MA3A6 ALGEBRAIC NUMBER THEORY SAMIR SIKSEK Abstract. This is an incomplete set of lecture notes for for Algebraic Number Theory. I do not know yet if it will be completed, so you are advised to continue taking notes in the lectures. Please send comments, misprints and corrections to [email protected] Contents 1. Orientation 2 1.1. Local Arguments 2 1.2. Infinite Descent 2 1.3. Descent 3 2. Beginnings of Algebraic Number Theory 4 3. Preliminaries on Rings and Ideals 5 3.1. Rings 5 3.2. Ideals 6 3.3. Quotient Rings 6 3.4. Units 6 3.5. Fields 7 3.6. Fields of Fractions 8 3.7. Prime and Maximal Ideals 8 3.8. Unique Factorization Domains 8 4. Algebraic Numbers and Algebraic Integers 9 5. Minimal Polynomials 10 6. Conjugates 12 7. Factorization of Polynomials 13 8. Eisenstein’s Criterion For Irreducibility 14 9. Symmetric Polynomials 15 10. Algebraic Integers form a Ring 17 11. Algebraic Numbers form a Field 19 12. Number Fields 19 13. Fields Generated by Conjugate Elements 21 14. Embeddings 22 15. Field Polynomial 23 16. Ring of Integers 25 17. Determinants and Discriminants 28 18. Ideals 30 18.1. Quotient Rings 31 Date: September 26, 2006. 1 2 SAMIR SIKSEK 19. Prime and Maximal Ideals 34 20. Towards Unique Factorization for Ideals I 35 21. Towards Unique Factorization for Ideals II 36 22. Unique Factorization Proof—A Summary so Far 37 23. A Special Case of the Cancellation Lemma 37 24. Ideal Classes 38 25. Unique Factorization Proof—Summary So Far (again) 40 26. What are the prime ideals of K? 40 1. Orientation Algebraic number theory arose out of the study of Diophantine equations. A Diophantine equation is a polynomial equation in several variables with integer coefficients and one desires the solutions in integers. The study of Diophantine equations seems as old as human civilization itself; they are however named after Diophantus of Alexandria who lived in the 3rd century BC. Soon we will give our motivating example for algebraic number theory. Before that we look at three methods for attacking Diophantine equations which should be close to the heart of every student of number theory: Local Methods, Infinite Descent, Descent. 1.1. Local Arguments. This means that we look at the equation modulo some positive integer m and try to get information about integral solutions. Example 1.1. Show that the equation x2 + 1 = 4y2 does not have solutions in integers. Answer: Suppose x, y is an integer solution. Reducing the equation modulo 4 we get x2 + 1 ≡ 0 (mod 4). However, the squares modulo 4 are x2 ≡ 0 or 1 (mod 4). So x2 + 1 ≡ 1 or 2 (mod 4), giving a contradiction. Included in ‘local arguments’ is the idea that if an equation does not have real solutions then it does not have integeral solutions. For example, the equation x2 + 5 = −2z2 does not have integral solutions because it does not have real solutions. 1.2. Infinite Descent. ‘infinite descent’ is used to show that equations do not have solutions, or that they do not have non-trivial solutions. The idea is to suppose that a Diophantine equation has solutions, take the smallest one and show that there must be a smaller one, giving a contradiction. Example 1.2. Show that the equation X2 = 2Y 2 does not have non-trivial solu- tions in integers. √ You will recognize this as essentially the proof that 2 is irrational. Suppose that we have non-trivial solutions. Let (x, y) be a non-trivial solution with the 2 2 value of |x| minimal. Since x = 2y we get that x is even. So x = 2x1 for some ALGEBRAIC NUMBER THEORY 3 2 2 x1 ∈ Z. So 2x1 = y and we show that y = 2y1. Now (x1, y1) is a non-trivial 2 2 solution to X = 2Y and |x1| < |x|, giving a contradiction. The name ‘infinite descent’comes from the original way in which the method is applied. We start with a non-trivial solution (x, y) and construct another one (x1, y1), and then from (x1, y1) we get another one (x2, y2) and so on. We get an infinite list of non-trivial solutions satisfying |x| > |x1| > |x2| > ··· > 0; but we cannot squeeze infinitely many integers between |x| and 0 and we have a contradiction. Here is another example due to Euler. Example 1.3. Show that the equation (1) X2 + 2Y 3 + 4Z3 = 0 has no non-trivial solutions. Answer: Suppose it has non-trivial solutions. Let (x, y, z) be a non-trivial solution with the value of |x| minimal. We see that x3 is even and so x is even. Write x = 2x1. Then 3 3 3 4x1 + y + 2z = 0. Thus y = 2y1 and applying the same trick again z = 2z1. We see that (x1, y1, z1) is a non-trivial solution to equation (1) with |x1| < |x| giving a contradiction. In both the above examples, we could have obtained a contradition by explaining that we may assume that x, y are coprime. Then the argument shows that x, y are not coprime. This is not always the case with infinite descent examples. 1.3. Descent. Descent is not the same as infinite descent. It is a technique invented by Fermat. It uses variants of the following very simple Lemma: Lemma 1.1. Suppose that U, V , W are non-zero integers, with U, V coprime. n n Suppose also that UV = W where n is a positive integer. Then U = ±W1 and n n V = ±W2 for some integers W1, W2. Moreover, if n is odd, we can take U = W1 n and V = W2 . Proof. Let p1, . pr be the distinct prime divisors of U and q1, . , qs the distinct prime divisors of V . Since U, V are coprime, these lists are disjoint (this is a very important point). Notice that p1, . , pr, q1, . , qs are the distinct prime divisors of W . By the Fundamental Theorem of Arithmetic, we can write a1 ar b1 bs c1 cr d1 ds U = ±p1 . pr ,V = ±q1 . qs ,W = ±p1 . pr q1 . qr . Substituting in UV = W n we get a1 ar b1 bs nc1 ncr nd1 nds p1 . pr q1 . qs = ±p1 . pr q1 . qr . From the uniqueness of factorization we deduce that ai = nci, bj = ndj; it is here that we need the fact that the ps and qs are distinct. Hence n c1 cr n d1 ds U = ± (p1 . pr ) ,V = ± q1 . qs . 4 SAMIR SIKSEK This completes the proof for n even. For n odd, simply absorb the ± inside the n-th power. We are ready to give a first example of descent. Example 1.4. Show that the equation 4X2 − 1 = Y 3 has no solutions in integers apart from (X, Y ) = (0, −1). Answer: First notice that if (X, Y ) is a solution then so is (−X, Y ). So we may suppose that we have a solution with X ≥ 0. Write (2X + 1)(2X − 1) = Y 3. Let d be the gcd of 2X +1 and 2X −1. Since d divides them both, it must divide their difference; thus d | 2 and so d = 1 or d = 2. However, d divides 2X + 1 which is odd, so d is odd and so d = 1. In other words, 2X + 1 and 2X − 1 are coprime. Applying Lemma 1.1 we see that 3 3 (2) 2X + 1 = Y1 , 2X − 1 = Y2 , where Y = Y1Y2 (this step is called ‘descent’). Subtracting we get 3 3 2 2 2 = Y1 − Y2 = (Y1 − Y2)(Y1 + Y1Y2 + Y2 ). Recall our assumption that X ≥ 0. Hence 2X + 1 > 2X − 1 which gives Y1 > Y2; i.e. Y1 − Y2 > 0. We deduce that either 2 2 Y1 − Y2 = 1,Y1 + Y1Y2 + Y2 = 2, or 2 2 Y1 − Y2 = 2,Y1 + Y1Y2 + Y2 = 1. But, from (2) we know that Y1, Y2 are both odd, and so Y1 − Y2 is even. Hence we have only to deal with the last case: Y1 − Y2 = 2. Write Y1 = Y2 + 2 and 2 2 2 substitute in to Y1 + Y1Y2 + Y2 = 1. We get Y2 + 2Y2 + 1 = 0. Thus Y2 = −1 and Y1 = Y2 + 2 = 1. Hence Y = Y1Y2 = −1 and we see that X = 0 as required. 2. Beginnings of Algebraic Number Theory The following example is the beginning of algebraic number theory. Fermat claimed that he has shown that the only solutions to the equation X2 + 2 = Y 3 is (X, Y ) = (±5, 3). Euler ‘proved’Fermat’s assertion in 1770. The ‘proof’ mimics the standard factorization or descent argument used in the above example. We say ‘proof’ in quotes because what Euler wrote down was not rigorous, but can be made rigorous. Let us see Eulers argument: simply factor the left-hand side to get: √ √ (X + −2)(X − −2) = Y 3. √ √ We leave√ the usual integers Z and work with Z[ −2]. Now the ‘integers’ X + −2, X − −2 are ‘coprime’and so each must be a cube. So √ √ X + −2 = (a + b −2)3, where a, b are in Z. Once we get past the dodgy step above, the rest of the argument is respectable. Expand the brackets to get √ √ X + −2 = (a3 − 6ab2) + (3a2b − 2b3) −2. ALGEBRAIC NUMBER THEORY 5 √ Comparing the coefficients of −2 we get X = a3 − 6ab2, 1 = 3a2b − 2b3. Hence b | 1 and so b = ±1 which gives 3a2 − 2 = ±1. Therefore a = ±1 and we get 3 2 X = a − 6ab = ±5. Hence (X, Y ) = (±5, 3) as required.