F is the root of a monic polynomial f (X) ∈ DF [X]. By exercise 3.1, α ∈ Zalg , so α ∈ Zalg ∩ F = DF as required. Before beginning the proof of the theorem, let’s see how “unique factorisation of ideals” compares with “unique factorisation of elements”: √ √ √ Example 5.15. Back in the first lecture we saw that Z[ −5] is not a UFD, because 3, 7, 1+2 −5, 1−2 −5 are non-associated irreducibles which satsify √ √ 21 = 3 · 7 = (1 + 2 −5)(1 − 2 −5). Here is how the ‘ideal version of unique factorization’ fixes the problem: We have the following identities concerning products of ideals: √ √ h3, 1 + 2 −5ih3, 1 − 2 −5i = h3i √ √ h7, 1 + 2 −5ih7, 1 − 2 −5i = h7i √ √ √ h3, 1 + 2 −5ih7, 1 + 2 −5i = h1 + 2 −5i √ √ √ h3, 1 − 2 −5ih7, 1 − 2 −5i = h1 − 2 −5i 32
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(Prove two of these). Also need to know that all ideals on the left are prime! Let’s√just check q = h3, 1 + √ 2 −5i; all the others are the same argument. Firstly, it is impossible that h3,√1 + 2 −5i is the whole ring, for that would √ imply (by one of the product identities above) that h3, 1 − 2 −5i = h3i, which is absurd / h3i. because 1 − 2 −5 ∈ √ √ Now let α = a + b −5 be a typical element of Z[ −5]. Then √ √ a := α − 3 −5 + (1 + 2 −5)b ∈ Z, and α ≡ a mod q. Consider the ring homomorphism √ √ f : Z → Z[ −5] → Z[ −5]/q; we have just shown that f is surjective. Its kernel contains 3Z, which is a maximal ideal, so its kernel is equal to 3Z. By an isomorphism theorem for rings, we get √ Z/3Z ∼ = Z[ −5]/q; √ this is a field, so q is a prime ideal (even√a maximal√ideal) of Z[ −5]. Thus the identity 21 = 3 · 7 = (1 + 2 −5)(1 − 2 −5) is really an identity concerning products of prime ideals: √ √ √ √ h21i = h3, 1 + 2 −5ih3, 1 − 2 −5ih7, 1 + 2 −5ih7, 1 − 2 −5i This is enough to prove part of the main theorem: Theorem 5.16. Every non-zero ideal of DF is a product of prime ideals. Proof. Let S be the set of all non-zero ideals of DF which cannot be written as a product of prime ideals. For a contradiction, suppose that S is not empty, and let I1 ∈ S. Now pick I2 ∈ S strictly containing I1 (if such an I2 exists), then pick I3 ∈ S strictly containg I2 (if such an I3 exists), etc. This produces a strictly increasing chain of elements of S: I1 ⊂ I2 ⊂ I3 ⊂ . . . Since DF is Noetherian, this chain cannot go on forever, meaning that we eventually find an ideal I ∈ S such that no other ideal in S properly contains I. Now, I 6= DF , since DF is the empty product of ideals, so I is a proper ideal of DF . Recall that given a proper ideal of any commutative ring, there always exists a maximal ideal containing it. So there is a maximal ideal (= prime ideal since DF is a Dedkind domain) p ⊆ DF containing I. The previous proposition implies there exists a non-zero ideal J such that I = pJ. Next, I ⊆ J, but I 6= J (else cancellation of ideals would imply p = DF ), so the choice of I means that J 6∈ S. But therefore J is a product of prime ideals, so I = pJ is also a product of prime ideals. All that remains is to prove uniqueness of the decomposition into prime ideals, for which the following tool is used: Definition 5.17. Let I be a non-zero ideal of DF , and p a prime ideal. Let ordp I be the unique integer r ≥ 0 such that I ⊆ pr , I 6⊆ pr+1 . We should check that the definition makes sense: If p a non-zero prime ideal of DF then DF = p0 ⊃ p1 ⊃ p2 · · · is a strictly descending chain of ideals of DF (for if pi = pi+1 then p = DF by the cancellation of ideals, which is absurd). Moreover, we claim that ∞ \ pi = {0}; i=1
T∞
i
for if not, then J := i=1 p is a non-zero ideal of DF which is easily checked to satisfy pJ = p, which would again imply p = DF . Therefore the definition makes sense, and next we check the properties of ordp 33
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Lemma 5.18. ordp has the following properties: (i) ordp p = 1; (ii) ordp DF = 0; (iii) if q is a non-zero prime ideal of DF distinct from p, then ordp q = 0; (iv) if I, J are non-zero ideals of DF , then ordp (IJ) = ordp I + ordp J. Proof. (i): Certainly ordp p ≥ 1, so we must only prove that p2 6⊆ p; but if this happened then p2 = p, whence p = DF by cancellation of ideals (ii) is clear. (iii): We must show that q 6⊆ p. Since DF is a Dedekind domain, the prime ideal q is also maximal, and so if q ⊆ p then q = p, which we have assumed is not the case. (iv): This is the most important part. Let s = ordp I, t = ordp J; thus ps ⊇ I, so ps I 0 = I for some ideal I (by proposition ??). Since ps+1 6⊇ I, we see that p 6⊇ I 0 . Similarly, pt J 0 = J for some ideal J 0 not contained inside J. Clearly ps+t ⊇ IJ = ps+t I 0 J 0 , so we must show that ps+t+1 6⊇ ps+t I 0 J 0 . Well, if then ps+t+1 ⊇ ps+t I 0 J 0 then there is an ideal K such that ps+t+1 K = ps+t I 0 J 0 , whence proposition ?? implies that pK = I 0 J 0 . So I 0 J 0 ⊆ p. But neither I 0 or J 0 is contained in p, so there exist a ∈ I 0 \ p, b ∈ J 0 \ p; since p is prime, we see that ab 6∈ p, contadicting I 0 J 0 ⊆ p. Now we may prove the uniqueness of prime decomposition: Theorem 5.19. Let pr11 . . . prmm = qs11 . . . qskk be two products of prime ideals, with r1 , . . . , rm , s1 , . . . , sk ≥ 1. Then n = m and, after reordering, ri = si for all i. Proof. By reordering the product and inserting terms like p0 = DF , it is enough to prove that if we have two expressions pr11 . . . prmm = ps11 . . . psmm , with r1 , . . . , rm , s1 , . . . , sm ≥ 0 then ri = si for all i. But this follows immediately from the previous lemma by applying ordpi to both sides. In other words, I=
Y
pordp (I) ,
p
where the product is taken over all non-zero prime ideals of DF , and where ordp (I) = 0 for all but finitely many p.
5.3
Norms of ideals
Our final theory concerning rings of integers of number fields analyses norms of ideals. Recall from definition 4.24 that the norm of a non-zero ideal I ⊆ DF is defined to be the size of the quotient ring DF /I: N (I) := #DF /I In this section we establish the following two fundamental properties of norms of ideals: (N1) If I = hαi is a principal ideal, then N (I) = |NF/Q (α)|. (N2) If I, J ⊆ DF are non-zero ideals, then N (IJ) = N (I)N (J). 34
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We begin with (N1), whose proof is essentially some tricky linear algebra: Proposition 5.20. Let F be a number field and let α ∈ DF be non-zero. Then N (hαi) = |NF/Q (α)|. Proof. We begin by recalling some linear algebra. If P ∈ Mn (Z) is a matrix with non-zero determinant, then Gaussian elimination, the theory of Echelon forms, or the structure theory of finitely generated abelian groups implies that there exist elementary matrices E1 , E2 ∈ GLn (Z) (i.e., matrices obtained by applying row and column operations to the identity matrix), and a diagonal matrix D = diag(d1 , . . . , dn ) ∈ Mn (Z) such that P = E1 DE2 . Then we have isomorphisms of abelian groups as follows: Zn /P Zn = Zn /E1 DEn Zn = E1 Zn /E1 DEn Zn ∼ = Zn /DEn Zn = Zn /DZn ∼ = Z/d1 Z ⊕ · · · ⊕ Z/dn Z, and so #Zn /P Zn = d1 · · · dn = det D = | det P |, since det Ei = ±1. Now let ω1 , . . . , ωn ∈ DF be an integral basis for F/Q and let P ∈ Mn (Z) be the matrix for “multiplication for α” with respect to this basis. Then, by definition of an integral basis, we have an isomorphism '
Zn → DF ,
(a1 , . . . , an ) 7→
n X
ai ωi
i=1
and this takes P Z to αDF = hαi; therefore Z/P Z ∼ = DF /hαi as abelian groups. So, |NF/Q (α)| = | det P | = #Z/P Z = #DF /hαi = N (hαi). Remark 5.21. In the special case that α = m ∈ Z, the proposition states that N (hmi) = |m|n , where n = |F : Q|. We have already proved this directly in proposition 4.25. To prove property (N2) of norms of ideals, we will check it first for powers of prime ideals, and then extend it to all ideals via a Chinese Remainder type argument; we begin with powers of prime ideals: Lemma 5.22. Let F be a number field and let p ⊆ DF be a prime ideal. Then N (pe ) = N (p)e for all e ≥ 0. Proof. The claim is trivial when e = 0, 1, so assume that e > 1 and proceed by induction. A standard isomorphism theorem for rings implies that I := pe−1 /pe is an ideal of R := DF /pe , with quotient R/I ∼ = D/pe . Counting elements, we see that #DF /pe = #D/pe−1 , #pe−1 /pe i.e., N (pe ) = N (pe−1 ) #pe−1 /pe . Therefore the proof by induction will be complete if we can show that #pe−1 /pe = N (p); in fact, we will prove the stronger statement that there is an isomorphism of abelian groups pe−1 /pe ∼ = DF /p. Since pe ⊂ pe−1 is a strict inclusion (by uniqueness of prime factorization of ideals, or simply by cancellation of ideals), we may pick an element α ∈ pe−1 \ pe . From this element we define a group homomorphism φ : DF → pe−1 /pe , 35
β 7→ αβ mod pe ,
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which we claim is surjective with kernel p. Proof of surjectivity: Let q be a prime ideal in the prime factorization of the ideal hαi + pe ; then e p ⊆ hαi + pe ⊆ q, whence p ⊆ q and so p = q (by the usual prime ideals = maximal ideal trick). So the only prime occuring in the prime factorization of hαi + pe is p itself, and therefore hαi + pe = pr for some r ≥ 0. But pe ⊂ hαi + pe−1 ⊆ pe−1 and so r = e − 1; i.e., hαi + pe = pe−1 . This means φ is surjective. Proof that Ker φ = p: Let β ∈ DF . Then φ(β) = 0 if and only if αβ ∈ pe , which is equivalent to ordp hαβi ≥ e. But ordp hαβi = e − 1 + ordp hβi, which is ≥ e if and only if ordp hβi ≥ 1; so Ker φ = {β ∈ DF : ordp hβi ≥ 1} = p, as required. Therefore φ induces a group homomorphism DF /p ∼ = pe /pe−1 , as desired. Proposition 5.23. Let F be a number field and let I, J ⊆ DF be non-zero ideals. Then N (IJ) = N (I)N (J). Proof. We reduce the proof to powers of prime ideals using two intermediate results. First we recall the (Generalised) Chinese Remainder Theorem from algebra, which we do not prove: Let R be a commutative ring, let I1 , . . . , Im be ideals of R such that Ii + Ij = R whenever i 6= j, and set I := I1 · · · Im . Then there is a natural isomorphism of rings '
R/I → R/I1 ⊕ · · · ⊕ R/Im . Secondly we claim that this is valid for DF in the following sense: if p, q are distinct non-zero prime ideals of DF , and r, s ≥ 0, then pr + qs = DF . If r or s equals 0 then the claim is obvious, so assume r, s ≥ 1. For a contradiction, assume pr + qs is a proper ideal of DF ; then there exists a maximal ideal P ⊂ DF containing it. In particular, pr ⊆ pr + qs ⊆ P, which implies p ⊆ P (since P is prime); but p is a maximal ideal (since prime ideal = maximal ideal in DF ), so this forces p = P. But the same argument reveals that q = P, contradicting p 6= q. This completes the proof of the claim. Now we are equipped to prove the proposition. We may write I = pr11 · · · prmm ,
J = ps11 · · · psmm
rm +sm for some prime ideals p1 , . . . , pm ⊂ DF and integers r1 , . . . , rm , s1 , . . . , sm ≥ 0. Then IJ = pr11 +s1 · · · pm and so
N (IJ) = |D/IJ| = |D/pr11 +s1 | · · · |D/prmm +sm | = N (p1 ) =
r1 +s1
. . . N (pm )
N (pr11 ) . . . N (prmm )
= N (I)N (J).
(by CRT)
rm +sm
· N (ps11 ) . . . N (psmm ) (by CRT again)
Corollary 5.24. Suppose that p is a non-zero ideal of DF such that N (p) is a prime number; then p is a prime ideal. Proof. Certainly p 6= DF , so let q be any prime ideal containing p. Then “containment equals division” implies p = qI for some ideal I; so N (p) = N (q)N (I). But N (p) is prime and N (q) 6= 1; so N (I) = 1, whence I = DF and p = q. 36
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Explicitly constructing ideals in DF and generators of ClF
6
In the previous section we proved the main theoretic properties of rings of integers of number fields. To be able to apply this theory to a concrete number field F , we need to be able to calculate ClF ; to do this we must describe the ideals of DF , for which it is enough to describe the prime ideals (by factorisation of ideals). To do this, we begin by partitioning the prime ideals of DF according to which usual prime number they contain: Definition 6.1. Let F be a number field and p ∈ N a prime number. Then a prime ideal p ⊂ DF is said to sit over p (or that p sits under p) if and only if p occurs in the prime ideal factorisation of the principal ideal hpi ⊂ DF . The following exercise offers alternative definitions, and the main properties, of this relationship; in particular, it shows that each prime ideal of DF sits over a unique prime number. Exercise 6.1. Let F be number field, p ∈ N a prime number, and p ⊂ DF a prime ideal. Show that the following are equivalent: (i) p sits over p. (ii) p ∈ p. (iii) N (p) is a positive power of p. Prove that for any fixed p, there is a unique p sitting under it. Then prove that if p is fixed, there are only finitely many p sitting over it. Therefore, to describe all prime ideals of DF , it is necessary and sufficient to explicitly factor hpi, for all prime numbers p ∈ N. This will be achieved by theorem 6.3 in a moment, under the assumption that DF = Z[α] for some α ∈ DF ; this is true in all cases of interest to us, but can fail in general. We begin by explaining how to construct prime ideals sitting over p: Lemma 6.2 (Constructing prime ideals of DF ). Let F be a number field such that DF = Z[α] for some α ∈ DF , and let f (X) ∈ Z[X] be the minimal polynomial of α; let p ∈ N be a fixed prime number. Suppose that g(X) ∈ Fp [X] is a monic irreducible polynomial (6= 1) which divides f (X) := f (X) mod p. Let ge(X) ∈ Z[X] be any lift of g(X) to Z[X], and put pg := hp, ge(α)i ⊆ DF . Then (i) pg is a prime ideal of DF depending only on g(X), not on the choice of lift ge(X). (ii) N (pg ) = pdeg g . (iii) If g1 (X) ∈ Fp [x] is a different monic irreducible polynomial dividing f (X), then pg1 6= pg . Proof. Firstly, any other lift of g(X) to Z[X] has the form g(X) + pA(X) for some A(X) ∈ Z[X]; then it is clear that hp, g(α)i = hp, g(α) + pA(αi, so pg really does only depend on g(X) and not on the chosen lift. Next we prove that pg is prime and that N (pg ) = pdeg g . We start with the ring homomorphism φ : Z[X] → DF ,
h(X) 7→ h(α),
which is surjective by our assumption that DF = Z[α]. We claim that Ker φ = hf (X)i. The inclusion ⊇ is obvious, so suppose h(X) ∈ Ker φ: then h(α) = 0 and so h(X) = h1 (X)f (X) for some h1 (X) ∈ Q[X]; but 37
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Gauss’ lemma implies that actually h1 (X) ∈ Z[X], and so h(X) ∈ hf (X)i, proving the claim. Therefore φ induces an isomorphism ' φ : Z[X]/hf (X)i → DF . g (X)) = ge(α), so there is another induced isomorphism Furthermore, φ(p) = p and φ(e '
Z[X]/hf (X), p, ge(X)i → DF /pg , which we will use to prove (ii) and the rest of (i). Since g(X) divides f (X), we easily see that f (X) ∈ hp, ge(X)i and so hf (X), p, ge(X)i = hp, ge(X)i. Thus DF /pg ∼ = Z[X]/hp, ge(X)i = Fp [X]/hg(X)i; since g(X) is an irreducible polynomial of degree deg g, the right hand ring is indeed an integral domain with pdeg g elements (by Algebra). Hence pg is a prime ideal of DF with norm pdeg g . Finally, we prove (iii): suppose g1 (X) 6= g(X) is a different monic irreducible polynomial in Fp [X]. Then the Euclidean algorithm lets us write G(X)g(X) + G1 (X)g1 (X) = 1 for some G, G1 ∈ Fp [X]; lifting to Z[X], e e 1 (X)e we may write G(X)e g (X) + G g1 (X) = 1 + pE(X) for some E(X) ∈ Z[X]. If pg1 = pg then this ideal would contain e g (α) + G e 1 (α)e G(α)e g1 (α) − pE(α) = 1, whence pg1 = pg = DF . But this contradicts N (pg ) = pdeg g . The next theorem says that not only does the previous lemma give all the prime ideals of DF sitting over p, but it even offers the precise prime ideal factorisation of the ideal hpi; the moral of the theorem is that factoring hpi is equivalent to factoring f (X) mod p. Theorem 6.3 (How to factor hpi). Let F be a number field such that DF = Z[α] for some α ∈ DF , and let f (X) ∈ Z[X] be the minimal polynomial of α; let p ∈ N be a fixed prime number. Let f (X) = f1 (X)e1 · · · fr (X)er be the factorisation of f (X) := f (X) mod p into distinct monic irreducible polynomials f1 (X), . . . , fr (X) ∈ Fp [X], where e1 , . . . , er ≥ 1. Then the principal ideal hpi of DF has prime ideal factorization hpi = pe11 . . . perr , where pi := pfi in the notation of the previous lemma. Proof. According to the previous lemma, pe11 · · · perr is a product of powers of distinct prime ideals. Moreover, the inclusion pe11 . . . perr ⊆ hpi is quite clear: pe11 . . . perr is generated by terms all but one of which are obviously multiples of p, and the remaining generator is fe1 (α)e1 . . . fer (α)er , which belongs to f (α) + hpi = hpi. To complete the proof it is now enough to show that N (hpi) = N (pe11 · · · perr ), since this forces the above inclusion to be an equality. Well, N (hpi) = p|F :Q| , while the norm of the right hand side, using proposition 5.23 is r Y pei deg fi = pdeg f . i=1
But deg f = |F : Q|, completing the proof. Corollary 6.4. Let F, α, f (X) be as in the theorem, and let p ∈ N be a prime number. Then hpi is a prime ideal of DF if and only if f (X) mod p is irreducible in Fp [X]. Proof. Immediate from the theorem. 38
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√ √ √ Example 6.5. Let F = Q( −5). Since DF = Z[ −5], the theorem may be applied with α = −5 and f (X) = X 2 + 5: √ • p = 2 : X 2 + 5 factors in F2 [X] as X 2 + 5 = (X + 1)2 , so the theorem implies that h2, 1 − −5i is the unique prime ideal √ of DF containing 2, that it has norm 2, and that the prime ideal factorisation of h2i is h2i = h2, 1 − −5i2 . 5 2 • p =√3 : X 2 + 5 factors √ in F3 [X] as X + 5 = X − 1 = (X − 1)(X + 1), so the theorem implies that h3, −5 − 1i and h3, −5 + 1i are the unique prime ideals of DF containing 3,√that they are √ distinct and both have norm 3, and that the prime ideal factorisation of h3i is h3i = h3, −5 + 1ih3, −5 − 1i. √ √ • p = 5 : X 2 + 5 factors in F5 [X] as X 5 + 5 = X 2 , so the theorem implies that h5, −5i = h −5i is the unique prime√ideal of DF containing 5, that it has norm 5, and that the prime ideal factorisation of h5i is h5i = h −5i2 . 2 2 • p =√7 : X 2 + 5 factors √ in F7 [X] as X + 5 = X − 2 = (X − 3)(X + 3), so the theorem implies that h7, −5 − 3i and h7, −5 + 3i are the unique prime ideals of DF containing 7,√that they are √ distinct and both have norm 7, and that the prime ideal factorisation of h7i is h7i = h7, −5 + 3ih7, −5 − 3i. √ √ √ √ (Note that h7, 1 + 2 −5i = h7, −5 − 3i and h7, 1 − 2 −5i = h7, −5 + 3i, so this agrees with an example we proved by hand.)
• p = 11 : X 2 + 5 = X 2 − 6 in Z/11Z[X], which is irreducible since 6 is not a quadratic residue mod 11. Therefore h11i is a prime ideal of DF of norm 112 . Next we use this analysis to describe all ideals J of DF having norm ≤ 10. Well, if N (J) ≤ 10, then the only possible primes dividing N (J) are 2, 3, 5, 7, so by exercise 6.2 below the only prime ideals occurring in the factorisation of J are prime ideals sitting over 2, 3, 5, 7; i.e., J is a product of powers of the following ideals: √ √ √ √ √ √ p2 := h2, 1− −5i, p3 := h3, −5+1i, q3 := h3, −5−1i, p5 := h −5i, p7 := h7, 3+ −5i, q7 := h7, 3− −5i These ideals respectively have norms 2, 3, 3, 5, 7, 7; since norms of ideals are multiplicative, we see that the following is an exhaustive list of all ideals J such that N (J) ≤ 10, together with their norms: J = DF N (J) = 1
p2 2
p22 4
p32 8
p2 p3 6
p2 q3 6
p2 p5 10
p3 3
p23 9
q3 3
q23 9
p3 q3 9
p7 7
q7 7
Notice that these ideals are all distinct, by unique factorisation of ideals. √ √ Example√6.6. It is a fact that the ring of integers of Q[ 3 5] is Z[ 3 5]; so we may apply the previous theorem with α = 3 5 and f (X) = X 3 − 5. • p = 2 : X 3 − 5 factors in F2 [X] as X 3 − 5 = X 3 − 1 = (X − 1)(X 2 + X + 1) (notice that X 2 + X + 1 is irreducible mod 2 since it doesn’t have√a root√mod 2), so the theorem implies that the prime ideal √ factorisation of h2i is h2i = h2, 3 5 − 1ih2, 3 25 + 3 5 + 1i, and that these prime ideals on the right hand side have norm 2, 4 respectively. • p = 3 : X 3 − 5√factors in F3 [X] as X 3 − 5 = X 3 − 23 = (X − 2)3 , so the prime ideal factorisation of h3i is h3i = h3, 3 5 − 2i3 . Example 6.7. Suppose that DF = Z[α], where the minimal polynomial f (X) ∈ Z[X] satisfies Eisenstein’s criterion for some prime number p ∈ N. Then f (X) ≡ X n mod p, so hpi = pn where p is the prime ideal p = hp, αi = hαi, which has norm N (p) = p. 39
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MATH 242 Algebraic Number Theory
Example 6.5 shows, at least in principle, how we can exhaustively list all the ideals J of the ring of integers of a number field of norm N (J) less than some fixed bound. But what is a reasonable bound if we wish to list all ideals up to principal equivalence, thus offering a representative of every element of the class group? This is offered by the next lemma. We must first recall the constant C defined in lemma 5.6. If F is a number field and ω1 , . . . , ωn ∈ DF is an integral basis, then let (br,i,j )i,j be the matrix for multiplication by ωr with respect to this basis, and set C :=
X
n X n Y
|br,i,σ(i) |
σ∈Sym(n) i=1 r=1
From now on C will be called the Hurwitz constant of F (don’t forget that it depends on the choice of integral basis). Proposition 6.8. Let F be a number field; fix an integral basis ω1 , . . . , ωn ∈ DF and let C ∈ N be the Hurwitz constant above. Then any non-zero ideal I ⊆ DF is principally equivalent to a non-zero ideal J such that N (J) ≤ C. In other words, ClF = {[J] : J ⊆ DF a non-zero ideal s.t. N (J) ≤ C} Proof. Let I 0 be a non-zero ideal such that I 0 I is principal (e.g. there exists k ≥ 1 such that I 0 = I k−1 works, by corollary 5.10). Now take c ∈ N large enough so that (1 + c−1 N (I 0 )−1/n )n < 1 +
1 C
and put I 00 = cI 0 , which also has the property that I 00 I is principal. Put n X S = {α = ci ωi : bi ∈ N, 0 ≤ bi ≤ N (I 00 )1/n } ⊂ DF , i=1
which has size #S = ( N (I 00 )1/n + 1)n > N (I 00 ). Since DF /I 00 has N (I 00 ) elements, the pigeonhole principle implies that there must be distinct α, β ∈ S such that α ≡ β mod I 00 ; put γ := α − β 6= 0. So hγi ⊆ I 00 and therefore hγi = I 00 J for some non-zero ideal J ⊆ DF . Multiply both sides by I, noting that II 00 is principal, to see that J ∼ I. It remains to prove that |N (J)| ≤ C. Pn Write γ = i=1 bi ωi with ci ∈ Z; since α, β ∈ S, we see that |bi | ≤ N (I 00 )1/n + 1 for all i; therefore |NF/Q (γ)| ≤ C max{|c1 |, . . . , |cn |}n ≤ C(N (I 00 )1/n + 1)n , where the first inequality follows just as it did for δ in the proof of lemma 5.6. Moreover, |NF/Q (γ)| = N (I 00 J) = N (I 00 )N (J), so |N (J)| ≤ CN (I 00 )−1 (N (I 00 )1/n + 1)n = C(1 + N (I 00 )−1/n )n = C(1 + c−1 N (I 00 )−1/n )n < C + 1. Thus N (J) ≤ C, as required. Exercise 6.2. Let F be a number field and I ⊆ DF a non-zero ideal. Prove that if p is a prime ideal occurring in the ideal factorisation of I, then p sits over a prime number p which divides N (I). Corollary 6.9. Let F be a number field, and let C be the Hurwitz constant for some choice of integral basis. Then ClF is generated by the classes of prime ideals sitting over (positive) primes numbers which are ≤ C. 40
MATH 242 Algebraic Number Theory
41
Proof. By the previous proposition, any element of the class group can be represented as [J], where N (J) ≤ C. Let J = pe11 · · · perr be the prime ideal factorisation of J. Then [J] = [p1 ]e1 · · · [pr ]er in ClF , and the previous exercise implies that each pi sits over a prime number which divides N (J), hence is ≤ C. Remark 6.10 (Minkowski’s bound). As we shall see, the Hurwitz constant C works well for computing class group of small number fields, but a much better result is called Minkowski’s bound, which we will not prove in the course but which will be useful several times. It states that if F is a number field, then proposition 6.8, and hence corollary 6.9, remain true if C is replaced by the real number r n! 4 2 |∆F |1/2 , nn π where • n = |F : Q|. • ∆F is the absolute discriminant of F . • r2 is half the number of field embeddings of F into C which do not have image in R (i.e., if F = Q(α) then r2 is half the number of conjugates of α which are not in R).
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MATH 242 Algebraic Number Theory
Calculations of class groups of quadratic extensions, and applications
√ Let F = Q( d) for a square-free integer d ∈ Z \ {0, 1}. What do we know about the class group of F ? In fact, all we have proved so far is that: • If d = −1, −2, −3, −7, −11, then |NF/Q | is a Euclidean norm on DF and so hF = 1. See exercise 5.1. • If d = −5, −10, −14 then DF if not a PID and so hF > 1. In this section we will improve the state of this list! As we do so, we will obtain applications to Diophantine equations. However, to illustrate the difficulty of the problem, we mention the following results and conjectures: • |NF/Q | is a Euclidean norm on DF if and only if d is one of the following values: −11, −7, −3, −2, −1, 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57, 73 We will prove most of the “if” direction in section 7.11. • If d = 69 then DF is a Euclidean domain, but the Euclidean norm is not |NF/Q |. (See D. Clark, A quadratic field which is Euclidean but not norm-Euclidean, Manuscripta Math. 83 (1994), no. 3–4, 327–330.) • hF → ∞ as d → −∞ (“Gauss conjecture”, proved in 1934 by H. Heilbronn). √ In other words, for any fixed value of h, there are only finitely many square-free d < 0 for which Q( d) has class number h. For actual lists of values of d, given h in the range 1 ≤ h ≤ 100, see M. Watkins, Class numbers of imaginary quadratic fields, Math. Comp. 73 (2004), no. 246, 907–938, which required seven months of computer run time on an ordinary desktop machine. • Conversely, Gauss also conjectured that there are infinitely many square-free d > 1 for which DF is a PID; this conjecture remains wide open. In the 80s this prediction was made more precise via the so-called Cohen–Lenstra heuristics; for example, it is now believed that DF will be a PID for about 75.446% values of positive, square-free d. Here is a table of the class numbers of quadratic number field when |d| < 20; we will prove almost all of these: d = −19 −17 −15 −14 −13 −11 −10 −7 −6 −5 −3 −2 −1 2 3 5 6 7 10 11 13 14 15 17 19 hF = 1 4 2 4 2 1 2 1 2 2 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 (The only cases where the structure of the class group is not obvious is d = −14, −17. If d = −14 then ClF is cyclic of order 4, generated by either of the prime ideals sitting over 3. If d = −17 I don’t know yet...) To prove these and give applications we need to start by knowing the Hurwitz bound C for F : √ Lemma 7.1. Let F = Q( d) for a square-free integer d ∈ Z \ {0, 1}. Then: √ • If d ≡ 2, 3 mod 4 and we take 1, d as integral basis, then C = |d| + 1. √ + 2. • If d ≡ 1 mod 4 and we take 1, 1+2 d as integral basis, then C = d−1 4 Proof. √ We only treat the case that d ≡ 2, 3 mod 4. The matrices for multiplication by 1 and multiplication by d are respectively given by 1 0 0 d (b1,i,j ) := , (b2,i,j ) := . 0 1 1 0 P Qn Pn Recalling that C = σ∈Sym(n) i=1 r=1 |br,i,σ(i) |, we make the following calculations: 42
MATH 242 Algebraic Number Theory
• σ = id, i = 1: Then
Pn
• σ = id, i = 2: Then
Pn
r=1
43
|br,i,σ(i) | = |b1,1,1 | + |b2,1,1 | = 1 + 0 = 1.
|br,i,σ(i) | = |b1,2,2 | + |b2,2,2 | = 1 + 0 = 1. Pn • σ = transpose, i = 1: Then r=1 |br,i,σ(i) | = |b1,1,2 | + |b2,1,2 | = 0 + |d| = |d|. Pn • σ = transpose, i = 2: Then r=1 |br,i,σ(i) | = |b1,2,1 | + |b2,2,1 | = 0 + 1 = 1. r=1
Therefore C = 1 · |d| + 1 · 1 = |d| + 1, as claimed. Remark 7.2. For comparative purposes, we also calculate Minkowski’s bound from remark 6.10. Recall from exercise 4.2 that ∆F = 4d if d ≡ 2, 3 mod 4 and = 4 if d ≡ 1 mod 4. Moreover, the field embedding number r2 which appears in Minkowski’s bound is given by r2 = 1 if d < 0 and = 0 if d > 0. Therefore Minkowski’s bound is ( ) ( p ) r 4 if d < 0 2 |d| if d ≡ 2, 3 mod 4 1 n! 4 2 1/2 |∆F | = × π × p nn π 2 1 if d > 0 |d| if d ≡ 1 mod 4 p For example, if d is negative and ≡ 2 or 3 mod 4, then Minkowski’s bound is π4 |d|, which is (much) less, hence better, than Hurwitz’s bound of |d| + 1. We will avoid using Minkowski’s bound as much as possible, √ since we have not proved it, but will resort to it to compute the class group of F ( d) for certain d 0.
7.1
d = −5
We may finally calculate our first example of a non-trivial class group: √ Theorem 7.3. Let d = −5 and F = Q( −5). √ Then ClF is a cyclic group of order 2, generated by the class of any non-principal ideal, such as h3, 1 + −5i. Therefore, if I is a non-zero ideal of DF , we deduce that: √ (i) I is either principal or principally equivalent to h3, 1 + −5i. (ii) I 2 is princpial Any two non-principal ideals are principally equivalent. Proof. The Hurwitz bound is C = 6, so corollary 6.9 tells us that the class group ClF is generated by the classes of prime ideals sitting over 2, 3 and 5. By examples 6.5, we know that √ h2i = p22 , where p2 := h2, 1 − −5i2 √ √ h3i = p3 q3 , where p3 := h3, 1 + −5i, q3 := h3, 1 − −5i √ h5i = p25 , where p5 := h −5i. So ClF is generated by p2 , p3 , q3 and p5 ; the final of these is principal, so represents the trivial element in the class group and may be discarded. Also, [p3 ][q3 ] = 1 in ClF , so [p3 ] = [q3 ]−1 . Therefore ClF is generated by p2 and p3 . But moreover p2 ∼ p3 , since √ √ (1 − −5)p3 = h3 − 3 −5, 6i = 3p2 ; hence ClF is cyclic, generated by p3 (or by p2 , if you prefer). Finally we claim that p23 is principal, for which we point out a general principle:
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MATH 242 Algebraic Number Theory
If F is a number field, I ⊆ DF is a non-zero ideal, and α ∈ I satisfies |NF/Q (α)| = N (I), then I = hαi. (Proof: |NF/Q (α)| = N (hαi), so hαi ⊆ I is an inclusion of ideals with the same norm, whence an equality.) So to prove that I is principal, you only need to find an element α ∈ I with |NF/Q (α)| = N (I).
Since N (p3 )2 = 9, in order to prove p23 is principal we must merely√show that it contains an element α with |NF/Q (α)| = 9. The only elements of DF with norm ±9 are ±2 ± −5 and ±3. Well, √ √ p23 = h9, 3 + 3 −5, −4 + 2 −5i, √ √ which is easily checked to contain 2 − −5; hence p23 = h2 − −5i. Since DF has no element of norm 3, the ideal p3 is not principal; so this completes the proof that ClF is cyclic of order two generated by [p3 ]. If I ⊆ DF is a non-zero ideal, then either [I] = 1 in ClF , i.e. I is principal, or [I] = [p3 ], i.e. I is principally equivalent to p3 . In any case, [I 2 ] = [I]2 = 1, so I 2 is principal. We have just computed our first example of a non-trivial class group, and so it is a good moment to ask “Why?”. The following lemma, whose proof if very important since we will use it as an opportunity to point out several tools, demonstrates how the class group can be used. We will also show in the proof how, in the special case that DF is a UFD, the theory of ideals and the class group can be completely avoided. This accurately depicts the historical development of algebraic number theory: under the erroneous belief that DF was alway a UFD, it was seen to be a useful too for solving Diophantine equations, and when mathematicians realised their error, they studied ideals and introduced the class group to remedy the problem. Lemma 7.4. Let d < 0 be a negative square-free integer √ such that d ≡ 2 or 3 mod 4, and suppose that there exist integers x, y satisfying y 3 = x2 − d. Let F = Q( d). If hF is not divisible by 3 then there exists α ∈ DF such that √ x + d = α3 . Hence there exists an integer n ∈ Z such that x = n(n2 + 3d) and 3n2 = ±1 − d. Proof. We claim that y is odd and that x, y are coprime. Firstly, if y were even then x2 ≡ 2 or 3 mod 4, which is impossible. Secondly, if a prime number p were to divide both x and y, then p2 |x2 − y 3 = d, which contradicts d being square-free. So y is odd and x, y are coprime. Let’s now make the wild assumption that DF is actually √ a UFD √ (equivalently, a PID) and √ show how√to easily finish the proof. In DF we may write y 3 = (x + d)(x − d). We claim that x + d and x − d are coprime, which in a UFD means that the√two elements√have no common (non-unit) divisor. So, for a contradiction, suppose that we can write x + d = αt, x − d = βt, where α, β, t ∈ DF , and t is not a unit. Notice that 2x = (a + b)t and y = abt2 , and take norms: 4x2 = NF/Q (a + b)NF/Q (t),
y 6 = NF/Q (ab)NF/Q (t)2
Since t is not a unit, its norm is not ±1. Since y is odd, the right equation shows that NF/Q (t) is divisible by an odd prime number p, and then also p|y; but then the left equation shows √ √ that p|x, contradicting the coprimality of x and y. This contradiction shows that indeed x + d and x − d are coprime. But if a product of two coprime elements in a UFD is a cube, then each element must be an associate of a cube: this is easily proved by looking √ at the √ unique factorisation of the two elements, and √ the reader should check if uncertain. Since (x + d)(x − d) = y 3 , we apply this to deduce that x + d = uα3 for some u, α ∈ DF , with u a unit. But D× F = {v ∈ DF : NF/Q (v) = ±1} = {±1}; if u = −1 replace α by −α. √ We have written x + d = α3 , as desired. 44
MATH 242 Algebraic Number Theory
45
√ Finally, writing α = n + m d for some n, m ∈ Z, we obtain √ √ x + d = α3 = n(n2 + 3m2 d) + m(3n2 + m2 d) d, so we see that m(3n2 + m2 ) = 1,
n(n2 + 3m2 d) = x.
Hence m = ±1, and so −d = 3n2 ± 1, x = n(n2 + 3d). Now we drop the assumption that DF is a UFD, and show how to rescue the result under the much weaker assumption that 3 - hF , √ using the √ theory of ideals and the class group which we have developed. √ 3 −5)(x − −5) in DF , and now we claim that the ideals hx + −5i and Again, we may write y = (x + √ hx − −5i are coprime, i.e. that they √ are contained √ in no common proper ideal. If not, then there is a prime ideal p ⊆ DF such that p 3 (x + −5), (x − −5); so p contains 2x and y. But y is odd, so if 2 is also in p then 1 ∈ p, which is impossible; therefore 2 ∈ / p, so x ∈ p. Hence x, y ∈ p ∩ Z = pZ for whichever prime√number p sits√under p; this contradicts coprimality of x and y. This contradiction shows that indeed hx + di and hx − di are coprime ideals. √ √ Under the wild assumption that DF was a UFD, we deduced that the elements x + d and x − d were cubes. Now we must work with ideals:
If D is a Dedekind domain, and I1 , I2 ⊆ D are coprime ideals such that I1 I2 = J k for some ideal J and k > 0, then there exist ideals J1 , J2 such that I1 = J1k and I2 = J2k . That is, if a product of coprime ideals is an k th power, then each ideal is already a k th power. (See the exercise after this lemma.) √ Therefore we may write hx + di = I 3 for some ideal I ⊆ DF . The next observation is the fundamental application of the theory of the class group:
Since 3 - hF , the class group ClF contains no element of order 3. So, since I 3 is principal, I must already be principal! √ √ Hence I = hαi for some α ∈ DF , and so hx + di = hα3 i; this means x + d = uα3 for some unit u ∈ DF , and the final line of the proof is just as it was in the UFD case. Exercise 7.1. Carefully check the (relatively easy but important!) omitted claims in the previous proof: (i) Show that if R is a UFD, and that x, y ∈ R are coprime elements such that xy is a k th power, then x and y are associates of k th power. (ii) Prove the first of the boxed claims in the previous proof. (iii) Prove the following general version of the second boxed claim: if F is a number field and I ⊆ DF is a non-zero ideal such that I k is principal for some integer k which is coprime to hF , then I is already principal. Theorem 7.5. There are no integer solutions to Y 3 = X 2 + 5. √ Proof. Let F = Q( −5). For a contradiction, suppose that x, y ∈ Z satisfy y 3 = x2 + 5. By the previous theorem, hF = 2 is not divisible by 3, so the previous lemma plies that 3n2 = 5 ± 1 for some integer n. But this is clearly impossible, so such x, y cannot not exist. Exercise 7.2. Check that this theorem could not have been proved using congruence arguments, by showing that Y 3 = X 2 + 5 has integer solutions mod n for any n ∈ N. 45
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7.2
MATH 242 Algebraic Number Theory
d = −6
√ Let F = Q( −6). The Hurwitz bound is C = 7, so ClF is generated by the primes ideals sitting over 2, 3, 5, 7. √ h2i = h2, −6i2 since X 2 + 6 ≡ X 2 mod 2 √ h3i = h3, −6i2 since X 2 + 6 ≡ X 2 mod 3 √ √ h5i = h5, 2 + −6ih5, 2 − −6i since X 2 + 6 ≡ (X + 2)(X − 2) mod 5 √ √ h7i = h7, 1 + −6ih7, 1 − −6i since X 2 + 6 ≡ (X + 1)(X − 1) mod 7 √ √ √ √ The −6i = √ prime ideals over 7 are principal, generated respectively by 1+ −6 and 1− −6. Also, h2, −6ih3, √ h√ −6i, so the prime ideals over 2 and 3 are principally equivalent to one another. Finally, h2, −6ih5, 2+ √ −6i contains 2 − −6, which has norm 10, and hence this element generates the product. √ In conclusion, ClF is generated by h2, −6i, which is non-principal by norm considerations. So ClF is cyclic of order 2. Imitating theorem 7.5, one shows that Y 3 = X 2 + 6 has no integer solutions.
7.3
d = −7
√ Let F = Q( −7). The Hurwitz bound is C = 4, so we factor (using α = √
√ 1+ −7 , 2
f (X) = X 2 − X + 2)
√
h2i = h2, 1+ 2 −7 ih2, 1− 2 −7 i h3i is prime since f (X) = (X + 1)2 − 2 mod 3 The prime ideals over 2 are principal, each generated by their second generator. Hence hF = 1, i.e. DF is a PID. Indeed, this calculation was unnecessary since we already know it is a Euclidean domain. Imitating theorem 7.5, one shows that Y 3 = X 2 + 7 has no integer solutions.
7.4
d = −10
√ Let F = Q( −10). The Hurwitz bound is C = 11, so we factor √ h2i = h2, −10i2 h3i is prime since 2 is not a quadratic residue mod 3 √ h5i = h5, −10i2 h7i is prime since 3 is not a quadratic residue mod 3 √ √ h11i = h11, 1 + −10ih11, 1 − −10i The prime ideals sitting √ over 11 are principal, √ each generated by their second generator. √ Hence ClF is generated by p2 = h2, −10i and p5 = h5, −10i. But it is easy to see that p2 p5 = h −10i, so ClF is generated by [p2 ], which has order 2. Hence ClF is again a cyclic group of order 2, and the usual argument shows that Y 3 = X 2 + 10 has no integer solutions. 46
MATH 242 Algebraic Number Theory
7.5
47
d = −13
√ Let F = Q( −13). The Hurwitz bound is C = 14, so we factor √ h2i = h2, 1 + −13i2 h3i is prime since 2 is not a quadratic residue mod 3 h5i is prime since 2 is not a quadratic residue mod 5 √ √ h7i = h7, 1 + −13ih7, 1 − −13i √ √ h11i = h11, 3 + −13ih11, 3 − −13i √ h13i = h −13i2 √ √ √ Hence ClF is generated by p2 = h2, 1 + −13i, p7 = h7, 1 + −13i and p11 = h11, 3 + −13i, and we must now check relations between √ these. √ √ √ p2 p7 contains both 7(1 + −13) and 2(1 + −13), hence contains 1 + −13; since NF/Q (1 + −13) = √ 14 = N (p2 p7 ), we deduce that p2 p7 = h1 + −13i is principal. Next we look at √ √ √ p2 p11 = h22, 6 + 2 −13, 11 + 11 −13, −1 + 4 −13i, √ √ which has norm √ 22. The only√elements of DF with norm ±22 are√±3 ± −13, and indeed 3 + −13 = 22 + (11 + 11 −13) − 5(6 + 2 −13) ∈ p2 p11 , so that p2 p11 = h3 + −13i. This proves that ClF is cyclic, generated by p2 . But p22 = h2i is principal, and p2 is not principal since DF contains no element of norm 2. Therefore ClF is cyclic of order 2 and hF = 2. Theorem 7.6. The only integer solutions to the equation Y 3 = X 2 + 13 are x = ±70, y = 17. Proof. Suppose that x, y ∈ Z satisfy y 3 = x2 + 13. Then lemma 7.4 implies (since we now know that 3 - hF ) that there exists n ∈√ Z such that x = n(n2 − 39) and 3n2 = 13 ± 1. This obviously forces n = ±2, whence 3 x = ±70 and so y = 702 + 13 = 17. Notice that the theorem not only told us that (±70, 17) are the only solutions to the equation, but the proof actually constructed the solutions for us.
7.6
d = −14
√ Let F = Q( −14). The Hurwitz bound is C = 15, so we factor √ h2i = h2, −14i2 √ √ h3i = h3, 1 + −14ih3, 1 − −14i √ √ h5i = h5, 1 + −14ih5, 1 − −14i √ h7i = h7, −14i2 h11i is prime since 8 is not a quadratic residue mod 11 h13i is prime since 12 is not a quadratic residue mod 13 √ √ √ √ So ClF is generated by p2 = h2, √ −14i, p3 = h3, 1 + −14i, p5 = h5, 1 + −14i, √ and p7 = h7,√ −14i. It is easy √ to see that p2 p7 = h −14i. Moreover, p5 p3 contains both 3(1 + −14) and 5( −14), hence √ contains 1 + −14; comparing norms we deduce p5 p3 = h1 + −14i. Exercise 7.3. Considering norms, show that p3 , p23 , and p33 are not principal ideals. On the other hand, show that p23 p2 is principal. √ Theorem 7.7. F = Q( −14). Then ClF is cyclic of order 4, generated by the class of p3 . Proof. The proceeding calculations show that ClF is generated by [p3 ], and that this generator has order ≥ 4. But the exercise also shows that p23 p2 is principal; hence p43 p22 = p43 2 is principal, and so [p2 ]4 = 1. The usual argument now shows that Y 3 = X 2 + 14 has no integer solutions. 47
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7.7
MATH 242 Algebraic Number Theory
d = −15
√ F = Q( −15). Since −15 ≡ 1 mod 4, the Hurwitz bound is | −15−1 |+2 = 6, so we factor (using α = 4 f (X) = X 2 − X + 4)
√ 1+ −15 , 2
h2i = h2, αih2, 1 + αi h3i = h3, 1 + αi2 h5i = h5, 2 + αi2 To finish: ClF will be cyclic of order 2.
7.8
d = −17, −19
These are going to be tough with Hurwitz’s bound.
7.9
d = −23
√ F = Q( −23). The Hurwitz bound is C = | −23−1 | + 2 = 8, so ClF is generated by the classes of prime 4 √ 1+ −23 ideals which sit over 2, 3, 5, 7. Let α = , which has norm NF/Q (α) = 6 and minimal polynomial 2 f (X) = X 2 − X + 6. We factor: p 2 3 5 7
X 2 − X + 6 mod p hpi = X(X − 1) h2, αih2, 1 − αi X(X − 1) h3, αih3, 1 − αi (X + 2)2 − 3 is irred. h5i (X + 3)2 − 3 is irred. h7i
Therefore ClF is generated by the classes of p2 = h2, αi and p3 = h3, αi. But p2 p3 contains α, which has norm 30 = N (p2 p3 ); so p2 p3 = hαi and [p3 ] = [p2 ]−1 . Hence ClF is cyclic, generated by [p2 ], whose order we must now compute. m2 2 2 2 Given integers n, m, we have NF/Q (n + mα) = (n + m 2 ) + 23 4 = n + nm + 6m , which is ≥ 6 if m ≥ 1 2 and is = n if m = 0. So there is no element in DF of norm ±2, and the only element with norm ±4 is 2 itself, which does not generate p22 (otherwise p2 = hα, 2 − αi, which is not true). Hence p2 and p22 are not principal. However, p32 = h8, 4α, 2α2 = −12 + 2α, α3 = −6 − 5αi, which contains 2 − α. Since NF/Q (2 − α) = 8 = N (p32 ), we deduce p32 = h2 − αi. √ Theorem 7.8. The class group of F = Q( −23) is cyclic of order 3, generated by [p2 ].
7.10
d = −30
√ F = Q( −30). The Hurwitz bound if C = 31, so ClF is generated by the prime ideas sitting over 2, 3, 5, . . . , 31, which is far too √many to check by hand quickly! Instead we use Minkowski’s bound, which we know from remark 7.2 is π4 30 ≈ 6.79 in this case; hence ClF is actually generated by the prime ideals sitting over 2, 3, 5. p X 2 + 30 mod p hpi √ = 2 X2 h2, √−30i2 3 X2 h3, √−30i2 2 5 X h5, −30i2 In the obvious notation, we now know that ClF is generated by the classes of p2 , p3 , and p5 , which have norm 2, 3, 5 respectively. Since DF has no elements of norm ±2, ±3, or ±5, we see that [p2 ], [p3 ], [p5 ] are each non-trivial elements of order 2 in ClF . 48
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√ √ √ √ √ both −30 = 3 −30 − 2 −30 and 6, √ so p2 p3 p5 contains −30 = 6 −30 − √ √ Moreover, p2 p3 contains 5 −30; since NF/Q ( −30) = 30 = N (p2 p3 p5 ), we deduce p2 p3 p5 = h −30i. Therefore [p5 ] = [p5 ]−1 = [p2 ][p3 ], so we now know ClF is generated by [p2 ] and [p3 ], which are non-trivial and have order 2. We claim that [p2 ] 6= [p3 ], or equivalently that p2 p3 is not principal; but DF contains no element of order 6, proving the claim. √ Theorem 7.9. The class group of F = Q( −30) is isomorphic to Z/2Z × Z/2Z, with generators [p2 ] and [p3 ] Proof. We have proved that ClF is generated by two distinct elements, each having order 2. The usual argument shows that Y 3 = X 2 + 30 has no integer solutions.
7.11
d = 2, 3, 5, 6, 7, 11, 13, 17, 21, 29, 33, 37, 41
Reference: A. Oppenheim, Quadratic fields with and without Euclid’s algorithm, Math. Ann. 109 (1934), no. 1, pp. 349–352. Lemma 7.10. Let d = 2, 3, 6, or 7, and fix rational numbers 1 ≤ a, b ≤ 21 . Consider the following inequalities, for varying integers x, y: P (x, y) :
(x − a)2 ≥ 1 + d(y − b)2
N (x, y) :
d(y − b)2 ≥ 1 + (x − a)2
Then P (x, y) and N (x, y) are simultaneously false for some pair of integers x, y. Proof. P (0, 0) is clearly false, so if N (0, 0) is also false then we are done; so henceforth assume N (0, 0) is true, i.e. db2 ≥ 1 + a2 . It now easily follows that P (1, 0) is false (for else (1 − a)2 ≥ 1 + db2 ≥ 2 + a2 ≥ 2, which is absurd); so if N (1, 0) is also false then we are done; so henceforth assume N (1, 0) is true, i.e. db2 ≥ 1 + (1 − a)2 . Next, if P (−1, 0) were true, then (1 + a)2 ≥ 1 + db2 ≥ 2 + (1 − a)2 ,
(†)
whence 4a ≥ 2 and so a = 21 ; this would force the left and right sides of (†) to both be 94 , whence 1 + db2 = 94 5 and so b2 = 4d , which is easily seen to be absurd. Hence P (−1, 0) is not true. But since 1 ≤ d < 8 and 1 0 ≤ b ≤ 2 , we have db2 < 2 ≤ 1 + (1 + a)2 , i.e. N (−1, 0) is false. Therefore both P (−1, 0) and N (−1, 0) are false. In conclusion, there exists a pair (x, y) = (0, 0), (1, 0), or (−1, 0) for which P (x, y) and N (x, y) are simultaneously false. √ Theorem 7.11. Let d be any of 2, 3, 5, 6, 7, 13, 17, 21, 29, 33, 37, 41, and put F = Q( d). Then |NF/Q | is a Euclidean norm on DF . Proof. According to exercise 5.1, we must prove that if γ ∈ F then there exists ω ∈ DF such that |NF/Q (α − ω)| < 1. We will only √ give the full proof in the cases that d 6≡ 1 mod 4, i.e. d = 2, 3, 6, or 7. Let γ = p + q d for any p, q ∈ Q; let m, n ∈ Z be the integers which are closest to p, q respectively, so that a := |p − m| and b := |q − n| are ≤ 12 . By the previous lemma there are integers x, y such that −1 + d(y − |q − n|)2 < (x − |p − m|)2 < 1 + d(y − |q − n|)2 49
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Set
( y0 =
(
x − m if p ≥ m m − x if p ≤ m
x0 =
y−n n−y
if q ≥ n if q ≤ n
Then −1 + d(y 0 − q)2 < (x0 − p)2 < 1 + d(y 0 − q)2 , i.e., −1 < NF/Q (γ − ω) < 1, as desired. This completes the proof for those values of d which are 6≡ 1 mod 4. If d ≡ 1 mod 4 then replace the inequalities of the previous lemma by P (x, y) :
(x + 21 y − a)2 ≥ 1 + d(y − b)2
N (x, y) : d(y − b)2 ≥ 1 + (x + 21 y − a)2 Repeating the previous lemma and argument verbatim works whenever 1 ≤ d4 < 8, i.e. d = 5, 13, 17, 21, 29. To extend the proof to d = 33, 37, 41, one must continue the argument of the previous lemma to pairs (x, y) with |x|, |y| ≤ 2. Remark 7.12. The previous theorem remains true for d = 19, 57, and 73, but these are much harder to prove. An even deeper theorem is that this is then the complete list of √positive values of square-free d > 1 for which |NQ(√d)/Q | is a Euclidean norm on the ring of integers of Q( d). For a survey see §4.2 of Lemmermeyer, F., The Euclidean algorithm in algebraic number fields, 2004, http://www.fen.bilkent.edu.tr/~franz/publ/survey.pdf
8
Cyclotomic extensions and Fermat’s Last Theorem
Our main source of examples so far has been the quadratic number fields. In this section we investigate the number field F = Q(ζ) where ζ = ζm = e2πi/n is a primitive nth root of unity. In fact, we will restrict attention to the easiest case, namely n being a prime number, and describe both the ring of integers DF and its group of units. We will then apply our results to prove Fermat’s Last Theorem in the so-called “weak, regular” case.
8.1
Q(ζ) and its ring of integers.
Let p ≥ 3 be a fixed prime number and set ζ = e2πi/p and F = Q(ζ). Then ζ is an algebraic integer with minimal polynomial Φp (X) = X p−1 + · · · + X + 1, by proposition 3.15. So |F : Q| = p−1 and Z[ζ] is a subring of DF . Our first goal is to prove that DF = Z[ζ]; it requires some preliminary results: Lemma 8.1. Suppose that r, s are integers, neither divisible by p. Then ζr − 1 ∈ Z[ζ]× . ζs − 1 Proof. We may write r ≡ st mod p for some t ∈ Z. Then ζr − 1 ζ st − 1 = s = 1 + ζ s + ζ 2s + · · · + ζ (t−1)s ∈ Z[ζ]. s ζ −1 ζ −1 The same argument, after swapping r and s shows that
ζ s −1 ζ r −1
is also in Z[ζ], which completes the proof.
Lemma 8.2. There is a unit u ∈ Z[ζ]× such that p = u(1 − ζ)p−1 . 50
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Proof. We have X p−1 + · · · + X + 1 = Φp (X) =
p−1 Y
(X − ζ i ),
i=1
Qp−1
and we evaluate at 1 to get p = i=1 (1 − ζ i ). But the previous lemma implies that ui := Z[ζ], and so u = u1 . . . up−1 has the desired property: p=
p−1 Y
1−ζ i 1−ζ
is a unit of
ui (1 − ζ) = u(1 − ζ)p−1 .
i=1
It is common to use the notation π = 1 − ζ, and we will do so in the proof of the next result. For the moment notice just two things: 1, π, . . . , π p−2 are a basis for F/Q, and Z[π] = Z[ζ]. Theorem 8.3. As above, let p ≥ 3 be a prime number, ζ := e2πi/p , and F := Q(ζ). Then DF = Z[ζ]. Proof. The inclusion ⊇ has already been noted, so we suppose that α ∈ DF and we will prove α ∈ Z[ζ]. The proof proceeds by two steps, firstly using Z[ζ] = Z[π], then DF . Since 1, ζ, . . . , ζ p−2 is a basis for F/Q, we may write α = c0 + c1 ζ + · · · + cp−2 ζ p−2 for some c0 , . . . , cp−2 ∈ Q (which we don’t yet know are in Z). For i an integer not divisible by p, the algebraic integer ζ i generates F and has minimal polynomial Φp (X), whence TrF/Q (ζ i ) = −1 by the standard formula for the trace of an element in terms of its minimal polynomial. Therefore, for 0 ≤ r ≤ p − 2, TrF/Q (αζ −r ) = TrF/Q (c0 ζ −r + · · · + cr−1 ζ −1 + cr + cr+1 ζ + · · · + cp−2 ζ p−2−r ) = −c0 − · · · − cr−1 + (p − 1)cr − cr−1 − · · · − cp−2 , which is in Z since αζ −r ∈ DF . The same argument show that TrF/Q (αζ) = −c0 − · · · − cp−2 is also in Z. Comparing these we see that pcr ∈ Z for r = 0, . . . , p − 2, and so pα ∈ Z[ζ]. Since Z[ζ] = Z[π], we may therefore write pα = b0 + b1 π + · · · + bp−2 π p−2 for some b0 , . . . , bp−2 ∈ Z. The next step of the proof takes place in DF . To complete the proof it is enough to show that b0 , . . . , bp−2 are divisible by p: (i) The previous lemma implies p ∈ hπi and so b0 = pα − (b1 π + · · · + bp−1 π p−2 ) ∈ hπi ∩ Z = pZ. (ii) Proceeding by induction, suppose b0 , . . . , bs−1 can be written as pb00 , . . . , pb0s−1 for some b00 , . . . , b0s−1 ∈ Z. We now have bs π s = pα − p(b00 + · · · + b0s−1 π s−1 ) − (bs+1 π s+1 + · · · + bp−2 π p−2 ). But p ∈ hπip−1 (by the previous lemma) and α ∈ DF (by assumption), so every term on the right of this expression belongs to hπis+1 , and therefore bs π s ∈ hπis+1 . So bs ∈ hπi, whence bs ∈ hπi ∩ Z = pZ completing the proof of our claim. 51
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We now know that F = Q(ζ) has ring of integers DF = Z[ζ], so an integral basis is 1, ζ, ζ 2 , . . . , ζ p−2 . Example 8.4. Since ( p−1 TrF/Q (ζ ) = −1 i
p|i p - i,
we deduce that
p−1 −1 −1 p−2 ∆F = ∆(1, ζ, . . . , ζ ) = −1 .. . −1 −1
−1 −1 −1 ···
−1 ··· ··· −1 . ..
··· −1 −1 −1 . ..
−1 −1 −1 p−1
−1 −1 p−1 −1 .. .
−1 −1
−1 p−1
p−1 ···
··· −1
−1 −1
(i.e. −1s everywhere except for at the top left and along a diagonal strip). Some rather tedious combinatorics allows one to evaluate this as ∆F = (−1)(p−1)/2 pp−2 . Our next goal is to describe some of the units in DF . The proofs of the following lemma requires the use of some mild Galois theory. Notice that DF is closed under complex conjugation x 7→ x. Lemma 8.5. (i) Let α ∈ Zalg be an algebraic integer with the following property: all the conjugates of α have complex absolute value 1. Then α is a root of unity. (ii) If u ∈ D×F then u/u is a root of unity. Proof. For (i), set S = {β ∈ Z[α] : all conjugates of β have complex absolute value ≤ 1}, which is finte by exercise 4.1. We will show that αr ∈ S for all r ≥ 0, which obviously forces αr = αs for some s > r ≥ 1; but then αs−r = 1, which will complete the proof of the first claim. Well, if β is a conjugate of αr then there is a field embedding σ : Q(α) ,→ C such that σ(αr ) = β. Then σ(α) is a conjugate of α, hence has complex absolute value 1; so β = σ(α)r also has complex absolute value 1. So all conjugates of α have complex absolute value 1, and therefore α ∈ S, as desired. For (ii), since F/Q is a Galois extension, the conjugates of α := u/u ∈ DF are precisely {σ(α) : σ ∈ Gal(F/Q)}. The Galois group Gal(F/Q) not only contains complex conjugation, but it is an abelian group (isomorphic to (Z/pZ)× ); this means that σ(u) = σ(u) for all σ ∈ Gal(F/Q). So each conjugate of u/u has the form σ(u/u) = σ(u)/σ(u) for some σ ∈ Gal(F/Q): this number obviously has complex absolute value 1. The first part of the lemma now implies that u/u is a root of unity. Proposition 8.6. Let u ∈ D×F . There there exist v ∈ D×F , such that v = v, and r ∈ Z such that u = ζ r v. 52
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Proof. By the previous lemma u/u is a root of unity contained in DF , hence equal to ±ζ s for some s ∈ Z (by Homework). For a contradiction, suppose that u/u = −ζ s for some s ∈ Z. Since we now know DF = Z[ζ], we may write u = a0 + · · · + ap−2 ζ p−2 for some a0 , . . . , ap−2 ∈ Z, and so u ≡ a0 + · · · + ap−2 mod h1 − ζi Similarly, u = a0 + a1 ζ −1 · · · + ap−2 ζ −(p−2) ≡ a0 + · · · + ap−2 ≡ u = −ζ s u ≡ −u mod h1 − ζi. Therefore 2u ∈ h1−ζi, which is a prime ideal by lemma 8.2; since 2 ∈ / h1−ζi∩Z = pZ, we deduce u ∈ h1−ζi. But u is a unit, so this is impossible. So it must be the case that u/u = ζ s for some r ∈ Z. Let r ∈ Z satisfy 2r ≡ s mod p, and put v = ζ −r u. Then ζ r v = ζ r ζ −r u = u and v = ζ r u = ζ r ζ −s u = ζ r ζ −2r u = ζ −r u = v, as desired.
8.2
Fermat’s Last Theorem
In this section we will prove a special case of Fermat’s Last Theorem; we must begin with the following definition: Definition 8.7. A prime number p ≥ 3 is said to be regular if and only if the class number of Q(ζp ) is not divisible by p, where ζp = e2πi/p . Remark 8.8. The following primes are known to be regular: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, while 37, 59, 67, 101, 103, 131, 149 are not. 1 C. Siegel conjectured in 1964 that e− 2 of all primes should be regular (≈ 61%), but it remains unknown even whether there are infinitely many regular primes. On the other hand, it is know that there are infinitely many irregular (i.e., not regular) primes. The arithmetic of the cyclotomic fields Q(ζp ) is closely connected to the Bernoulli numbers; these are the sequence of rational numbers B0 , B1 , . . . defined by the generating function expression ∞
X Bk X = Xk, exp X − 1 k! k=0
P∞
k
where exp X = k=0 Xk! is the formal exponential. In particular, Kummer showed that a prime p is regular if and only if it does not divide the denominator of B2 , B4 , . . . , Bp−3 . Since the Bernoulli numbers can be easily calculated, either directly from the definition we have given or from various closed form expressions, this gives a straightforward way of checking whether any given prime number is regular (but does not offer much information on their theoretic properties!). We may now state our special case of FLT: Theorem 8.9 (Fermat’s Last Theorem: weak, regular case. Due to E. Kummer). Assume that p is a regular prime and that x, y, z are integers, none of which are divisible by p (this is sometimes called the “weakness” hypothesis). Then xp + y p 6= z p . 53
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Outline of proof and historical remarks. To prove the theorem we will work in the number field F = Q(ζ) where ζ = e2πi/p for some fixed prime number p ≥ 5 (it is easy to prove the theorem when p = 3, using mod 9 congruences). Using very similar arguments to lemma 7.4, which in particular require regularity of p and our theory of the class group, we will show that if x, y, z are integers which contradict the statement of the theorem, then (possibly after slightly modifying x, y, z – see lemma 8.11) we can write x + ζy = uαp
(†)
D× F,
for some u ∈ α ∈ DF . This is the key step, from which it will be easy to complete the proof. As in lemma 7.4, if we were to simply erroneously assume that the ring Z[ζ] were a UFD, then we would be able to prove (†) very quickly (without any mention of rings of integers, ideals, or the class group). This was the famous incorrect proof of FLT presented by G. Lam´e in 1847 which apocryphally led E. Kummer to develop a theory of “ideal numbers” which later become ideals and the class group. Having finished the outline, we now turn to proving the theorem properly. We first need some small lemmas: Lemma 8.10. Let p ≥ 5 and F = Q(ζ) be as above. (i) Suppose that a0 , . . . , ap−1 are integers, at least one of which is zero. Suppose that m is an integer such that a0 + · · · + ap−1 ζ p−1 ∈ hmi (= mDF ). Then ai ∈ mZ for i = 0, . . . , p − 1. (ii) Let α ∈ DF ; then there exists a ∈ Z such that αp ≡ a mod hpi. Proof. (i). Suppose ar = 0 for some 0 ≤ r ≤ p − 1. Multiply through by ζ p−1−r to reduce to the case ap−1 = 0, which then easily follows from the fact that 1, ζ, . . . , ζ p−2 was shown to be an integral basis for F . (ii). Since DF = Z[ζ] we may write α = a0 + a1 ζ + · · · + ap−2 ζ p−2 for some a0 , . . . , ap−2 ∈ Z. In any ring R, one has (x + y)p ≡ xp + y p mod pR. Applying this repeatedly we get αp ≡ ap0 + ap1 ζ p + · · · + app−2 ζ p(p−2) mod hpi = ap0 + ap1 + · · · app−2 , so that a := ap0 + ap1 + · · · app−2 works. Lemma 8.11. Suppose that theorem 8.9 is false. Then there are integers x, y, z such that xp + y p = z p , p does not divide any of x, y, z, and x 6≡ y mod p, x, y are coprime. In other words, we can modify our (non-existent) solution to also satisfy the second two conditions. Proof. Suppose that x, y, z are integers such that xp + y p = z p and p does not divide any of x, y, z. We claim it is impossible that x ≡ y ≡ −z mod p. For, if not, then −2z p ≡ z p mod p, whence 3z ≡ 3z p ≡ p mod 0; but p - 3 and p - z, so this is impossible. Therefore at least one of the triples x, y, z,
x, −z, −y
satisfies the first three desired properties. Divide through by the gcd to ensure that the fourth condition also holds. 54
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Lemma 8.12. Suppose that x, y, z satisfy the conditions of the previous lemma. If i, j are integers such that i 6≡ j mod p then the ideals hx + ζ i yi, hx + ζ j yi of DF are coprime. Proof. For a contradiction, suppose that there is a prime ideal p of DF containing both x + ζ i y and x + ζ j y. After swapping i and j we may suppose that i > j. Then p 3 ζ i y − ζ j y = ζ i (1 − ζ i−j )y = ζ i
1 − ζ i−j (1 − ζ)y 1−ζ
and so (1 − ζ)y ∈ p (the other terms are units, by lemma 8.2). Therefore either 1 − ζ ∈ p or y ∈ p. Similarly, 1 − ζ i−j (1 − ζ)x p 3 (x + ζ i y) − ζ i−j (x + ζ j y) = 1−ζ and so either 1 − ζ ∈ p or x ∈ p. We claim that 1−ζ ∈ p. Otherwise we deduce that x, y ∈ p, whence x, y ∈ p∩Z, contradicting coprimality of x and y. So, 1 − ζ ∈ p, whence p = h1 − ζi since h1 − ζi is a prime ideal by lemma 8.2 (recall that prime ideals are maximal ideals). Therefore x + y = x + ζ i y + (1 − ζ i )y ∈ p, and so x + y ∈ p ∩ Z = pZ. Finally, z p = xp + y p ≡ (x + y)p ≡ 0 mod p, so we see that p|z, which is the required contradiction. Now we can finish the proof of theorem 8.9: Proof of FLT’s in the weak, regular case. For a contradiction suppose that theorem 8.9 is false. Then we have shown that there are integers x, y, z satisfying the conditions of lemma 8.11, and the previous lemma tells us that the ideals hx + yi, hx + ζyi, · · · , hx + ζ p−1 yi are pairwise Qp−1coprime. But i=0 (x + ζ i y) = xp + y p = z p , so the product of these coprime ideals is a pth -power: p−1 Y
hx + ζ i yi = hzip .
i=0
By exercise 7.1(ii), each ideal is already a pth -power: hx + ζ i yi = Iip for some non-zero ideal Ii ⊆ DF . We only care about i = 1: since I1p is principal, our assumption that p is regular, i.e. that p - hF , implies that I1 is already principal by exercise 7.1(iii); so I1 = hαi for some α ∈ DF . Therefore x + ζy = uαp for some u ∈ D×F : as explained in the outline to the proof, this is the key step. By proposition 8.6 we may write u = ζ r v for some r ∈ Z and some unit v ∈ D× F such that v = v. By lemma 8.10(ii) there is a ∈ Z such that αp ≡ a mod hpi. So x + ζy = ζ r vαp ≡ ζ r va mod hpi Similarly, taking complex conjugates, x + ζ −1 y = ζ −r vαp ≡ ζ −r va mod hpi, 55
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and so ζ −r (x + ζy) ≡ ζ r (x + ζ −1 y) mod hpi. In other words, x + ζy − ζ 2r x − ζ 2r−1 y ∈ hpi
(†)
Finally, we will analyse several cases to get a contradiction from (†): (i) Case: 1, ζ, ζ 2r , ζ 2r−1 are all distinct. Then lemma 8.10(i) implies that x, y ∈ pZ, contradicting our original assumption that p - x and p - y. (ii) Case: 1 = ζ 2 r. Then (†) implies that ζy −ζ −1 y ∈ hpi, whence lemma 8.10(i) implies (using ζ −1 = ζ p−1 ) that y ∈ pZ, contradicting our original assumption that p - y. (iii) Case: 1 = ζ 2r−1 . Then (†) implies that (x − y) − (x − y)ζ ∈ hpi, whence lemma 8.10(i) implies x − y ∈ pZ, contradicting our original assumption that x 6≡ y mod p. (iv) Case: ζ = ζ 2r−1 . Then (†) implies that x − ζ 2 x ∈ hpi, whence lemma 8.10(i) implies x ∈ pZ, contradicting our original assumption that p - x. These cases are the only possibilities, since 1 6= ζ and ζ 2r 6= ζ 2r−1 , so the proof is complete.
9
Ramification theory
If F is a number field and p ∈ Z is a prime number then in section 6 we have already studied the factorisation of the principal ideal hpi in DF (especially in the case in which DF = Z[α] for some α ∈ DF ). The name given to the study of such factorisations is ramification. Whereas we previously studied this from a hands-on perspective of computing factorisations, now we will examine it theoretically. Definition 9.1. Let p ⊂ DF be a prime ideal, and let p be the prime number sitting under it. Recalling that N (p) is a power of p, we define the inertia degree, fp ≥ 1, of p by the formula N (p) = pfp . Meanwhile, the ramification degree, ep ≥ 1 is the exponent of p in the prime ideal factorisation of hpi (or equivalently, ep := ordp p). Definition 9.2. This definition is a little long, but explains all the notation used for how the prime factorisation of hpi may look. Let F be a number field of degree n over Q. Let p ∈ Z be a prime number, let hpi be the principal ideal of DF generated by p, and let hpi = pe11 . . . perr (e1 , . . . , em ≥ 1) be its prime factorization in DF . Then ei is called the ramification degree of p at pi . Then, p is said to be ramified in F , or to ramify in F if and only if ei > 1 for some i (in other words, if and only if there exists a prime ideal p of DF such that p2 3 p) and otherwise p is said to be unramified in F when ei = 1 for all i. When p is unramified there are two extreme possibilities (notice that these conditions really do imply that p is unramified): p is said to be inert in F when hpi is a prime ideal of DF . p is said to split completely when e1 = · · · = em = 1 and fp1 = · · · = fpm = 1 (by the previous theorem, p splits completely if and only if hpi is a product of n distinct prime ideals). 56
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Remark 9.3. Suppose that F = Z[α] for some α ∈ DF , and let f (X) ∈ Z[X] be the minimal polynomial of α. Then our hands-on factorisation result implies that the prime ideal factorisation of hpi in DF is described exactly by the factorisation of f (X) mod p into irreducibles. Thus the study of how prime numbers ramify in F is exactly the study of how the fixed polynomial factors mod p, as we vary p. Exercise 9.1. Let F be a number field and suppose that there is α ∈ DF such that DF = Z[α]; let f (X) ∈ Z[X] be the minimal polynomial of α. Let ` ∈ Z be a prime number. Suppose that there is a polynomial g(X) ∈ Z[X] with the following two properties: (i) f (X) divides g(X); (ii) g(X) mod ` and g 0 (X) mod ` are coprime in F` [X]. Show that ` is unramified in F . Exercise 9.2. Let p ∈ Z be an odd prime number, ζ = e2πi/p , and F = Q(ζ). In this exercise you will prove that the only roots of unity in F have the form ±ζ n for some n ∈ Z. (We will need this to prove a special case of Fermat’s Last Theorem.) For a contradiction, suppose that F contains a root of unity which cannot be written in the form ±ζ n for some n ∈ Z; show that F contains one of the the following: (i) e2πi/` for some prime number ` ∈ Z not equal to p; or 2
(ii) e2πi/p ; or √ (iii) −1. In each case, arrive at a contradiction by following the hints: (i) Show that the prime ideal factorization of h`i has the form h`i = l(`−1)e1 . . . l(`−1)er , for some prime ideals l1 , . . . , lr of DF and some integers e1 , . . . , er ≥ 1. Apply the previous question to get a contradiction. 2
(ii) Put p = h1 − ζi; show that e2πi/p ≡ 1 mod p and then show 1 − ζ ∈ p2 . Derive a contradiction. (iii) Use a similar argument as in (a): show that the prime ideal factorization of h2i has the form 2er 1 h2i = q2e 1 . . . qr
for some prime ideals q1 , . . . , qr of DF and some integers e1 , . . . , er ≥ 1; then apply the previous question to get a contradiction. Proposition 9.4. Let F be a number field of degree n over Q, and assume that F/Q is Galois (If you don’t know what this means, this is the only case we need: F/Q will be Galois if F = Q(α) for some α ∈ F such that all the conjugates of α also belong to F ). Let p ∈ Z be a prime number, let hpi be the principal ideal of DF generated by p, and let hpi = pe11 . . . pemm (e1 , . . . , em ≥ 1) be its prime factorization in DF . Then e1 = · · · = em and fp1 = · · · = fpm . 57
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Sketch of proof. For any i = 1, . . . , r and σ ∈ Gal(F/Q), the set σ(pi ) := {σ(x) : x ∈ pi } is a prime ideal containing σ(p) = p, hence σ(pi ) = pj for some other j. In other words, Gal(F/Q) acts on {p1 , . . . , pr }. The key point is to prove that this action is transitive. So fix indices i, j and suppose for a contradiction that σ(p S i ) 6= pj for all σ ∈ Gal(F/Q). Then pj 6⊆ σ(pi ), so the prime avoidance lemma implies that pj 6⊆ σ∈Gal(F/Q) σ(pi ). So there exists an element α ∈ pj such that σ(α) 6∈ pi for any σ (here we have replaced σ by σ −1 , since we are running over all σ ∈ Gal(F/Q)). Then Y Y NF/Q (α) = σ(α) = α σ(α) ∈ pj ∩ Z = pZ ⊆ pi , σ
σ6=id
implying that at least one of the terms in the product is in pi , which is the desired contradiction. So, we now know that for any i, j = 1, . . . , m we can find σ such that σ(pi ) = pj . Then ei = ordpi hpi = ordσ(pi ) σ(hpi) = ordpj hpi = ej . Also, σ induces an isomorphism
DF /pi ∼ = DF /pj ,
whence fpi = fpj .
9.1
Ramification in quadratic extensions and quadratic reciprocity
We have factored many prime numbers in particular quadratic extensions; now we return to the subject from a theoretic point of view and relate it to quadratic reciprocity. √ For this whole section fix a quadratic extension F = Q( d), where d ∈ Z \ {0, 1} is a square-free integer as usual. Recall that the absolute discriminant of F is ( 4d d ≡ 2, 3 mod 4 ∆F = d d ≡ 1 mod 4 If p ∈ Z is a prime number, then hpi ⊂ DF factors into a product of at most two prime ideals; therefore it must factor in exactly one of the following ways: (i) hpi is a prime ideal of DF (i.e. p is unramified and inert). (ii) hpi = p1 p2 for some non-zero, distinct, prime ideals p1 , p2 ⊆ DF (i.e. p is unramified and splits completely). (iii) hpi = p2 for some non-zero prime ideal p ⊆ DF (i.e. p ramifies). The aim of this section is to prove the following theorem: √ Theorem 9.5. Let F = Q( d) be as above. Then The way that hpi factors depends only on the value of p mod ∆F . In other words, if p, p0 ∈ Z are prime numbers and p ≡ p0 mod ∆F , then p is inert/completely split/ramified in F if and only if the same is true of p0 . In particular, a prime number ramifies in F if and only if it divides ∆F . The proof will crucially depend on all the main theorems about quadratic residues: the descriptions of 2 −1 p , p , and the Law of Quadratic reciprocity itself. The theorem will be proved in several steps. We begin by showing that a Legendre symbol determines the ramification, a result we have essentially already used when factoring hpi in particular quadratic number fields. √ Lemma 9.6. Let F = Q( d) be as above, and p an odd prime number. Then (i) p is inert in F ⇐⇒ dp = −1. 58
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(ii) p splits completely in F ⇐⇒ (iii) p ramifies in F ⇐⇒
d p
d p
= 1.
= 0 (which happens ⇐⇒ p | ∆F ).
Proof. To avoid separately considering the cases d ≡ 2, 3 mod 4 and d ≡ 1 mod 4, put ( √ √ 2d + d if d ≡ 2, 3 mod 4 ∆F + ∆F α := = d−1 1+√d 2 + if d ≡ 1 mod 4 2 2 Then DF = Z[α] and α has minimal polynomial f (X) = X 2 − ∆F X + Let s = p+1 2 so that 2s ≡ 1 mod p. Then
∆F (∆F −1) 4
∈ Z[X].
f (X) ≡ (X − s∆F )2 − s2 ∆F mod p, so the correspondence between the factorisation of hpi and that of f (X) mod p immediately gives: 2 F = −1. (i) p is inert in F ⇐⇒ f (X) is irreducible ⇐⇒ s2 ∆F is not a square in Z/pZ ⇐⇒ s ∆ p 2 (ii) p splits completely in F ⇐⇒ f(X) is a product of two distinct irreducible polynomials ⇐⇒ s ∆F is a 2 non-zero square in Z/pZ ⇐⇒ s p∆ = 1. 2 (iii) p ramifies in F ⇐⇒ f (X) is the square of an irreducible polynomial ⇐⇒ s ∆F is zero in Z/pZ ⇐⇒ 2 s ∆F = 0. p
To complete the proof notice that
s2 ∆F p
=
d p
, regardless of whether ∆F = 4d or d.
Next we give an analogue of the previous result for p = 2: √ Lemma 9.7. Let F = Q( d) be as above. Then (i) 2 is inert in F ⇐⇒ d ≡ 5 mod 8. (ii) 2 splits completely in F ⇐⇒ d ≡ 1 mod 8. (iii) 2 ramifies in F ⇐⇒ d ≡ 2 or 3 mod 4 (which happens ⇐⇒ 2 | ∆F ). Proof. Since the left and right sides constitute all possibilities, it is enough to prove ⇐= in each case. √ 1+ d (i): Then d ≡ 1 mod 4, so DF = Z[α], where α = 2 , with minimal polynomial f (X) = X 2 − X − d−1 4 . Then f (X) ≡ X 2 − X − 1 mod 2, which is well-known to be irreducible in Z/2Z[X] (since it doesn’t have a root) so h2i is a prime ideal in DF ; i.e. 2 is inert in F . √ (ii): Again d ≡ 1 mod 4 and DF = Z[α], where α = 1+2 d , with minimal polynomial f (X) = X 2 − X − d−1 4 . But now f (X) ≡ X(X − 1) mod 2, so h2i = h2, αih2, α − 1i is a product of distinct prime ideals in DF ; i.e. 2 splits completely in F . √ (iii): In this case DF = Z[ d] and we put f (X) = X 2 − d. Then ( X 2 + 1 = (X + 1)2 d ≡ 3 mod 4 f (X) = X2 d ≡ 2 mod 4 so h2i is a square of a prime ideal in DF ; i.e. 2 ramifies in F . Notice that parts (iii) of the previous two lemmas already prove the claim in the main theorem that a prime ramifies if and only if it divides the ∆F . Now we may prove the rest of the main theorem: 59
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Proof of theorem 9.5. We first obtain a formula for a certain Legendre symbol. The prime factorisation of d has the form d = 2ε q1 · · · qm where ε = 0 or 1, and q1 , . . . , qm are distinct odd prime numbers, which are allowed to be negative to accommodate for the fact that d may be negative. (Being careful, notice that if d = −2 then it cannot be expressed in this form, but the reader will have no difficulty introducing an extra piece of notation to fix this problem.) So, for any odd prime number p ∈ Z, ε 2 q1 qm d = ... p p p p ε qm −1 p−1 q1 −1 2 p p = ... (−1) 2 ( 2 +···+ 2 ) p q1 qm ε p−1 2 p p = (†) ... (−1) 2 r p q1 qm by quadratic reciprocity, where r is the number of the primes among q1 , . . . , qm which are ≡ 3 mod 4. Now let p, p0 ∈ Z be prime numbers which are congruent modulo ∆F ; we will prove that p, p0 ramify in the same way in DF , thereby completing the proof of the theorem. We will separately consider two cases: either p, p0 are both odd, or p is odd and p0 is even (there is nothing to prove if both p and p0 are even!!). 0 0 Suppose first that both p and p are odd. Then lemma 9.6 tells us that the ramification-type of p, p is d d d d 0 determined by p , p0 , so we must prove p = p0 . Using formula (†) and the fact that p ≡ p mod qi for all i, it remains only to prove that ε ε p−1 p0 −1 2 2 r 2 (−1) = (‡) (−1) 2 r . 0 p p Well, if d ≡ 3 mod 4 then ε = 0 and 4|∆F so that 0
d ≡ 2 mod 4 then ε = 1 and 8|∆F , so p ≡ p mod 8, p−1 2
p0 −1 2
p0 −1 (‡) in this case. Next, 2 mod 2, proving 2 2 whence p = p0 by the Legendre symbol of
p−1 2
≡
if 2
0
theorem, and still ≡ mod 2 since p ≡ p mod 4; again (‡) follows. Finally, if d ≡ 1 mod 4 then ε = 0 and r must beeven, whence of (‡) are 1 (moreover, this shows that if d ≡ 1 mod 4 and p is sides both p p d an odd prime, then p = q1 . . . qm ; we will need this in a moment). This completes the proof when both p and p0 are odd. The remaining case is to assume that p is odd and that p0 = ±2. This forces d ≡ 1 mod 4, for otherwise 4|∆F , implying that p ≡ ±2 mod 4, which is absurd. So d ≡ 1 mod 4, and we just noticed that this implies d p p = ... p q1 qm q−1 But moreover p ≡ ±2 mod qi for all i, so the Legendre symbol of 2 theorem and the identity −1 = (−1) 2 q now give 2 −1 2 −1 q1 qm q1 −1 qm −1 d = (±1) 2 +···+ 2 (−1) 8 +···+ 8 p The exponent of the ±1 is congruent mod 2 to r, which is even since d ≡ 1 mod 4, so this factor is = 1. 2 Next, the exponent of the −1 is congruent mod 2 to d 8−1 (this follows inductively from the simple identity 2 2 2 2 that a 8−1 + b 8−1 ≡ a b8 −1 mod 2 whenever a, b are odd integers). In conclusion, ( d2 −1 1 d ≡ 1 mod 8 d = (−1) 8 = p −1 d ≡ 5 mod 8 60
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Finally, looking at the previous two lemmas, we see that if d ≡ 1 mod 8 then both p and 2 completely split, while if d ≡ 5 mod 8 then both p and 2 are inert. This completes the proof of the main theorem. Remark 9.8. The previous theorem is probably the most important consequence of quadratic reciprocity: the way hpi factors in DF depends only on a certain congruence class of p. We will see similar results for other number fields; the most general abstract result of this type is an important consequence of so-called “class field theory”, which is the next step in algebraic number theory.
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9.2
MATH 242 Algebraic Number Theory
Ramification in cyclotomic extensions
In this section we will prove an analogue for cyclotomic extensions of theorem 9.5: the way in which a prime number ramifies is still determined by a congruence condition. If ζ = e2πi/m is a primitive mth root of unity for some integer m, then it is a fact that DF = Z[ζ]. We have proved this when m is a prime number and it was HW exercise ????? if m is a prime power; the case of general m can be deduced from the prime power case (non-trivially). Theorem 9.9 (“Cyclotomic reciprocity law” – a slightly misleading name!). Let m ≥ 1 be an integer and let F = Q(ζ), where ζ = e2πi/m is a primitive mth root of unity. Let p ∈ Z be a prime number. Then (i) p ramifies in F if and only if p divides m. (ii) Suppose p doesn’t divide m. Let s ≥ 1 be the smallest integer such that ps ≡ 1 mod m; in other words, s is the order of p mod m in the unit group (Z/mZ)× . Then the prime ideal factorisation of hpi has the form hpi = p1 · · · pr , where p1 , . . . , pr are distinct primes ideals with norm ps , and r = |F : Q|/s. In particular, the factorisation type of p depends only on p mod m. Proof. If p is a prime number dividing m, then ζp = ζ m/p is a primitive pth root of unity; so F contains Q(ζp ). Since p ramifies in Q(ζp ), it ramifies in F . Conversely, suppose that p is a prime number not dividing m. The minimal polynomial of ζ, which is commonly denoted Φm (X), divides X m − 1. Since X m − 1 and its formal derivative mX m−1 are coprime in Fp [X], the prime number p is unramified in F by exercise 9.1. We have proved (i). Continuing to suppose that p is a prime number not dividing m, let s ≥ 1 be as in (ii) and let p ⊂ DF be any prime ideal sitting over p. To complete the proof we must show N (p) = ps (the description of r comes from taking norms of both sides). Let f be the inertia degree of p; i.e., N (p) = pf . Put K = DF /p, which is a finite field with pf elements. Let z ∈ K denote the element ζ mod p. Noting first that z m = 1 in k, we claim that the order of z in the group K × is exactly m. For a contradiction, suppose not. Then z m/` = 1 for some prime number ` dividing m, which means that 1−ζ m/` ∈ p. But ζ` = ζ m/` is a primitive `th root of unity, so lemma 8.2 implies that ` = u(1 − ζ` )`−1 for some unit u ∈ Z[ζ` ] ⊆ DF . Therefore ` ∈ p. But ` and p are distinct primes, since only one of them divides m, and p is also in p, by choice of p; so this is a contradiction. This proves our claim that the order of z in the group K × is exactly m. Lagrange’s theorem from group theory now tells us that m divides #K × = pf − 1, i.e. pf ≡ 1 mod m; therefore f ≥ s. The theory of finite fields now tells us that s
M = {x ∈ K : xp = x} is a subfield of K such that |K : M | = f − s. Since m divides ps − 1 (by the definition of s), we see that s s z p −1 = 1 and so z p = z, i.e. z ∈ M . But DF = Z[ζ], so any element of K = DF /p is an integer linear combination of powers of z; but all these integer linear combinations lie in M , since ζ ∈ M , so this proves that K = M . Therefore f = s, completing the proof
10
A new proof of quadratic reciprocity
Let p ≥ 3 be a prime number, let ζ = e2πi/p be a primitive pth root of unity, and let F = Q(ζ). In this section we are going to present a proof of the Law of Quadratic Reciprocity using the algebraic number theory of F . The key tools will be the following two points, together with some basic Galois theory: 62
MATH 242 Algebraic Number Theory
• F contains
√
p∗ , where p∗ = (−1)
p−1 2
63
.
• The Galois group of F/Q is isomorphic to (Z/pZ)× , via '
(Z/pZ)× → Gal(F/Q),
a 7→ σa ,
where σa is defined by the rule σa (ζ) = ζ a . Although we have not proved this, many of you know it from Galois theory. We need two provisional lemmas: Lemma 10.1. If a is an integer which is coprime to p, then √ a √ ∗ p . σa ( p∗ ) = p √ √ √ √ √ Proof. Since the only conjugates of p∗ are ± p∗ , we see that σa ( p∗ ) is either p∗ or − p∗ . Therefore it is enough to prove that √ ∗ √ ∗ a σa ( p ) = p ⇐⇒ = 1. p √ √ √ Well, σa ( p∗ ) = p∗ if and only if σa fixes the subfield M = Q( p∗ ) ⊂ F , i.e. if and only if σa ∈ Gal(F/M ) ⊂ Gal(F/Q). Since Gal(F/Q) is cyclic of order p − 1, its index two subgroup Gal(F/M ) is ∼ precisely the set {σ 2 : σ ∈ Gal(F/Q)}. According to the isomorphism = (Z/pZ), this means that Gal(F/Q) √ ∗ √ ∗ a σa ( p ) = p if and only if a is a square mod p, i.e. if and only if p = 1. Lemma 10.2. If q ≥ 3 is a prime number distinct from p, then σq (α) ≡ αq mod hqi for all α ∈ DF . Proof. Put S = {α ∈ DF : σq (α) ≡ αq mod hqi}. Then S is obviously closed under multiplication, and the standard identity (α + β)q ≡ αq + β q mod hqi implies it is closed under addition. Clearly S contains 1 and ζ, so we now know that S is a subring of DF containing ζ. But DF = Z[ζ], so S = DF , as desired. Now we may give the new proof of quadratic reciprocity. Let q ≥ 3 be a prime number distinct from p. Combining the previous two lemmas gives us √ √ q q √ ∗ p = σq ( p∗ ) ≡ p∗ mod hqi p q−1 q−1 q−1 p−1 q−1 q−1 i.e., pq ≡ p∗ 2 mod hqi. So pq − p∗ 2 ∈ hqi ∩ Z = qZ. Also, p∗ 2 = (−1) 2 2 p 2 , giving p−1 q−1 q−1 q ≡ (−1) 2 2 p 2 mod q p q−1 Finally, recall from Euler’s lemma (lemma 2.6(i)) that p 2 ≡ pq mod q, and that a mod q congruence when both sides are ±1 is necessary an equality; hence p−1 q−1 q p = (−1) 2 2 , p q which is exactly the Law of Quadratic Reciprocity!
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