MATH 154. ALGEBRAIC NUMBER THEORY LECTURES BY BRIAN CONRAD, NOTES BY AARON LANDESMAN CONTENTS 1. Fermat’s factorization method 2 2. Quadratic norms 8 3. Quadratic factorization 14 4. Integrality 20 5. Finiteness properties of OK 26 6. Irreducible elements and prime ideals 31 7. Primes in OK 37 8. Discriminants of number fields 41 9. Some monogenic integer rings 48 10. Prime-power cyclotomic rings 54 11. General cyclotomic integer rings 59 12. Noetherian rings and modules 64 13. Dedekind domains 69 14. Prime ideal factorization 74 15. Norms of ideals 79 16. Factoring pOK: the quadratic case 85 17. Factoring pOK: the general case 88 18. Ramification 93 19. Relative factorization and rings of fractions 96 20. Localization and prime ideals 101 21. Applications of localization 105 22. Discriminant ideals 110 23. Decomposition and inertia groups 116 24. Class groups and units 122 25. Computing some class groups 128 26. Norms and volumes 134 27. Volume calculations 138 28. Minkowski’s theorem and applications 144 29. The Unit Theorem 148 References 153 Thanks to Nitya Mani for note-taking on two days when Aaron Landesman was away. 1 2 BRIAN CONRAD AND AARON LANDESMAN 1. FERMAT’S FACTORIZATION METHOD Today we will discuss the proof of the following result, and highlight some of its main ideas that will be important themes in the course: Theorem 1.1 (Fermat). For x, y 2 Z, the only solutions of y2 = x3 − 2 are (3, ±5). Remark 1.2. This is an elliptic curve (a notion we shall not define, essentially the set of solutions to a certain type of cubic equation in two variables) with infinitely many Q-points, a contrast with finiteness of its set of Z-points. Fermat’s equation can be rearranged into the form x3 = y2 + 2. Lemma 1.3. For any Z-solution (x, y) to x3 = y2 + 2, the value of y must be odd. Proof. Indeed, if y is even then x is even, so x3 is divisible by 8. But y2 + 2 = 4k + 2 is not divisible by 8. Fermat’s first great idea is to introduce considerations in the ring p p Z[ −2] := fm + n −2 j m, n 2 Zg. The point is that over this ring, the equation x3 = y2 + 2 can be expressed as p p x3 = (y + −2)(y − −2). To find solutions to this, we ask the following two questions: p p Question 1.4. Do y + −2 and y − −2 have “gcd = 1” (meaning no non- trivial factor in common, where “nontrivial” means “not a unit”)? Question 1.5. If we knowp the answer to thep above question is “yes,” can we conclude that both y ± −2 are cubes in Z[ −2]? Recall the definition of unit: Definition 1.6. For R a commutative ring, an element u 2 R is a unit if there exists u0 2 R, so that uu0 = 1. We let R× := f units in Rg; this is an abelian group. Example 1.7. (1) Z× = f±1g. (2) For F a field, F× = F − f0g by the definition of a field. MATH 154. ALGEBRAIC NUMBER THEORY 3 p p (3) We have that 1 + 2 is a unit in Z[ 2] because p p (1 + 2)(−1 + 2) = 1. p p Thus, likewise (1 + 2)n 2 Z[ 2]× for all n 2 Z. Note that if u 2 R× then for any x 2 R we have x = (xu−1)u. Hence, for any consideration of “unique” factorization we must allow for adjust- ing factors by unit multiples (absorbing the inverse unit elsewhere in the factorization). Definition 1.8. A domain (sometimes also called an integral domain) is a nonzero commutative ring R such that if ab = 0 with a, b 2 R then either a = 0 or b = 0. For a domain R, if a, b 2 R − f0g and a j b and b j a then a = bu for u 2 R×. (Indeed, if a = bs and b = at then a = a(ts), so ts = 1 since a 6= 0 and hence s, t 2 R×.) Away from Z+,“ajb, bja ) a = b” is very rare. Example 1.9. Later, we’ll see p p Z[ 2]× = f±1g × (1 + 2)Z. This is non-obvious! p Question 1.10. What is Z[ −2]×? In fact, we’ll answer this more generally: Lemma 1.11. Let d be any non-square integer > 1. We have p Z[ −d]× = f±1g. Proof. Consider p a := u + v −d for u, v 2 Z. Let p a := u − v −d, so 2 2 aa = u + dv 2 Z≥0. Since ab = ab, if ab = 1 then ab = 1 = 1, which implies (aa)(bb) = 1, 4 BRIAN CONRAD AND AARON LANDESMAN so aa = 1 2 2 2 2 because aa = u + dv 2 Z≥0. The equality u + dv = 1 with d > 1 forces v = 0 and then u = ±1. p Remark 1.12. If we try the same argument for Z[ d], we get u2 − dv2 = ±1 with u, v 2 Z (and v 6= 0 for units distinct from ±1). The question of finding such non-trivial units then becomes Pell’s equation, which is u2 − dv2 = 1 and its variant with −1 on the right side. For the case d = 2, Pell’s equation has infinitely many Z-solutions with u, v > 0 (as we will recover later in the course from a more sophisticated point of view), beginning with: (3, 2), (17, 12),... p p by considering powers (1 + 2)2n = (3 + 2 2)n for n > 0. 2 2 The variant equation u − 2v =p−1 also has infinitely many Z-solutions with u, v > 0 by considering (1 + 2)2n+1 with n ≥ 0, such as: (1, 1), (7, 5),... 3 2 We’ll now resume ourp goal of finding the Z-solutions to x = y + 2, first addressing if y ± −2 have a non-unit common factor. The answer is negative: p Lemma 1.13. If d 2 Z[ −2] satisfies p p d j (y + −2) and d j (y − −2) then d is a unit. Proof. First, recall from Lemma 1.3 that y is odd. Observe that p p d j (y + −2) − (y − −2) p = 2 −2 p = ( −2)3. Then, we claim the following sublemma: p p Lemma 1.14. The element −2 2 Z[ −2] is irreducible (i.e., it is a nonzero non-unit such that any factorization as ab must have a or b a unit). p p Proof. Assuming −2 = ab, we have − −2 = ab by conjugating both sides. Multiplying these relations, 2 = (aa)(bb) 2 Z≥0, so aa = 1 or bb = 1. This implies either a or b is a unit, as desired. MATH 154. ALGEBRAIC NUMBER THEORY 5 p In HW1 it will be shown that Z[ −2] is a UFD, so the irreducibility of p p e p −2 forces d = u −2 for somep 0 ≤ e ≤ 3 and some unit u 2 Z[ −2]. Thus, if d is not a unit then −2 j d. Hence,p to getp a contradictionp (and conclude d is a unit) it is enough to show −2 - (y + −2) in Z[ −2]. Suppose for some u, v 2 Z that p p p y + −2 = −2(u + v −2) p = 2v + u −2. This forces y = 2v to be even, but y is odd by Lemma 1.3. This concludes the proof that d is a unit. p We conclude that in the UFD Z[ −2] we have p p gcd(y + −2, y − −2) = 1. This completes the preparations for: Proof of Theorem 1.1. In any UFD, any nonzero element is a finite product of irreducibles, and by lumping together any irreducibles that agree up to unit multiple we can rewrite such a product as ei u · ∏ pi i with pi pairwise non-associate irreducible elements (non-associate means one is not a unit times another; by irreducibility of the pi’s, this amounts to saying pi - pj for any i 6= j). Since p p (y + −2)(y − −2) = x3 with p p gcd(y + −2, y − −2) = 1, it follows from expressing each of p y ± −2 as a unit multiple of a productp of pairwise non-associate irreducibles that all irreduciblep factors of y ± −2 occur with multiplicity divisible by 3. Therefore, y ± −2 is the productp of a unit and a cube. But now a miracle occurs: we know the units of Z[ −2] by Lemma 1.11, and from this we see that all units are themselves cubes (as −1 = (−1)3)! Hence, p p y + −2 = (a + b −2)3 for some a, b 2 Z. 6 BRIAN CONRAD AND AARON LANDESMAN Therefore, p p y + −2 = (a + b −2)3 p = a(a2 − 6b2) + b(3a2 − 2b2) −2, so 1 = b(3a2 − 2b2), from which we see that b = ±1, so b2 = 1. This implies ±1 = 3a2 − 2 so 3a2 = 2 ± 1. Then 3a2 equals either 3 or 1. The latter is impossible because we cannot have 3a2 = 1 with a 2 Z. Thus, 3a2 = 3, so a = ±1. This implies b = 1, so y = a(a2 − 6b2) = ±(1 − 6) = ±5 and x3 = y2 + 2 = 27, so x = 3, concluding the proof. The above considerations yield the following lessons: (1) For studying Z-solutions to polynomial equations, it’s useful to con- sider arithmetic in larger numberp systems. For example, in this case, it was useful to work in Z[ −2].