MA3A6 Algebraic Number Theory David Loeffler Term 2, 2014–15 Chapter 0 Introduction Lecture 1 0.1 What is this module about? This is a module about algebraic number fields. An algebraic number field is a special kind of field, which contains the rational numbers Q, but is a little bit bigger. We’ll give a formal definition soon enough, but a good example to bear in mind is the Gaussian field Q(i) = fa + bi : a, b 2 Qg, which comes with its subring of Gaussian integers Z[i] = fa + bi : a, b 2 Zg. Exercise. Why is the Gaussian field a field? (Most of the axioms are straightforward, but why is it closed under inverses?) In Algebra 2 you saw that Z[i] was a unique factorization domain, and you used this to show that any prime number p = 1 mod 4 could be written as the sum of two squares, p = x2 + y2. So rings like Z[i] have some interesting structure; and they tell us new things about Z. 0.2 Logistics • There will be 4 problem sheets, which will be distributed as we go along. These count for 15% of your grade. The deadlines will be – Sheet 1: distributed Thursday, week 2; deadline 3pm Monday, week 4. – Sheet 2: distributed Thursday, week 4; deadline 3pm Monday, week 6. – Sheet 3: distributed Thursday, week 6; deadline 3pm Monday, week 8. – Sheet 4: distributed Thursday, week 8; deadline 3pm Monday, week 10. • Weekly office hour: Tuesdays 13.30–14.30, Zeeman B1.25. • Support classes with Heline Deconinck: Fridays 11–12, MS.04, from week 2 onwards. 1 • Books: see list on Undergraduate Handbook page. The main reference is Stewart & Tall, which is also probably the friendliest of the books on the list; Swinnerton-Dyer’s book is harder going, but was the book which inspired me to become a number theorist. • Most of you have done Galois theory, and about half of you are doing Commutative Algebra. 2 Chapter 1 Algebraic number fields 1.1 Extensions of fields Notation 1.1.1. Let K and L be fields. If K is a subfield of L, we say L is a field extension of K, and we write L j K. For instance, C j Q is a field extension, as is C j R. Definition 1.1.2. Let L j K be a field extension, and let a 2 L. We say a is algebraic over K if there exists a nonzero polynomial g 2 K[X] such that g(a) = 0. Example 1.1.3. In the extension C j R, the element ip is algebraic over R (it’s a root of X2 + p2). However, it is not algebraic over Q. Proposition 1.1.4. Let a be algebraic over K. Then there is a unique polynomial f 2 K[X] such that f (a) = 0 and f is irreducible and monic (its leading coefficient is 1). We call this the minimal polynomial of f over K. Proof. Recall from Algebra 2 the concept of an ideal and a principal ideal. The set I ⊂ K[X] of polynomials g such that g(a) = 0 is an ideal of K[X]; the ring K[X] is a Euclidean domain, so every ideal of this ring is principal, i.e. consists of the multiples of some polynomial f (which we can assume is monic, by multiplying it by an element of K× if necessary). To see that f is irreducible, we suppose that we can write f = gh. Then g(a)h(a) = 0; since L is a field, we must have either g(a) = 0 or h(a) = 0, and thus at least one of g and h is in I. So f divides one of g and h, WLOG g. Since g also divides f , we have deg(g) = deg( f ) and hence h is constant. Thus f is irreducible. Remark. For Commutative Algebra students: a slightly posher way of stating the last part is that I is the kernel of the homomorphism K[X] ! L g 7! g(a). L is an integral domain (being a field); the kernel of a homomorphism to an integral domain is a prime ideal; and a generator of a principal prime ideal is a prime element, and hence must be irreducible. Definition 1.1.5. Let L j K be an extension. We say L j K is algebraic if every a 2 L is algebraic over K. We say L j K is finite if L has finite dimension as a K-vector space. Example 1.1.6. The extension C j R is finite (of degree 2), since f1, ig is a basis of C over R. It is also algebraic, because every a + bi 2 C satisfies the polynomial (X − a)2 + b2 = X2 − 2aX + (a2 + b2) 2 R[X]. Notation 1.1.7. If L j K is finite, we define the degree [L : K] to be the dimension of L as a K-vector space. 3 p Example 1.1.8. Let a = i + 2 2 C. But a is also algebraic over Q: we have p p a − 2 = i ) a2 − 2 2a + 2 = −1 p ) a2 + 3 = 2 2a ) (a2 + 3)2 = 8a2 ) a4 − 2a2 + 9 = 0. We’ll see later that X4 − 2X2 + 9 is irreducible in Q[X], so it is the minimal polynomial of a over Q. Lecture p 2 On the other hand, the minimal polynomial of a over R is X2 − 2 2X + 3, by the previous example. This shows that the minimal polynomial of a over K really depends on which K we use! Remark. I forgot to point out in the last lecture that in Proposition 1.1.4, the minimal polynomial f of a over K has the property that any polynomial g 2 K[X] such that g(a) = 0 is necessarily a multiple of f . This is clear from the proof. We’ll use this fact a lot, so make sure it’s in your notes! Proposition 1.1.9. Let L j K be a field extension. An element a 2 L is algebraic over K if and only if there exists a finite extension of K inside L which contains a. (In particular, any finite extension is algebraic, and any algebraic extension is a union of finite extensions. There are algebraic extensions which aren’t finite, as we’ll see later.) Proof. Firstly, let’s prove the “if” part. It suffices to show that if L j K is a finite extension and a 2 L, then a is algebraic. Suppose [L : K] = d < ¥. Then the powers 1, a, a2, ... , ad are d + 1 elements of a d-dimensional vector space over K, so they must be linearly dependent: that is, we can find elements c0, ... , cd of K, not all zero, such that d c0 + c1a + ··· + cda = 0. i Thus a satisfies the non-zero polynomial g(X) = ∑ ciX 2 K[X] of degree ≤ d. Thus a is algebraic over K. The “only if” part is a little harder. Let f be the minimal polynomial of a over K, and d its degree. We will show that the K-subspace M of L spanned by the powers of a is d-dimensional over L, with basis S = f1, ... , ad−1g, and is a subfield of L. Since S is a finite set, this shows that M is a finite field extension of K inside L which contains a. Claim 1: M is a subring of L. N By definition, M is exactly the elements of L which are of the form a0 + a1a + ··· + aNa for some a0, ... , aN 2 K; that is, L is the image of the ring homomorphism K[X] ! L given by mapping g to g(a). But the image of a ring homomorphism is always a subring (Algebra 2). Claim 2: M is spanned by S. By the division algorithm for polynomials, for each g 2 K[X] we can write g(X) = a(X) f (X) + b(X) where a, b 2 K[X] and deg(b) ≤ d − 1. But this implies that g(a) = a(a) f (a) + b(a) = b(a), since f (a) = 0. As b has degree ≤ d − 1, b(a) is a K-linear combination of the elements of S. Claim 3: M is closed under taking inverses of nonzero elements. This is the most difficult bit! There are many possible proofs, but here’s one. We know by this stage that M is finite-dimensional over K. Let x 2 M be non-zero, and consider the map mx : M ! M given by mx(y) = xy. (This is called the “multiplication-by-x map”). I claim that mx is injective. If not, there would be some nonzero y such that xy = 0; but this equality takes place inside L, which is a field, so either x = 0 or y = 0, which is a contradiction. 4 By the rank–nullity theorem, it follows that mx is surjective. In particular, 1 2 image(mx), which shows that 1/x 2 M. This concludes the proof that M is a field. As a by-product of the proof of the “only if” part, we get two interesting pieces of information. Corollary 1.1.10. (i) An element a 2 L is algebraic over K if and only if the powers of a span a finite-dimensional K-subspace of L. (ii) If a is algebraic over L, then there is a unique smallest extension of K in L which contains a, namely the K-subspace spanned by the powers of a; and this has a K-basis 1, a, ..