COMPLEMENTS ON COMMUTATIVE ALGEBRA

Contents 1. Rings 1 2. Modules and algebras 6 3. Noetherianity 9 4. Modules of finite type over PIDs 10 5. Tensor product 14 6. Tensor, symmetric and exterior algebras 16 7. Flatness 21 8. Localization 22 9. Integral extensions 26 10. Discriminants 28 11. Inverse limits 36 12. Appendix: Zorn’s lemma 43 13. Exercises 43 References 45

In what follows, A denotes a commutative ring with unit. Ring homomorphisms map units to units.

1. Rings 1.1. Ideals. Definition 1.2. An ideal of A is a subset I Ă A such that: (1) p@x, y P Iq x ` y P I (so that I is a subgroup of pA, `q); (2) p@a P Aq p@x P Iq ax P I. Given an ideal I Ă A, the quotient group A{I is endowed with a unique ring structure such that the canonical map A Ñ A{I is a ring homomorphism. Example 1.3. (0) t0u is an ideal. (1) A is an ideal (called the unit ideal). We say that an ideal I Ă A is strict if I ‰ A. (2) Ideals of Z are of the form n Z for a unique n P Zě0. (3) Similarly, if K is a field, nonzero ideals of KrXs are of the form P pXqKrXs for a unique monic polynomial P pXq P KrXs. Definition 1.4. Let I Ă A be a strict ideal. (1) I is maximal if it is maximal (for the inclusion) among strict ideals in A. (2) I Ă A is prime if p@x, y P Aq pxy P I ñ px P I or y P Iqq. Example 1.5. ‚ The ring A is an integral domain if and only if t0u is prime. ‚ n Z is prime in Z if and only if n Z is maximal if and only if n is a prime integer. Similarly, P pXqKrXs is prime in KrXs if and only if P pXqKrXs is maximal if and only if P pXq is irreducible. Remark 1.6. (1) A maximal ideal is prime. (2) An ideal I Ă A is maximal (reps. prime) if and only if A{I is a field (resp. an integral domain). (3) Let Λ be a set and pIλqλPΛ be ideals in A. Then Iλ is an ideal of A. λPΛ Ş Theorem 1.7. (Krull). Let I Ă A be a strict ideal. There exists(1) a maximal ideal m Ă A such that I Ă m.

Version: June 27, 2019 (1)This statement is equivalent to the axiom of choice. 1 2 Number theory

Proof. Let E be the set of strict ideals J Ă A containing I: it is non empty since I P E . We (partially) order E with the relation given by J1 ď J2 ô J1 Ă J2. The ordered set pE , ďq is inductive: if pJλqλPΛ is a chain (i.e. a totally ordered subset) of E , then J :“ Jλ is an element in E , and an upper bound of pJλqλPΛ. λPΛ By Zorn’s lemma, pE , ďq admits a maximal elementŤ m. If J Ă A is a strict ideal containing m, then J P E , hence J “ m by maximality. This shows that m is a maximal ideal, that contains I by definition.  Remark 1.8. One can show that Krull’s theorem is equivalent to the axiom of choice. Definition 1.9. ‚ If X Ă A, the ideal generated by X is the smallest ideal of A that contains X, this is nothing but the intersection of all ideals(2) of A that contain X. ‚ If I Ă A is an ideal and X Ă I, we say that X generates I if the ideal generated by X is I. We sometimes denote it by xXy. ‚ A principal ideal of A is an ideal generated by one element. The ring A is called principal (PID) if it is an integral domain and its ideals are all principal. Example 1.10. (1) Z and KrXs are principal, more generally euclidean rings are principal. (2) If K?is a field, xX,Y y is not principal in KrX,Y s. Similarly, x2,Xy is not principal in ZrXs. (3) Zri 5s is not principal.

Definition 1.11. ‚ Let Λ be a set and pIλqλPΛ be ideals in A. Their sum is the ideal generated by Iλ. λPΛ n Ť This is nothing but the set of finite A-linear combinations aixi with r P Zě0, a1, . . . , ar P A and xi P Iλi i“1 for all i P t1, . . . , ru. ř ‚ Let I,J Ă A be ideals. Their product IJ is the ideal generated by txyuxPI . yPJ Definition 1.12. Two ideals I,J Ă A are coprime (or I is prime to J) when I ` J “ A. Proposition 1.13. (1) Two distinct maximal ideals are coprime. (2) If I1,...,In are prime to J, then I1 ¨ ¨ ¨ In is prime to J. n m (3) If I,J Ă A are coprime and n, m P Zą0, then I and J are coprime. Proof. (1) As I Ĺ I ` J Ă A, we have I ` J “ A. (2) As Ik ` J “ A for all k P t1, . . . , nu, we have A “ pI1 ` JqpI2 ` Jq ¨ ¨ ¨ pIn ` Jq “Ă I1 ¨ ¨ ¨ In ` J, whence I1 ¨ ¨ ¨ In ` J “ A. n (3) Applied to Ik “ I for all k P t1, . . . , nu, (2) implies that I and J are coprime. After replacing I by J n n m and J by I , we deduce that I and J are coprime. 

Theorem 1.14. (Chinese remainder theorem). Assume I1,...,In Ă A are pairwise coprime ideals (i.e. i ‰ j ñ Ii ` Ij “ A). Then: (1) I1 X I2 X ¨ ¨ ¨ X In “ I1I2 ¨ ¨ ¨ In; n (2) the canonical ring homomorphism A{I1I2 ¨ ¨ ¨ In Ñ A{Ik is an isomorphism. k“1 Proof. By proposition 1.13, the case n “ 2 implies theś general case: let I,J Ă A be coprime ideals. There exist eI P I and eJ P J such that eI ` eJ “ 1. (1) We have always IJ Ă I X J. Let a P I X J: we have a “ apeI ` eJ q “ aeI ` aeJ . As a P J and eI P I, we have aeI P IJ. Similarly aeJ P IJ, hence a P IJ, proving the equality. (2) Let ϕ: A Ñ pA{Iq ˆ pA{Jq be the natural map. If x, y P A, we have ϕpxeJ ` yeI q “ px ` I, y ` Jq, so ϕ „ is surjective. As Kerpϕq “ I X J “ IJ, it induces an isomorphism A{IJ ÑpA{Iq ˆ pA{Jq.  1.15. UFDs. Definition 1.16. Assume that A is an integral domain. ‚ An element α P AzpAˆ Y t0uq is prime (resp. irreducible) if the ideal αA is prime (resp. p@a, b P Aq pab “ α ñ pa P Aˆ or b P Aˆqq). A prime element is always irreducible(3), but the converse is not true in general. ‚ The ring A is a unique factorization domain (UFD) if it is an integral domain in which every non-zero element can be written as a product of or irreducible elements, uniquely up to order and multiplication by units. More precisely, for any α P Azt0u, there exist n P Zě0 and irreducible elements p1, . . . , pn such that

αA “ p1 ¨ ¨ ¨ pnA

(2)This makes sense by Remark 1.6 (3). (3)Because A is a domain. Note that 2 is prime in Z {6 Z, but not irreducible since 2 “ 2 ˆ 4. Number theory 3

and if αA “ q1 ¨ ¨ ¨ qmA with m P Zě0 and q1, . . . , qm irreducible, then m “ n and there exists σ P Sn such that qk “ pσpkq for all k P t1, . . . , nu. ˆ There exists u P A such that α “ up1 ¨ ¨ ¨ pn: such an quality is called a prime decomposition of α. Example 1.17. (0) A field is a UFD. (1) Z and KrXs (where? K is a field)? are UFD. ? ? (2) The subring Zri 5s “ tx ` iy 5 P C ; x, y P Zu of C?is not a? UFD, because 2, 3, 1 ` i 5 and 1 ´ i 5 are irreducible, the unit are 1 and ´1, but 2.3 “ p1 ` i 5qp1 ´ i 5q (i.e. there is no unicity for a prime decomposition of 6). Lemma 1.18. In a PID, irreducible element are prime.

Proof. Let p P A be an irreducible element. Let m Ă A be a maximal ideal such that p P m (cf Krull theorem, or use the noetherianity of A). As A is a PID, there exists α P A such that m “ aA, so p “ αa for some a P A. As p is irreducible, we must have a P Aˆ (because α R Aˆ since m “ aA ‰ A). Thus pA “ m is maximal.  Definition 1.19. Assume A is a UFD, and let p P A be an irreducible element. If α P Azt0u, the p-adic valuation of α is k vppαq “ maxtk P Zě0 ; p | αu This is well defined and only depends on the ideals pA and αA.

Proposition 1.20. (Properties of valuations). Assume A is a UFD and let a, b P A. (1) vppabq “ vppaq ` vppbq ; (2) a | b if and only if vppaq ď vppbq for all irreducible element p P A; ˆ (3) a P A if and only if vppaq “ 0 for all irreducible element p P A. (4) vppa ` bq ě mintvppaq, vppbqu with equality when vppaq ‰ vppbq. Proof. (1)-(3) follow from the definition and the unicity of decomposition into a product of prime elements. v v v For (4), if v “ inftvppaq, vppbqu, then p | a and p | b, so p | a ` b, thus vppa ` bq ě v. Assume v 1 1 v 1 1 vppaq ‰ vppbq: we may assume that v “ vppaq ă vppbq. Write a “ p a with p - a and b “ p b with p | b , v 1 1 1 1 so that a ` b “ p pa ` b q and p - a ` b : we have vppa ` bq “ v.  Proposition 1.21. Assume A is a UFD and let p P Azt0u. Then p is irreducible if and only if p is prime.

Proof. If p is irreducible and p | ab, then vppaq ` vppbq “ vppabq ě 1 so vppaq ě 1 or vppbq ě 1 i.e. p | a or p | b. Conversely, a prime element is always irreducible.  Remark 1.22. In fact it can be shown that an integral domain is a UFD if and only if it is noetherian (cf definition 3.3) and all its irreducible elements are prime.

Definition 1.23. Assume A is a UFD and let a, b P Azt0u. The gcd (greatest common divisor) and the lcm (least common multiple) of a and b are the greatest lower bound (resp. smallest upper bound) of the set ta, bu for the divisibility relation. They are denoted gcdpa, bq and lcmpa, bq respectively. We say that a and b are coprime when gcdpa, bq “ 1. Remark 1.24. (1) Strictly speaking, gcdpa, bq and lcmpa, bq are only defined up to multiplication by a unit: only the ideal they generate are well defined. (2) Let a, b P Azt0u and an irreducible element p P A. Then vppgcdpa, bqq “ mintvppaq, vppbqu and vpplcmpa, bqq “ maxtvppaq, vppbqu. Note that gcdpa, bq lcmpa, bqA “ abA. (3) By induction, one can easily extend the definition and consider gcd and lcm of afinite family in Azt0u. Lemma 1.25. (Gauss lemma). Assume that A is a UFD and let a, b, c P Azt0u be such that gcdpa, bq “ 1. If a | bc, then a | c.

Proof. If p P A is irreducible and divides a, then vppbq “ 0 since p - b (because a and b are coprime). This implies that vppaq ď vppbcq “ vppcq. As this holds for any irreducible element p dividing a, we have a | c (cf proposition 1.20 (2)).  Proposition 1.26. A PID is a UFD. Lemma 1.27. Assume that A is an integral domain in which irreducible elements are prime (cf proposition 1.21). If an element admits a prime decomposition, the latter is unique (in the sense of definition 1.16). 4 Number theory

ˆ Proof. Assume α “ up1p2 ¨ ¨ ¨ pn “ vq1q2 ¨ ¨ ¨ qm, with n, m P Zě0, u, v P A and p1, . . . , pn, q1, . . . , qm irreducible elements. Possibly after exchanging the decompositions, we may assume n ď m. We proceed ˆ by induction on n. If n “ 0, then α “ u P A : the product vq1q2 ¨ ¨ ¨ qm is invertible so all its factors are: we must have m “ 0. assume n ě 1. As p1 is irreducible and divides the product vq1q2 ¨ ¨ ¨ qs, it divides ˆ one of the factors (since it is prime). As v P A , it is not divisible by p1: after renumbering the qi, we may assume that p1 | q1 i.e. p1A “ q1A. Dividing α by p1, we reduce to the case n ´ 1, and use induction hypothesis.  Proof of proposition 1.26. Assume A is a PID. By lemmas 1.18 and 1.27, it is enough to shows that any nonzero element in A admits at least one prime decomposition. Let E be the set of elements in Azt0u that do not admit a prime decomposition. Assume E is not empty. As A is noetherian, the set E admits a minimal element α (for the divisibility relation). The element α can be nor a unit, nor irreducible: it can ˆ be written α “ α1α2 with α1, α2 P AzpA Y t0uq. Then α1 and α2 are strict divisors of α, so α1, α2 R E by minimality of α: they admit prime decomposition. This implies that their product α admits a prime decomposition: contradiction.  Remark 1.28. ‚ When A is a PID, there is an other characterization of gcd and lcm of two element a, b P A: we have gcdpa, bqA “ aA ` bA and lcmpa, bqA “ aA X bA. Let’s prove it for the gcd (the proof for the lcm is similar). As A is principal, there exists d P A such that aA ` bA “ dA. As x P A divides a and b if and only if aA Ă xA and bA Ă xA i.e. dA Ă xA, so gcdpa, bq “ d. ‚ This characterization does not hold in any UFD. For instance, QrX,Y s is a UFD (cf theorem 1.41). As X and Y are irreducible and coprime, we have gcdpX,Y q “ 1, though X QrX,Y s ` Y QrX,Y s ‰ QrX,Y s (the LHS is the ideal of polynomials vanishing at p0, 0q). Of course, this follows from the fact that QrX,Y s is not a PID.

Example 1.29. If K is a field and n P Zą0, the ring KrX1,...,Xns is a UFD (cf theorem 1.41) but not a PID (cf remark above). Similarly the ring ZrXs is a UFD (cf loc. cit.) but not a PID (the ideal generated by 2 and X is not principal). Proposition 1.30. When A is principal, its nonzero prime ideals are all maximal. Proof. Let p Ă A be a nonzero prime ideal. By Krull theorem, there exists a maximal ideal m Ă A such that p Ă m. As A is principal, one can write p “ pA and m “ aA: we have p “ ab with b P A. As p is prime hence irreducible, we have b P Aˆ (because a R Aˆ since m is maximal), which implies that p “ m is maximal.  Definition 1.31. Assume that A is a integral domain. ‚ An euclidean function is a map φ: Azt0u Ñ Zě0 such that if b | a in Azt0u, then φpbq ď φpaq. ‚ An euclidean function φ defines an euclidean division if for all pa, bq P A ˆ Azt0u, there exist q, r P A such that a “ bq ` r and (r “ 0 or φprq ă φpbq). “The” element q is called the quotient and r the remainder of the division. ‚ The ring A is euclidean if it admits an euclidean function that defines an euclidean division. Remark 1.32. If A is an euclidean domain, there is not unicity for an euclidean function. Moreover, unicity of quotient and remainder is not required. Exemples 1.33. A field is an euclidean domain. The ring Z is euclidean domain, an euclidean function being given by φpaq “ |a| (absolute value). In that case, euclidean division is the usual one. When K is a field, the ring of polynomials KrXs is euclidean, an euclidean function being given by φpP q “ degpP q. Here again, euclidean division is the usual one. The ring Zris “ ta ` ib P C ; a, b P Zu of Gauss integers is euclidean, endowed with the euclidean function given by φpa ` ibq “ a2 ` b2. Proposition 1.34. An euclidean domain is a PID.

Proof. Assume A is an euclidean domain, let φ: Azt0u Ñ Zě0 be an euclidean function and I Ă A an ideal. To prove that I is principal, we may assume that I ‰ 0. In that case, φpIzt0uq is an nonempty subset of Zě0, so it admits a smallest element: let b P Izt0u be such that φpbq is minimal. One has bA Ă I. Conversely, let a P I. There exist q, r P A such that a “ qb ` r with r “ 0 or φprq ă φpbq. Assume r ‰ 0, so that φprq ă φpbq. As r “ a ´ qb P I and r ‰ 0, we have φpbq ď φprq by minimality of φpbq, which s absurd: we must have r “ 0, i.e. a “ qb P bA. Thus I “ bA is principal.  ? 1`i 19 Remark 1.35. There are PID that are not euclidean domains, for instance Z 2 . Corollary 1.36. Let K be a field, the rings Z and KrXs are PID, hence UFD“ (cf proposition‰ 1.26). Number theory 5

A ring homomorphism f : A Ñ B induces a ring homomorphism ArXs Ñ BrXs. If A is a subring of B, then ArXs is a subring of BrXs. n Definition 1.37. Assume that A is a UFD and let P “ a0 ` a1X ` ¨ ¨ ¨ ` anX P ArXszt0u. The content of P is cpP q “ gcdtai ; ai ‰ 0u. Lemma 1.38. (Gauss Lemma). If A is a UFD and P,Q P ArXszt0u, then cpPQq “ cpP qcpQq. Remark 1.39. As gcd is defined up to multiplication by a unit, one should write cpPQqA “ cpP qcpQqA. In what follows, we will keep this abusive notation to avoid heaviness. Proof. Write P “ cpP qP and Q “ cpQqQ with cpP q “ 1 and cpQq “ 1: we have PQ “ cpP qcpQqP Q. Replacing P and Q by P and Q respectively, we may assume that cpP q “ 1 and cpQq “ 1: we have to show that cpPQq “ 1. r r r r r r Assume instead that therer existr a prime element p P A such that p | cpPQq. Denote by P and Q the images of P and Q in pA{pAqrXs respectively, this implies that P Q “ 0 in pA{pAqrXs. As p is prime, the ring A{pA is an integral domain: so is the ring pA{pAqrXs. This implies that P “ 0 or Q “ 0, i.e. p | cpP q or p | cpQq, contradicting cpP q “ 1 and cpQq “ 1.  Proposition 1.40. Assume that A is a UFD. Let K “ FracpAq and P P ArXs such that cpP q “ 1. Then P is irreducible in ArXs if and only if P is irreducible in KrXs.

Proof. ‚ Assume that P is irreducible in KrXs and write P “ Q1Q2 with Q1,Q2 P ArXs. As P is irreducible in KrXs, possibly after exchanging Q1 and Q2, the polynomial Q1 is constant so Q1 “ cpQ1q. By lemma ˆ 1.38, we have 1 “ cpP q “ cpQ1qcpQ2q, so Q1 P A . Thus P is irreducible in ArXs. ‚ Conversely, assume that P is irreducible in ArXs write P “ Q1Q2 with Q1,Q2 P KrXs. There exist a1, a2 P Azt0u such that a1Q1 P ArXs and a2Q2 P ArXs. We have a1a2 “ cpa1a2P q “ cpa1Q1qcpa2Q2q by lemma 1.38, because cpP q “ 1. Write a1Q1 “ cpa1Q1qQ1 and a2Q2 “ cpa2Q2qQ2 with Q1, Q2 P ArXs: we have a1a2P “ cpa1Q1qQ1cpa2Q2qQ2 “ a1a2Q1Q2 whence P “ Q1Q2 (the ring A is an integral domain). ˆ As P is irreducible in ArXs, we may assume, possiblyr after exchanging Q1 andr Q2, thatr Qr1 P A . Then ˆ Q1 P K and P is irreducibler in KrrXs. r r r r  r r r Theorem 1.41. If A is a UFD, then(4) ArXs is a UFD. Proof. ‚ If p P A is an irreducible element, the constant polynomial p is irreducible in ArXs. Indeed, A{pA is an integral domain: so is ArXs{pArXs » pA{pAqrXs and p is prime hence irreducible in ArXs. ‚ If P P ArXs is of degree ě 1 and irreducible, then cpP q “ 1. Indeed one can write P “ cpP qP with P P ArXs, providing a non trivial factorization if cpP q is not invertible. ‚ Existence of a prime decomposition. Let P P ArXszt0u. Write P “ cpP qP with P P ArXs suchr that crpP q “ 1. As A is a UFD, cpP q has a prime decomposition, so it is enough to show that P has a prime decomposition: we may assume that cpP q “ 1. If P P A, then P “ 1: we mayr assumer that degpP q ě 1. Putr K “ FracpAq. As KrXs is a UFD (cf corollary 1.36), we may write P “ P1P2 ¨ ¨ ¨ Pr withr Pi P KrXs irreducible for all i P t1, . . . , ru. For i P t1, . . . , ru, let ai P Azt0u be such that aiPi P ArXs, and Pi “ ´1 cpaiPiq paiPiq P ArXs. As Pi has content 1 and is irreducible in KrXs (because Pi is), it is irreducible in ArXs (cf proposition 1.40). We have a1a2 ¨ ¨ ¨ ar “ cpa1P1q ¨ ¨ ¨ cparPrq by lemma 1.38, because cpP qr “ 1, hence the prime decompositionr P “ P1P2 ¨ ¨ ¨ Pr. ‚ Unicity of prime decomposition. Let P P ArXszt0u and P “ P1P2 ¨ ¨ ¨ Pr and P “ Q1Q2 ¨ ¨ ¨ Qs two prime decompositions in ArXs. Possibly afterr r renumberingr the Pi (resp. the Qj), there exist r0 ď r (resp. s0 ď s) such that Pi P Azt0u for i ď r0 and degpPiq ą 0 for r0 ă i ď r (resp. Qj P Azt0u for j ď s0 and degpQjq ą 0 for s0 ă j ď s). By the second point above, we have cpPiq “ cpQjq “ 1 for r0 ă i ď r and s0 ă j ď s. Taking contents in the equality P1P2 ¨ ¨ ¨ Pr “ Q1Q2 ¨ ¨ ¨ Qs, we get P1P2 ¨ ¨ ¨ Pr0 “ Q1Q2 ¨ ¨ ¨ Qs0 , which is an equality of two prime decompositions in the UFD A: we have r0 “ s0, and after renumbering, we may assume that

PiA “ QiA for all i P t1, . . . , r0u. Dividing P by P1P2 ¨ ¨ ¨ Pr0 , we get Pr0`1 ¨ ¨ ¨ PrArXs “ Qr0`1 ¨ ¨ ¨ QsArXs. This is a prime decomposition in KrXs, which is a UFD: we have r “ s and after renumbering, we may assume that PiKrXs “ QiKrXs for all i P tr0 ` 1, . . . , ru. As cpPiq “ cpQiq “ 1, we have in fact PiArXs “ QiArXs for all i P tr0 ` 1, . . . , ru.  Remark 1.42. (1) During the proof, we showed that a complete family of representative of irreducible elements in ArXs is given by the union of a complete family of representative of irreducible elements in A

(4)The converse is true and easy. 6 Number theory and that of a family of polynomials in ArXs with content 1 that forms a complete family of representatives of irreducible elements in KrXs. (2) In general, A may be a UFD without ArrXss being one. To summarize the relationships between the classes of rings recalled above, we have the following implications (whose reverses are false): fields ñ Euclidean domains ñ PID ñ UFD ñ integrally closed domains ñ integral domains

2. Modules and algebras 2.1. Modules. Definition 2.2. An A-module is a triple pM, `, ¨q where pM, `q is an abelian group and ¨: A ˆ M Ñ M an external composition law such that : (1) p@a, b P Aq p@m P Mq pa ` bq ¨ m “ a ¨ m ` b ¨ m ; (2) p@a, b P Aq p@m P Mq pabq ¨ m “ a ¨ pb ¨ mq ; (3) p@a P Aq p@m1, m2 P Mq a ¨ pm1 ` m2q “ a ¨ m1 ` a ¨ m2 ; (4) p@m P Mq 1 ¨ m “ m This amounts to give a ring homomorphism A Ñ EndpMq. Remark 2.3. Elements in A are called scalars. As usual, we usually denote a module by the underlying set and write am instead of a ¨ m. Example 2.4. (1) A Z-module in nothing but an abelian group. (2) If K is a field, an K-module is just a K-vector space. (3) If K is a field, a KrXs-module is a K-vector space endowed with a K-linear endomorphism (correspond- ing to the multiplication by X). (4) If I Ă A is an ideal, then I and A{I are A-modules. Definition 2.5. Let M be an A-module. A sub-A-module of M is an additive subgroup N Ă M which is stable under multiplication by scalars, i.e. such that

p@a P Aq p@n1, n2 P Nq n1 ` an2 P N. Exemples 2.6. Submodules of A are nothing but its ideals. When A is a field, submodules are sub-vector spaces.

Operations on submodules of an A-module. Let M be an A-module and pMλqλPΛ a family of sub-A- modules of M. The intersection Mλ is a submodule of M. Put λPΛ Ş Mλ “ mλ pmλqλPΛ P Mλ λ Λ λPΛ ! λPΛ ˇ P ) ÿ ÿ ˇ à (the set of finite sums of elements in Mλ). This isˇ a sub-A-module of M, called the sum of pMλqλPΛ. λPΛ Definition 2.7. Let M be an A-module.Ť (1) Let X Ă M. There exists a smallest sub-A-module N of M such that X Ă N: it is called the sub-A- module of M generated by X (it is the intersection of all sub-A-modules of M that contain X). It is also the sum Ax (where Ax “ tax, a P Au). xPX (2) A subsetř X Ă M generates M when the sub-A-module of M generated by X is M itself. (3) The A-module M is of finite type if it is generated by a finite part. (4) The A-module M is called noetherian if all its sub-A-modules are of finite type.

Definition 2.8. Let Λ be a set and pMλqλPΛ a family of A-modules. (1) The product Mλ is the A-module of maps f :Λ Ñ Mλ such that fpλq P Mλ for all λ P Λ. λPΛ λPΛ (2) The (direct) sumś Mλ is the sub-A-module of MŤλ made of maps f :Λ Ñ Mλ such that the λPΛ λPΛ λPΛ set tλ P Λ, fpλq ‰ 0u isÀfinite. ś Ť Λ pΛq (3) If Mλ “ M for all λ P Λ, one writes M and M instead of M and M. When n P Zě0 and λPΛ λPΛ Λ “ t1, . . . , nu, one denotes it M n. ś À

Remark 2.9. When Λ is finite, the A-modules Mλ and Mλ are the same. λPΛ λPΛ ś À Number theory 7

Definition 2.10. (1) Let M and N be A-modules. An A-linear map from M to N is a group homomorphism f : M Ñ N such that fpamq “ afpmq for all a P A and m P M. The set of A-linear maps from M to N is an abelian group denoted HomApM,Nq. ´1 (2) The kernel of f P HomApM,Nq is the submodule Kerpfq “ f p0q of M, and the image of f is the submodule Impfq “ fpMq of N. The cokernel of f is Cokerpfq :“ N{ Impfq. (3) We say that f is an isomorphism when f is bijective (the inverse map f ´1 is then A-linear). This is equivalent to Kerpfq “ t0u (i.e. f is injective) and Impfq “ N (that is Cokerpfq “ t0u, i.e. f is surjective). Definition 2.11. Let M be an A-module and N a sub-A-module. The quotient group M{N is naturally endowed with a A-module structure (because apm ` Nq “ am ` aN Ă am ` N for all m P M and a P A). The A-module M{N is called the quotient of M by N. The canonical map π : M Ñ M{N; m ÞÑ m ` N is A-linear, and has the following universal property: for all A-linear map f : M Ñ M 1 such that N Ă Kerpfq, there exists a unique A-linear map f : M{N Ñ M 1 such that f “ f ˝ π.

f M / M 1 r < r π f  M N { r In particular, if f : M Ñ M 1 is A-linear, there is a canonical decomposition f “ ι˝f ˝π where ι: Impfq Ñ M 1 is the inclusion, f an isomorphism and π : M Ñ M{ Kerpfq the canonical projection. r Definition 2.12. (1)A free A-module is an A-module isomorphic to ApΛq for some set Λ. r pΛq (2) Let Λ be a set. For λ P Λ, let eλ P A be the element defined by eλpηq “ δλ,η (Kronecker symbol). pΛq The family peλqλPΛ is called the canonical basis of A . pΛq Proposition 2.13. (1) If a P A , then a “ apλqeλ (the sum is finite). λPΛ (2) If M is an A-module, the A-linear map ř pΛq Λ HomApA ,Mq Ñ M

f ÞÑ pfpeλqqλPΛ is an isomorphism. In other words, the data of an A-linear map f : ApΛq Ñ M is equivalent to that of the family pfpeλqqλPΛ.

Proof. (1) For η P Λ, one has apλqeλ pηq “ apηq. λPΛ ´ ¯ pΛq pΛq (2) Follows from fpaq “ apλqřfpeλq for all f P HomApA ,Mq and a P A .  λPΛ ř Definition 2.14. Let M be an A-module and tmλuλPΛ Ă M. Form proposition 2.13 (2), there exists a pΛq unique A-linear map f : A Ñ M such that fpeλq “ mλ for all λ P Λ. The A-module Impfq is the submodule of M generated by tmλuλPΛ. In particular, the family tmλuλPΛ generates M if and only if f is surjective. When f is injective, we say that tmλuλPΛ is free (or linearly independent). When f is an isomorphism (so that M is free), we say that pmλqλPΛ is a basis of M. In that case, any m P M can be pΛq uniquely written m “ aλmλ with paλqλPΛ P A . Such a family pmλqλPΛ is called a basis of M (this λPΛ generalizes the usual notionř of basis of a vector space over a field). Remark 2.15. When A is a field, any A-module is free (any vector space has a basis). This is not true if A is not a field: there exists a non zero ideal I Ă A such that I ‰ A, and the A-module A{I is not free (if e P A{I and a P Izt0u, then ae “ 0). For instance, Z {2 Z is a Z {4 Z-module, and it is not free. It can be shown (this in not obvious) that ZZě0 is not free over Z (though it has no torsion). Proposition 2.16. Bases of a free modules have all the same cardinality. Proof. We have to show that if Λ and Λ1 are sets such that the A-modules ApΛq and ApΛ1q are isomorphic, then 1 Λ and Λ1 have the same cardinality. Let f : ApΛq Ñ ApΛ q be an isomorphism, and I Ă A a maximal ideal A 1 (cf Krull’s theorem). As f is A-linear, it induces an isomorphism f : pA{IqpΛq Ñ pA{IqpΛ q. As I is maximal, pΛq pΛ1q 1 A{I is a field: the A{I-vector spaces pA{Iq and pA{Iq are isomorphic, so CardpΛq “ CardpΛ q.  n Definition 2.17. From the preceding proposition, if M is isomorphic to A with n P Zě0, the integer n is an invariant of M, called the rank of M and denoted by rkpMq. 8 Number theory

Remark 2.18. When M and N are free A-module of ranks m and n, proposition 2.13 (2), implies that the choice of bases of M and N provide an isomorphism m n HomApM,Nq » HomApA ,A q “ MnˆmpAq. As for vector spaces over a field, after the choice of bases, the data of a A-linear map between free A-modules of finite rank is equivalent to that of its matrix in the chosen bases.

Definition 2.19. Let M be an A-module and m P M. Put annApmq “ ta P A ; am “ 0u. This is an ideal of A, called annihilator of m. We say that m is torsion if annApmq ‰ t0u, i.e. if it exists a P Azt0u such that am “ 0. We denote Mtors the set of torsion elements in M, and we say that M is torsion-free (resp. has torsion) if Mtors “ t0u (resp. Mtors “ M). Put annApMq “ ta P A ; p@m P Mq am “ 0u “ annApmq (the annihilator of A): this is an ideal. The mPM A-module structure on M induces an A{ annApMqŞ-module structure on M. Note that M may have torsion even if annApMq “ t0u: for instance annZpQ { Zq “ t0u. Example 2.20. If I Ă A is a non zero ideal, the A-module A{I has torsion. For instance, Z {2 Z is a Z {6 Z-module with torsion. Idem for the Z-module Q { Z.

Proposition 2.21. If A is an integral domain and M is an A-module, then Mtors is a submodule of M and the quotient A-module M{Mtors is torsion-free.

Proof. If m1, m2 P Mtors and α P A, there exist a1, a2 P Azt0u such that a1m1 “ 0 and a2m2 “ 0. As A is an integral domain, we have a1a2 ‰ 0 and a1a2pm1 ` αm2q “ 0 so that m1 ` αm2 P Mtors. Let m P M whose image m`Mtors is torsion in M{Mtors: there exists a P Azt0u such that am`Mtors “ Mtors i.e. am P Mtors, so that there exists b P Azt0u such that bpamq “ 0. As A is an integral domain, we have ab ‰ 0, and m P Mtors.  Remark 2.22. (1) The previous statement does not hold if A is not an integral domain. For instance, if A “ M “ Z ˆ Z, then Mtors “ pZ ˆt0uq Y pt0u ˆ Zq is not a submodule of M. (2) A free A-module is torsion-free, but the converse is false in general (it holds for modules of finite type over principal rings). 2.23. Algebras. Definition 2.24. An A-algebra is a ring homomorphism f : A Ñ B (which may not be injective), whose image lies in the center of B. We will often denote it by the underlying ring B.A morphism between two A-algebras f1 : A Ñ B1 and f2 : A Ñ B2 is a ring homomorphism g : B1 Ñ B2 such that g ˝ f1 “ f2. g B1 / B2 a = f1 A f2 Remark 2.25. (0) Any ring is a Z-algebra, in a unique way. (1) If f : A Ñ B is an algebra, then B is naturally endowed with an A-module structure, and the multipli- cation law B ˆ B Ñ B is A-bilinear. Conversely, if B is a ring endowed with an A-module structure such that the multiplication B ˆ B Ñ B is A-bilinear, then the map f : A Ñ B; a ÞÑ a1B is an A-algebra. Example 2.26. (1) A field extension L{K is a K-algebra. (2) IF K is a field an V a K-vector space, the (non commutative) ring EndK pV q is a K-algebra. (3) If A is a ring, the polynomial ring ArXλsλPΛ is an A-algebra. (4) If B and C are A-algebras, so is their product B ˆ C. (5) If B is an A-algebra and I Ă B an ideal, then B{I is an A-algebra. Definition 2.27. Let f : A Ñ B an A-algebra. (1) A sub-A-algebra is a subring B1 Ă B such that f factors through a ring homomorphism A Ñ B1 (in other words such that the inclusion map B1 Ñ B is a morphism of A-algebras). (2) Let txλuλPΛ Ă B. There exists a smallest sub-A-algebra of B that contains X (this is nothing but the intersection of all the sub-A-algebras of B containing X). This subalgebra is denoted ArxλsλPΛ and is called the sub-A-algebra generated by X. If it is equal to B itself, we say that txλuλPΛ generates the A-algebra B. (3) An A-algebra is of finite type if it is generated by a finite set. This is equivalent to the existence of a surjective morphism of A-algebras ArX1,...,Xns Ñ B. (4) An A-algebra is finite if it is finite as an A-module. Number theory 9

Remark 2.28. (1) A finite A-algebra is of finite type, but the converse does not hold (for instance the polynomial A-algebra ArXs is not finite). (2) Let B a finite A-algebra and M a B-module of finite type. The M is an A-module of finite type. Indeed, r s one can write B “ biA and M “ Bmj, so that M “ Abimj. i“1 j“1 1ďiďr ř ř 1ďřjďs

3. Noetherianity Proposition 3.1. (1) Let M be an A-module. The following properties are equivalent: (i) M is noetherian; (ii) every ascending sequence of sub-A-modules of M is stationary; (iii) every non empty subset of submodules of M has a maximal element (for the inclusion). (2) Let M be an A-module and N Ă M a submodule. Then M is noetherian if and only if the A-modules N and M{N are.

Proof. (1) (i)ñ(ii). Let pMnqPZě0 be an ascending sequence of submodules. As the submodule Mn is of nPZě0 finite type, it is generated by a finite set tm1, . . . , mru: there exists N P Zě0 such that tm1, . . . ,ř mru Ă MN , so that MN Ă Mn Ă MN , hence Mn “ MN , and Mn “ MN for all n ě N. nPZě0 nPZě0 (ii)ñ(iii). Let Eřbe such a subset. If itř has no maximal element, one can inductively construct a strictly ascending (for the inclusion) sequence of elements in E , contradicting (ii). (iii)ñ(i). Let N Ă M be a submodule and E the set of submodules of finite type in N. As t0u P E , we have E ‰ ∅: by (iii), the set E contains a maximal element N0. Assume N0 ‰ N: there exists x P NzN0 1 1 and N “ N0 ` Ax Ă N P E . As N0 Ĺ N , this contradicts the maximality of N0: we have N0 “ N and N is of finite type. (2) ‚ If M is noetherian, then N is noetherian. If N 1 is a submodule of M{N, we can write N 1 “ N{N with N “ π´1pN 1q (where π : M Ñ M{N is the canonical map). As M is noetherian, N is of finite type, which implies that N 1 “ N{N is of finite type as well, and M{N is noetherian. r

‚ Assumer N and M{N are noetherian. Let pMnqnPZě0 be an ascending sequence of submodulesr of M. The sequences pMn X NqnPZě0r and ppN ` Mnq{NqnPZě0 are ascending in N and M{N respectively. As they are noetherian, those sequences are stationary: there exists n0 P Zě0 such that Mn X N “ Mn0 X N and pN ` Mnq{N “ pN ` Mn0 q{N i.e. N ` Mn “ N ` Mn0 for all n ě n0. If m P Mn, there exists x P N and y P Mn0 Ă Mn such that m “ x ` y. As x “ y ´ m P N X Mn “ N X Mn0 , we have m P Mn0 , hence

Mn Ă Mn0 i.e. Mn “ Mn0 . The A-module M is thus noetherian. 

Corollary 3.2. If M1 and M2 are noetherian, so is their product M1 ˆ M2.

Proof. As M1 » M1 ˆ t0u and M2 » pM1 ˆ M2q{pM1 ˆ t0uq are noetherian, this follows from proposition 3.1 (2).  Definition 3.3. The ring A is noetherian if it is as an A-module. By definition, this means that every ideal of A is of finite type. By proposition 3.1, this is equivalent to the fact that any ascending sequence of ideals in A is stationary. Proposition 3.4. If A is noetherian, every A-module of finite type is noetherian.

n Proof. Let M be an A-module of finite type: there exists n P Zě0 and a surjective A-linear map f : A Ñ M. n n As A is noetherian, so is A (corollary 3.2), and M “ A { Kerpfq (proposition 3.1 (2)). 

Example 3.5. (1) Let R be a ring and I an infinite set. The ring of polynomials A “ RrXisiPI is not noetherian: the ideal generated by tXiuiPI is not of finite type. (2) Let A “ Zr2X, 2X2, 2X3,...s “ Z `2X ZrXs Ă ZrXs. Then A is not noetherian: the ideal I generated i by t2X uiPZą0 is not finitely generated. Indeed, the ring homomorphism f : ZrXisiPZą0 Ñ ZrXs defined by i i´1 i fpXiq “ 2X factors through an injective morphism ZrXisiPZą0 {x2 Xi ´ X1yiPZą1 Ñ ZrXs, inducing an i´1 i „ 2 „ isomorphism ZrXisiPZą0 {x2 Xi ´ X1yiPZą1 Ñ A, hence an isomorphism F2rXisiPZą0 {xX1 y Ñ A{2A: the image of I in A{2A corresponds to the ideal generated by tXiuiPZą0 : it is not finitely generated. Moreover, the ideal x2Xy X x2X2y “ x4X2, 4X3,...y is not finitely generated: this gives an example of an intersection of two principal ideal which is not finitely generated (same reasoning as above). Theorem 3.6. (Hilbert) If the ring A is noetherian, so is ArXs. 10 Number theory

Proof. Let I Ă ArXs be an ideal. For n P Zě0, let Jn denote the set of leading coefficients of elements in I which are of degree n. As I is an ideal in ArXs, the set Jn is an ideal in A. If n ď m and a P Jn (so that there exists P P I of degree n whose leading coefficient is a), then a P Jm (since a is the leading m´n coefficient of X P ): the sequence of ideals pJnqnPZě0 is ascending. As A is noetherian, this sequence is stationary: let d P Zě0 be such that n ě d ñ Jn “ Jd. As A is noetherian, the ideal Jd is of finite type: choose α1, . . . , αr generators of Jd, these are the leading coefficients of P1,...,Pr P Jd respectively. On the other hand, denote by ArXsăd the sub-A-module of ArXs made of elements of degree ă d, and put M “ I X ArXsăd. As ArXsăd is an A-module of finite type, it is noetherian (cf proposition 3.4), hence M is of finite type: let Q1,...,Qs be generators of M. We have of course

α1ArXs ` ¨ ¨ ¨ ` αrArXs ` Q1ArXs ` ¨ ¨ ¨ ` QsArXs Ă I

If P P I has degree n ě d, its leading coefficient a belongs to Jd: there exists a1, . . . , ar P A such that r n´d a “ aaα1 ` ¨ ¨ ¨ ` arαr. The polynomial P ´ aiX Pi P I has degree ă n: after subtracting an i“1 element of α1ArXs ` ¨ ¨ ¨ ` αrArXs to P , we mayř assume degpP q ă d. Then P P M “ I X ArXsăd, hence P P Q1ArXs ` ¨ ¨ ¨ ` QsArXs, which shows that P P α1ArXs ` ¨ ¨ ¨ ` αrArXs ` Q1ArXs ` ¨ ¨ ¨ ` QsArXs. The I is of finite type, and ArXs is noetherian.  Corollary 3.7. Let A be a noetherian ring and B an A-algebra of finite type. Then B is a noetherian ring.

Proof. As B is of finite type, there exist b1, . . . , br P B such that B “ Arb1, . . . , brs: there is a surjective morphism of A-algebras f : ArX1,...,Xrs Ñ B defined by fpXiq “ bi for i P t1, . . . , ru. Put I “ Kerpfq: we have B » ArX1,...,Xrs{I. As A is noetherian, so is ArX1,...,Xrs (apply theorem 3.6 r times), so that B is a noetherian ArX1,...,Xrs-algebra: it is a noetherian ring. 

4. Modules of finite type over PIDs In this paragraph, we assume that A is a PID. The ring A is an integral domain: denote by K its fraction field. Recall that A is a UFD (cf proposition 1.26): there are gcd and lcm. Moreover, as ideals in A are generated by one element, A is noetherian. In what follows, empty entries in a matrix correspond to zeros. If n P Zą0 and a1, . . . , an P A, we put

a1 . . diagpa1, . . . , anq “ . P MnpAq. an ´ ¯ ˆ Definition 4.1. If n P Zą0, we put GLnpAq “ tM P MnpAq ; detpMq P A u. Cramer formulas imply that this is the group of units in the (non commutative) ring MnpAq (note that the detpAq ‰ 0 is not enough). Put SLnpAq “ tM P MnpAq ; detpMq “ 1u: this is a subgroup of GLnpAq.

Proposition 4.2. Let n P Zě2 and a1, . . . , an be elements in A generating the unit ideal. Then there exists a matrix in SLnpAq whose first row is pa1, . . . , anq.

´1 Proof. Put X “ pa1, . . . , anq: we have to build M P SLnpAq such that XM “ p1, 0,..., 0q. We work by induction on n ě 2. a1 a2 Case n “ 2. As A “ Aa1 ` Aa2, there exist u, v P A such that va1 ´ ua2 “ 1. The matrix M “ p u v q does the job. Case n ą 2. Let dA “ gcdpa2, . . . , anq and b2, . . . , bn P A such that dbi “ ai for i P t2, . . . , nu. We have 1 1´1 gcdpb2, . . . , bnq “ A: by induction, there exists M1 P SLn´1pAq such that YM1 “ p1, 0,..., 0q where Y “ pb2, . . . , bnq. Let 1 M1 “ 1 M1

1 ´1 ´ ¯ We have detpM1q “ detpM1q “ 1 and XM1 “ pa1, d, 0,..., 0q. Use case n “ 2: as gcdpa1, dq “ A, there 1 ´1 exists M2 P SL2pAq with pa1, dqM2 “ p1, 0q. Let

1 M2 M2 “ In´2

´ ¯ 1 ´1 ´1 (In´2 P SLn´2pAq is the unit matrix). We have detpM2q “ detpM2q “ 1 and XM1 M2 “ p1, 0,..., 0q, i.e. ´1 XM “ p1, 0,..., 0q with M “ M2M1 P SLnpAq.  Remark 4.3. This proof provides an effective procedure to construct the matrix provided one can deal with the case n “ 2 (which is the case, for instance, when A is euclidean). Number theory 11

Definition 4.4. If n, m P Ną0, we make the group SLnpAq ˆ SLmpAq act on the A-module MnˆmpAq by pP,Qq ¨ M “ P ´1MQ.

Two matrices M1,M2 P MnˆmpAq are equivalent if they are in the same orbit for this action. We write then M1 „ M2 (this defines an equivalence relation). Note that we may also make GLnpAq ˆ GLmpAq act in a similar way. Remark 4.5. When n “ m, one should not confuse this notion with the finer notion of similarity: two ´1 matrices M1,M2 P MnpAq are similar if there exists P P GLnpAq such that M2 “ P M1P . Definition 4.6. A reduced matrix is a matrix of the form α1 . . . P Mn mpAq αr ˆ ˆ ˙ with r P t0,..., mintm, nuu and α1, . . . , αr P Azt0u such that αi | αi`1 for all i P t0, . . . r ´ 1u.

Notation. (1) Fix a family ppλqλPΛ of representatives of irreducible elements in A. Any element a P Azt0u admits a unique decomposition as a product of irreducible factors:

nλ a “ u pλ λPΛ ź ˆ where u P A and pnλqλPΛ is a family of integers, all but finitely many being equal to zero. We put

`paq “ nλ P Zě0 λPΛ ÿ that we call the length of a. This is nothing but the number of irreducible factors in a (for instance, we ˆ have `paq “ 0 ô a P A and `paq “ 1 if and only if A is irreducible). If M “ rmi,js 1ďiďn P MnˆmpAqzt0u, 1ďjďm we put

`pMq “ min `pmi,jq 1 ď i ď n, 1 ď j ď m, mi,j ‰ 0 .

(2) If σ P Sn is a permutation, we put Pσ ˇ “ δσpiq,j P MnpAq (where( δi,j is the Kronecker ˇ 1ďi,jďn symbol). We have detpP q “ εpσq (where εpσq is the signature of σ), so that P P GL pAq. Put σ ` ˘ σ n Pσ “ diag 1,..., 1, εpσq Pσ P SLnpAq. If M P Mn mpAq, the matrix PσM is the element in Mn mpAq whose i-th row is the σpiq-th row of M. `ˆ ˘ ˆ Similarly,r if γ P Sm is a permutation, the matrix MPγ is deduced from M by permuting the columns according to γ. Multiplying M by Pσ on the left (resp. by Pγ on the right), permutes rows according to σ (resp. columns according to γ) and multiplies the last row (resp. column) by εpσq (resp. εpγq). r r Theorem 4.7. Every matrix M P MnˆmpAq is equivalent to a reduced matrix. Proof. We may assume M ‰ 0. We proceed by induction on d “ mintm, nu. Assume d “ 1. Transposing if necessary, we may assume n “ 1, so that M is a row. If m “ 1, there is nothing to do: assume m ě 2. Let α1 be the gcd of the coefficients of M: we have M “ α1X where X is a row vector whose entries generate the unit ideal. By proposition 4.2, there exists Q P SLnpAq such that the ´1 first row of Q is X. Then XQ “ p1, 0,..., 0q thus MQ “ pα1, 0,..., 0q is reduced. 1 1 Assume d ą 1. Recall that M ‰ 0. Let δ “ min `pM q ; M „ M P Zě0. Replacing M by an appropriate equivalent matrix, we may assume that `pMq “ δ. There exist i0 P t1, . . . , nu and j0 P t1, . . . , mu such ( that `pmi0,j0 q “ δ. Let τ1,i0 P Sn (resp. τ1,j0 P Sm) be the transposition of t1, . . . , nu (resp. t1, . . . , mu) 1 ´1 that exchanges 1 and i0 (resp. j0), and put M “ P MPτ P MnˆmpAq (where Pτ P SLnpAq and τ1,i0 1,j0 1,i0 P SL A are the modified permutation matrices, cf definition 4.1 (2)). We have M 1 M and τ1,j0 P mp q „ 1 1 r r r m1,1 “ mi0,j0 : replacing M by M , we may assume that `pm1,1q “ δ. Put α1 :“ m1,1. ‚rWe first show that α1 divides the coefficients of the first row and of the first column of M. Transposing if necessary, it is enough to deal with the first column. Assume there exists i P t2, . . . , nu such that α1 - mi,1. Exchanging the second and the i-th rows, we may assume i “ 2. Let α1 “ gcdpα1, m2,1q. As α1 strictly divides α1, we have `pα1q ă δ. On the other hand, there exist a, b P A such that α1 “ am1,1 ` bm2,1. Put a b r r r ´m2,1{α1 m1,1{α1 r P “ 1 . ¨ . . ˛ r r 1 ˝ 1 ‚ 1 We have detpP q “ 1 and the entry of index p1, 1q in M “ PM is α1: this implies that M „ M and 1 `pM q ď `pα1q ă δ, contradicting the definition of δ. r r 12 Number theory

‚ Multiplying M on the left by the matrix 1 ´m {α 1 2.,1 1 . . . . P SLnpAq ˜ ´mn,1{α1 1 ¸ on the left, and by

1 ´m1,2{α1 ¨¨¨ ´m1,m{α1 1 . SL A . . P mp q ˜ 1 ¸ on the right, we may assume that mi,1 “ 0 for P t2, . . . , nu and m1,j “ 0 for j P t2, . . . , mu. Indeed this provides an equivalent matrix, with same length (the entry of index p1, 1q was not modified). ‚ The matrix M is now of the form α1 M1 with M1 P Mpn´1qˆpm´1qpAq. By induction hypothesis,` there˘ exist P1 P SLn´1pAq, Q1 P SLm´1pAq, r P N, and elements α2, . . . , αr P Azt0u such that αi | αi`1 for all i P t2, . . . , r ´ 1u and α2 . ´1 . . P M1Q1 “ 1 αr ˆ ˙ 1 1 Multiplying M by ´1 P SLnpAq on the left and by P SLmpAq on the right, we may assume that P1 Q1

´ ¯ α1 . . ` ˘ M “ . αr ˆ ˙ 1 It remains to check that α1 | α2. Assume the contrary. Let α1 “ gcdpα1, α2q. As α1 - α2, we have 1 1 `pα1q ă `pα1q “ δ. There exist a, b P A such that aα1 ` bα2 “ α1. The equality α 1 α1 1 1 p q p q p q “ 1 a 1 α2 b 1 α1 α2 1 ´ ¯ 1 1 imply that there exists M “ pmi,jq 1ďiďn P MnˆmpAq equivalent to M and such that m2,1 “ α1: we have 1ďjďm 1 1 `pM q ď `pα1q ă δ, contradicting the definition of δ.  Remark 4.8. (1) When A is euclidean, it is possible to make this statement constructive, using elementary operations. (2) When A is a field, one recovers the well known fact that the orbits for the equivalence relation are 1 . . . characterized by the rank: every matrix M is equivalent to 1 (where the number of 1 is rkpMq). ˆ ˙ Notation. Let M P Mn,mpAq. If k P t0,..., mintn, muu, let IkpMq be the ideal generated by the minors of (5) order k of M (so this is the gcd of those minors). The sequence of ideals pIkpMqq0ďkďmintn,mu is decreasing , and IkpMq “ t0u if k ą rkpMq. These are called the invariant factors of M. Lemma 4.9. Two matrices that are equivalent have the same invariant factors.

1 ´1 1 Proof. Let M P Mn,mpAq and P P GLnpAq. Put M “ P M. Lines of M are A-linear combinations of those of M: by multilinearity of the determinant, a minor of order k of M 1 is an A-linear combination of 1 1 1 minors of M of order k. This implies that IkpM q Ă IkpMq. As M “ PM , we have also IkpMq Ă IkpM q, 1 i.e. IkpM q “ IkpMq. Similarly, we have IkpMQq “ IkpMq for all Q P GLmpAq (using the fact that columns of MQ are A-linear combinations of those of M). 

Theorem 4.10. With the notations of theorem 4.7, we have IkpMq “ α1 ¨ ¨ ¨ αkA for k P t1, . . . , ru (where r “ rkpMq). In particular, the sequence of ideals α1A Ą α2A Ą ¨ ¨ ¨ Ą αrA is unique.

Proof. By lemma 4.9, we have IkpMq “ Ikpdiagpα1, . . . , αr, 0,..., 0qq “ α1 ¨ ¨ ¨ αkA for k P t1, . . . , ru.  Theorem 4.11. (Adapted basis theorem). Let M be a sub-A-module of an A-module L free of finite rank n. Then M is free, and there exists a basis pe1, . . . , enq of L, an integer r ď n and α1, . . . , αr P Azt0u such that

αi | αi`1 for all i P t0, . . . r ´ 1u #pα1e1, . . . , αrerq is a basis of M.

(5)This follows from the fact that minors of order k a linear combinations of minors of order k ´1, as can be seen by developing determinant along the first row. Number theory 13

Proof. As A is a PID, it is noetherian. As L is of finite type, it is noetherian (proposition 3.4): its sub- A-module M is of finite type as well. Choose a generating family x1, . . . , xm P M: we have an A-linear map f : Am Ñ L m pa1, . . . , amq ÞÑ ajxj j“1 ÿ whose image is nothing but M. After the choice of a basis B of L, this map is given by an n ˆ m matrix (whose j-th column consists in the coordinates of xj in B). By theorem 4.7, this matrix is equivalent to a reduced matrix: after a change of bases in Am and L, it has the form

α1 . . . αr ˆ ˙ with r P t0,..., mintm, nuu and α1, . . . , αr P Azt0u such that αi | αi`1 for i P t0, . . . r ´ 1u. Denote by pe1, . . . , enq the new basis of L: the image M of f is then the free sub-A-module with basis pα1e1, . . . , αrerq. 

Remark 4.12. The previous result is obviously false when A is not a PID. For instance Z {2 Z is a sub- Z {4 Z-module of Z {4 Z. Similarly, the sub-Z ˆ Z-module Z ˆt0u of Z ˆ Z is not free. Theorem 4.13. (Invariant factor decomposition). Let M be an A-module of finite type. There exist ˆ integers d, r P Zě0 and a1, . . . , ad P Az t0u Y A such that ` ˘ ai | ai`1 for all i P t0, . . . d ´ 1u r #M » pA{a1Aq ˆ ¨ ¨ ¨ ˆ pA{adAq ˆ A

Moreover, the integers d, r and the ideals a1A, . . . , adA are unique. The integer r is called the rank of M and when r “ 0, the elements pa1, . . . , adq “the” invariant factors of M.

Proof. ‚ We start with the existence. As M is of finite type, we can choose a generating family m1, . . . , mn: we have a surjective A-linear map f : An Ñ M n pλ1, . . . , λnq ÞÑ λimi. i“1 ÿ n As A is free of finite rank, there is a basis pe1, . . . , enq such that s Kerpfq “ Aαiei i“1 à with s P t1, . . . , nu and α1, . . . , αs P Azt0u such that αi | αi`1 for all i P t0, . . . s ´ 1u (cf theorem 4.11). Taking the quotient, f induces an A-linear isomorphism s n n M » A { Kerpfq “ pA{αiAqei ‘ Aei i“1 i“s`1 ´ à ¯ ´ à ¯ ˆ ˆ Let t “ max i P t1, . . . , su ; αi P A (we have t “ 0 if α1 R A ). Put d “ s ´ t, r “ n ´ s and ai “ αt`i for i P t1, . . . , du. We have a , . . . , a P Az t0u Y Aˆ and a | a for all i P t0, . . . d ´ 1u. Moreover, as 1 d( i i`1 ` 0 ˘ if i ď t A{αiA “ #A{ai´tA if t ă i ď s we have r M » pA{a1Aq ˆ ¨ ¨ ¨ ˆ pA{adAq ˆ A . r ‚ We now prove the unicity. We have Mtors “ pA{a1Aq ˆ ¨ ¨ ¨ ˆ pA{adAq thus M{Mtors » A . The integer r thus depends only on M (cf proposition 2.16). We are thus reduced to the case where M is a torsion d module. We have M » pA{aiAq with a1 | a2 | ¨ ¨ ¨ | ad in Azt0u. Let P be the set of irreducible i“1 ś 14 Number theory elements in A. If p P P, the ideal pA is prime an non-zero, hence maximal(6): the A-module M{pM is an A{pA-vector space of finite dimension dppMq (we have dppMq “ #ti P t1, . . . , du ; p | aiu). This shows in particular that d “ dpMq :“ max dppMq only depends on M. pPP n n`1 For all n P Zě0, we have dppp M{p Mq “ #ti P t1, . . . , du ; vppaiq ě n ` 1u. This implies that for all n P Zą0, the integer n´1 n n n`1 #ti P t1, . . . , du ; vppaiq “ nu “ dppp M{p Mq ´ dppp M{p Mq only depends on M and p. As vppa1q ď vppa2q ď ¨ ¨ ¨ ď vppadq, this implies that for all p P P and all i P t1, . . . , du, the integer vppaiq only depends on M and p. This means that the ideals aiA only depend on M. Remark: an other way to conclude. Lemma 4.14. If a, b A 0 , we have a A bA A b A. P zt u p { q » { gcdpa,bq

Proof. Write a “ α gcdpa, bq and b “ β gcdpa, bq: we have gcdpα, βq “ 1. Let π : A Ñ A{bA be the canonical projection. Then apA{bAq is the image of the composite π ˝ ma, where ma : A Ñ A is the multiplication by a. We have x P Kerpπ ˝ maq ô ax P bA ô αx P βA ô x P βA (because „ gcdpα, βq “ 1). The surjective morphism π ˝ ma : A Ñ apA{baq thus induces an isomorphism A{βA Ñ apA{bAq. 

d d We prove the unicity of the ideals taiAu1ďiďd by induction on d, the case d “ 0 being empty. Assume that M » pA{aiAq » pA{biAq i“1 i“1 d ś d ś with a1 | a2 | ¨ ¨ ¨ | ad and b1 | b2 | ¨ ¨ ¨ | bd. Let s “ maxti P t1, . . . , du ; aiA “ a1Au. We have a1M » a1pA{aiAq » a1pA{biAq. i“s`1 i“1 ś ś d a d b b By lemma 4.14, this means that a M A i A A i A. By unicity of d a M , this implies that A i A 0 , 1 » { a » { gcdpa ,b q p 1 q { gcdpa ,b q “ t u i“s`1 1 i“1 1 i 1 i i.e. biA “ gcdpa1, biqA whence a1A Ă biAśfor all i P t1, . .ś . , su. Symmetrically, we also have biA Ă a1A, so aiA “ biA for i P t1, . . . , su. d ai d bi ai bi Moreover, we have a1M » A{ A » A{ A: the induction hypothesis implies that A “ A and thus aiA “ biA for all i“s`1 a1 i“s`1 a1 a1 a1 i P ts ` 1, . . . , du, finishingś the proof. ś  Corollary 4.15. A torsionfree A-module of finite type is free.

Corollary 4.16. The ideals α1A, . . . , αrA in theorems 4.7 and 4.11 are unique. r n r n´r Proof. If M “ Aαiei Ă Aei “ L, we have L{M » pA{αiAqeiˆA . Let s be the number of indices i“1 i“1 i“1 ˆ n´r i P t1, . . . , ru suchÀ that αiAÀ“ A (i.e. αi P A ). We haveÀL{M » pA{αs`1Aq ˆ ¨ ¨ ¨ ˆ pA{αrAq ˆ A . By theorem 4.13, the integers r´s and n´r and thus s only depend on L and M, and the ideals αs`1A, . . . , αrA as well, which implies unicity in theorem 4.11. This implies unicity in theorem 4.7.  5. Tensor product Let M and N be A-modules. Definition 5.1. Let L be an A-module. A map f : M ˆ N Ñ L is bilinear if it satisfies the following conditions:

(1) f is left-linear, i.e. p@a P Aq p@m1, m2 P Mq p@n P Nq fpam1 ` m2, nq “ afpm1, nq ` fpm2, nq ; (2) f is right-linear, i.e. p@a P Aq p@m P Mq p@n1, n2 P Nq fpm, an1 ` n2q “ afpm, n1q ` fpm, n2q.

The set BilApM,N; Lq of bilinear maps M ˆ N Ñ L is an A-module.

Proposition 5.2. There exists a pair pM bA N, ϕq where M bA N is an A-module and ϕ: M ˆN Ñ M bA N a bilinear map, having the following universal property: if f : M ˆ N Ñ L is a bilinear map, there exists a unique A-linear map f : M bA N Ñ L such that f “ f ˝ ϕ. f r M ˆ N r /7 L ϕ ) f M bA N r Remark 5.3. The universal property of the pair pM bA N, ϕq implies its unicity up to a unique isomorphism. Proof. Consider the A-module ApMˆNq of maps M ˆN Ñ A having a finite support, and its canonical basis e . Let K be the submodule of ApMˆNq generated by the following elements: pm,nq pm,nqPMˆN e e e for m , m M and n N ; ` ‚˘ pm1`m2,nq ´ pm1,nq ´ pm2,nq 1 2 P P

‚ epm,n1`n2q ´ epm,n1q ´ epm,n2q for m P M and n1, n2 P N ; ‚ epam,nq ´ aepm,nq and epm,anq ´ aepm,nq for a P A, m P M and n P N.

(6)If A is a PID and p Ă A is prime and non-zero, then p is maximal. Indeed, let m Ą p be a maximal ideal (cf Krull’s theorem, cf theorem 1.7). As A is a PID, there exist a, b P Azt0u such that p “ aA and m “ bA. As p Ă m, we have b | a: there exists c P A such that a “ bc. As p is prime, we have b P p or c P p. In the last case, there would exist d P A such that c “ ad, whence a “ abd i.e. bd “ 1 since A is a domain and a ‰ 0. This would imply that b P Aˆ i.e. m “ A which is not. We thus have b P p, hence m Ă p i.e. p “ m is maximal. Number theory 15

pMˆNq pMˆNq pMˆNq Put M bA N “ A {K. Let i: M ˆ N Ñ A ; pm, nq ÞÑ epm,nq and π : A Ñ M bA N the canonical projection. Put ϕ “ π ˝ i: by definition of K, the map ϕ is bilinear. If f : M ˆ N Ñ L is bilinear, pMˆNq we define an A-linear map f : A Ñ L by fpepm,nqq “ fpm, nq for all m P M and n P N. As f is bilinear, we have K Ă Kerpfq: the map f factors through a map f : M bA N Ñ L, so that f “ f ˝ ϕ (we p p have fpπpepm,nqqq “ fpm, nq for all m P M and n P N). p p r r f M N r ˆ ϕ ) f % i M bA N /7 L 5 r  π pMˆNq A f

p  Remark 5.4. (1) The universal property of tensor product means that there is a functorial isomorphism

BilpM,N; Lq » HomApM, HomApN,.qq » HomApM bA N,.q

(2) If M is an A-module and B an A-algebra, then B bA M is endowed with a B-module structure (base change).

Notation. With notations of proposition 5.2, put m b n “ πpepm,nqq P M bA N for all m P M and n P N. Elements in M bA N of this form are called simple tensors. They generate M bA N as an A-module, but in general, all elements in M bA N are not simple tensors. Proposition 5.5. Let M be an A-module. „ (1) If N is an A-module, there is an isomorphism M bA N Ñ N bA M sending x b y to y b x.

(2) If pNλqλPΛ is a family of A-modules, then M bA Nλ » pM bA Nλq (distributivity of the tensor λPΛ λPΛ product). ´ À ¯ À Proof. Follow from the universal property of the tensor product. 

Proposition 5.6. If M and N are free, with bases peλqλPΛ and pfδqδP∆ respectively, then M bA N is free, with base peλ b fδqpλ,δqPΛˆ∆.

Proof. Write N “ Afδ. By proposition 5.5 (2), we have M bA N “ M bA Afδ. Similarly, we have δP∆ δP∆ M bA Afδ “ AeÀλ bA Afδ. As Aeλ bA Afδ “ Aeλ b fδ, we get M bÀA N “ Aeλ b fδ, whence the λPΛ λPΛ À δÀP∆ result.  Functoriality of tensor product. Let f : M Ñ M 1 and g : N Ñ N 1 be two A-linear maps. They induce 1 1 a map M ˆ N Ñ M bA N ; pm, nq ÞÑ fpmq b gpnq. It is bilinear, so factors uniquely through an A-linear map 1 1 f b g : M bA N Ñ M bA N . fbIdN 1 In particular, if N an A-module, there is a natural A-linear map M bA N ÝÝÝÝÑ M bA N. An important 1 special case is base change: if B is an A-algebra, f induces a B-linear map B bA M Ñ B bA M .

1 fbIdN 1 Remark 5.7. If f : M Ñ M is an isomorphism, then M bA N ÝÝÝÝÑ M bA N is an isomorphism. If f fbIdN 1 is only injective, then M bA N ÝÝÝÝÑ M bA N may not be injective. If f is surjective, then f b IdN is surjective (even better, Cokerpf b IdN q » Cokerpfq bA N, see below).

Example 5.8. (1) pZ {a Zq bZ pZ {b Zq » Z { gcdpa, bq Z for all a, b P Zą0. (2) pQ { Zq bZ pQ { Zq “ 0. (3) Q bZ Q “ Q. 2 (4) The maps C bC C Ñ C; z1 b z2 ÞÑ z1z2 and C bR C Ñ C ; z1 b z2 ÞÑ pz1z2, z1z2q are isomorphisms. _ (5) Let K be a field, V and W be K-vector spaces, and let V “ HomK pV,Kq be the dual of V . The map _ _ W bK V Ñ HomK pV,W q sending w b α (with w P W and α P V ) to the rank 1 linear map given by x ÞÑ αpxqv is an isomorphism (because it is surjective since any element in HomK pV,W q can be written as _ a sum of rank 1 maps, and dimK pW bK V q “ dimK pV q dimK pW q “ dimK pHomK pV,W qq). In particular, _ „ _ one has V bK V Ñ EndK pV q. Note that the map V bK V Ñ K; v b α ÞÑ αpvq corresponds, via this isomorphism, to the trace map Tr: EndK pV q Ñ K. 16 Number theory

5.9. Tensor product of algebras. Let B and C be A-algebras. The multiplication on B (resp. C) provides maps mB : B bA B Ñ B; x b y ÞÑ xy and mC : C bA C Ñ C; x b y ÞÑ xy. Moreover, there is an „ isomorphism ε: C bA B Ñ B bA C; x b y ÞÑ y b x. Consider the composite

IdB bεbIdC mB bmC pB bA Cq bA pB bA Cq pB bA Bq bA pC bA Cq /1 B bA C µ „ (here we tacitly used the natural isomorphisms pB bA Cq bA pB bA Cq Ñ B bA pC bA Bq bA C and „ B bA pB bA Cq bA C ÑpB bA Bq bA pC bA Cq i.e. the associativity of tensor product).

Definition 5.10. The preceding map µ: pB bA Cq bA pB bA Cq Ñ B bA C endows the A-module B bA C with an A-algebra structure: the product law is simply given by

pb1 b c1q ¨ pb2 b c2q “ pb1b2q b pc1c2q on simple tensors. This A-algebra is called the tensor product of the A-algebras B and C. Remark 5.11. Note that this construction is functorial.

There are natural morphisms of A-algebras iB : B Ñ B bA C; b ÞÑ bb1C and iC : C Ñ B bA C; c ÞÑ 1B bc. Proposition 5.12. (Universal property of the tensor product of algebras). If X is an A-algebra, then

HomA -algpBbA C,Xq “ pf, gq P HomA -algpB,XqˆHomA -algpC,Xq ; p@b P Bq p@c P Cq fpbqgpcq “ gpcqfpbq

In particular, if B and C are commutative, the tensor product pB bA C, iB, iC q is the coproduct of B and( C in the category of commutative A-algebras.

f B 8 iB ) % A B bA C / X 5 9 & C iC g

Example 5.13. (1) ArX1,...,Xns bA B » BrX1,...,Xns. (2) If I Ă B is an ideal and I Ă B bA C the ideal generated by iBpIq, then pB{Iq bA C » pB bA Cq{I . For instance, assume that P1,P2 P A :“ CrX,Y s, and let B “ CrX,Y s{xP1y and C “ CrX,Y s{xP2y. Then B bA C » CrX,Y s{xP1,P2y. Geometrically, this corresponds to the functions on the intersection of the two 2 curves defined by P1 and P2 in the affine plane AC. (3) (Example 5.8 (4) continued) Let L{K is a finite Galois extension with group G, the natural map L bK L Ñ L; x b y ÞÑ pxσpyqqσPG is an isomorphism of L-algebras (for the left structure on the σPG LHS, and theÀ diagonal structure on the RHS). Indeed, choose a primitive element α P L (i.e. such that p1, α, α2, . . . , αd´1q is a K-basis of L, where d “ rL : Ks), and let P pXq “ pX ´ σpαqq P KrXs be σPG its minimal polynomial over K. Then L bK L » L bK KrXs{xP y » LrXs{xśP y » L, the last map σPG sending the class of X to pσpαqqσPG (this is nothing but the Chinese remainder theorem).À By L-linearity, i it is obvious that the composite maps x b y to pxσpyqqσPG (remark: in down to earth terms, p1 b α q0ďiăd i is an L-basis of L bK L, which is mapped to ppσpαq q0ďiădqσPG, which is an L-basis of L because the σPG i Vandermonde matrix pσpαq q0ďiăd P MdpLq is invertible). À σPG 6. Tensor, symmetric and exterior algebras 6.1. Graded algebras.

Definition 6.2. Let A Ñ B be an A-algebra. A grading on B is a collection of sub-A-modules tBnunPZě0 such that 8 ‚ B “ Bn; n“0 ‚ p@m, nÀP Zě0q BnBm Ă Bn`m. A graded A-algebra is an A-algebra endowed with a grading. 8 Remark 6.3. If B “ Bn is a graded A-algebra, then B0 is an A-algebra. n“0 À Number theory 17

Example 6.4. ‚ B “ ArX1,...,Xds has a natural grading, for which Bn is the sub-A-module made of 0 and homogeneous polynomials of degree n. ‚ Idem for ArrX1,...,Xnss.

Remark 6.5. By analogy with the previous example, elements in Bn are sometimes called homogeneous of degree n. 8 8 Definition 6.6. Let B “ Bn be a graded A-algebra. An ideal I Ă B is called graded if I “ pI X Bnq. n“0 n“0 À À Example 6.7. If B “ ArXs and I “ x1 ` Xy Ă B, then I is not graded (because I X Bn “ t0u for all n P Zě0). 8 Proposition 6.8. If B “ Bn is a graded A-algebra and I Ă B an ideal generated by homogeneous n“0 elements, then I is graded. À

Proof. Write I “ βλB with βλ homogeneous of degree nλ P Zě0 for all λ P Λ. Let x P I: there exists λPΛ ř r 8 λ1, . . . , λr P Λ and b1, . . . , br P B such that x “ βλk bk. For k P t1, . . . , ru, write bk “ bk,n with k“1 n“0 ř8 ř b P B , and b “ 0 for n " 0: we have x “ x with x “ β b P I X B , so that k,n n k,n n n λk k,n´nλk n n“0 kPZě0 n ďn ř λřk 8 I Ă pI X Bnq. The reverse inclusion is trivial.  n“0 À 8 Proposition 6.9. Let B “ Bn be a graded A-algebra and I Ă B a graded ideal. For n P Zě0, let n“0 À 8 pB{Iqn “ pBn ` Iq{I » Bn{pI X Bnq be the image of Bn in B{I. Then B{I “ pB{Iqn, so that B{I is a n“0 graded A-algebra. À 8 8 Proof. The map B “ Bn Ñ pB{Iqn is surjective (because Bn Ñ pB{Iqn » Bn{pI X Bnq is for each n“0 n“0 À 8 À n P Zě0) and its kernel is pI X Bnq “ I.  n“0 À8 8 Definition 6.10. Let B “ Bn and C “ Cn be graded A-algebras. n“0 n“0 ‚ A morphism of A-algebrasÀϕ: B Ñ C is gradedÀ if ϕpBnq Ă Cn for all n P Zě0. n ‚ The tensor product algebra B bA C is naturally graded by pB bA Cqn “ Bk bA Cn´k. k“0 8 8 À Remark 6.11. As B “ Bn and C “ Cn, we have B bA C “ Bn bA Cm (cf proposition 5.5 n“0 n“0 n,mPZě0 8 À À À 1 (2)), so B bA C “ pB bA Cqn. Moreover, if 0 ď k ď n and 0 ď ` ď m are integers, and x P Bk, x P B`, n“0 1 1 1 1 1 y P Cn´k and y P CÀm´`, we have px b yqpx b y q “ xx b yy P Bk`` bA Cn`m´pk``q Ă pB bA Cqn`m, so the previous definition makes sense.

6.12. Tensor, symmetric and exterior algebras. In this section M denotes an A-module. If n P Zě0, we put bn M “ M bA M bA ¨ ¨ ¨ bA M . n times b0 b1 (in particular M “ A and M “ M). loooooooooooooomoooooooooooooon Definition 6.13. The tensor algebra of M is 8 bn TApMq :“ M n“0 where the A-algebra structure is characterized by à

px1 b ¨ ¨ ¨ b xnq b py1 b ¨ ¨ ¨ b ymq ÞÑ x1 b ¨ ¨ ¨ b xn b y1 b ¨ ¨ ¨ b ym. It is a graded A-algebra, the n-th graded piece being M bn. 18 Number theory

Remark 6.14. In general, TApMq is not commutative. bn bn Example 6.15. ‚ If M “ Ax in free of rank 1, then M “ Ax is of rank 1 for all n P Zě0, and 8 bn TApMq “ Rx » ArXs is isomorphic to the ring of polynomials in one variable X corresponding to n“0 p0, x, 0,...qÀ P TApMq. ‚ If M “ Ax ‘ Ay is free of rank 2, then TApMq is isomorphic to the free A-algebra on two indeterminates X and Y (that correspond to p0, x, 0,...q and p0, y, 0,...q respectively).

Definition 6.16. Let IspMq Ă TApMq (resp. IapMq Ă TApMq) be the two-sided ideal generated by elements of the form x1 b ¨ ¨ ¨ xn ´ xσp1q b ¨ ¨ ¨ b xσpnq with n P Zą0, x1, . . . , xn P M and σ P Sn (resp. of the form x1 b ¨ ¨ ¨ xn where n P Zě2 and x1, . . . , xn P M are such that there exist 1 ď i ă j ď n such that xi “ xj).

Remark 6.17. As Sn is generated by transpositions, a set of generators for IspMq (resp. IapMq) is given by tx b y ´ y b xux,yPM (resp. tx b xuxPM ).

Being generated by homogeneous elements, the ideals IspMq and IapMq of TApMq are graded. Definition 6.18. The symmetric algebra (resp. exterior algebra) of M is

SymApMq :“ TApMq{IspMq presp. AltApMq “ TApMq{IapMqq 8 8 n n By proposition 6.9, these are graded A-algebras: SymApMq “ SymApMq and AltApMq “ AltApMq n“0 n“0 n bn bn n bn bn where SymApMq “ M {pIspMq X M q and AltApMq “ M {pÀIapMq X M q. À 1 1 Remark 6.19. (1) As A-algebras, SymApMq and AltApMq are generated by SymApMq “ AltApMq “ M. As A is commutative, this implies in particular that the ring SymApMq is commutative, and that the graded nm n m A-algebra AltApMq is anticommutative, which means that yx “ p´1q xy if x P AltApMq and y P AltA pMq. (2) These constructions are functorial: an A-linear map f : M Ñ M 1 induces morphisms of A-algebras 1 1 1 TApfq: TApMq Ñ TApM q, SymApfq: SymApMq Ñ SymApM q and AltApfq: AltApMq Ñ AltApM q. (3) Base change: if B is a commutative A-algebra and M an A-module, then TBpB bA Mq » B bA TApMq, SymBpB bA Mq » B bA SymApMq and AltBpB bA Mq » B bA AltApMq. n n Definition 6.20. The A-module SymApMq (resp. AltApMq) is called the n-th symmetric power (resp. exterior power) of M. n n Notation. ‚ Quite often, AltApMq is denoted by A M. n bn n n n n ‚ Let t: M Ñ M ; px1, . . . , xnq ÞÑ x1 b ¨ ¨ ¨ b xn and s: M Ñ Sym pMq (resp. a: M Ñ Alt pMq) be Ź A A the composite of t with the natural projection. Then one writes x1 ¨ x2 ¨ ¨ ¨ xk´1 ¨ xn instead of spx1, . . . , xnq and x1 ^ ¨ ¨ ¨ ^ xn instead of apx1, . . . , xnq.

Example 6.21. ‚ If M “ Ax in free of rank 1, then SymApMq “ TApMq » ArXs, and AltApMq “ A ‘ Ax. ‚ If M “ Ax ‘ Ay is free of rank 2, then SymApMq » ArX,Y s, and AltApMq “ A ‘ Ax ‘ Ay ‘ Ax ^ y is free of rank 4. n Definition 6.22. Let L be an A-module and n P Zą0. A map f : M Ñ L is n-linear if it is A-linear with respect to each of its variables. A n-linear map f : M n Ñ L is symmetric (resp. alternating) if fpxσp1q, . . . , xσpnqq “ fpx1, . . . , xnq for all x1, . . . , xn P M and σ P Sn (resp. fpx1, . . . , xnq “ 0 as soon as there are 1 ď i ă j ď n such that mi “ mj). Remark 6.23. If f : M n Ñ L is an alternating n-linear map, then f it is antisymmetric, i.e.

fpxσp1q, . . . , xσpnqq “ εpσqfpx1, . . . , xnq ˆ for all x1, . . . , xn P M and σ P Sn. When 2 P A , the converse holds, i.e. an antisymmetric map is alternating.

n bn n n n n Proposition 6.24. The n-linear map t: M Ñ M (resp s: M Ñ SymApMq, resp. a: M Ñ AltApMq) has the following universal property: if f : M n Ñ L is a n-linear map (resp. a symmetric, resp. an bn n alternating n-linear map), then there exists a unique A-linear map f : M Ñ L (resp. f : SymApMq Ñ L, n resp. f : AltApMq Ñ L) such that f “ f ˝ t (resp. f “ f ˝ s, resp. f “ f ˝ a), i.e. such that the diagram r r n f n f n f r M /9 L presp.rM r 7/ L , resp. rM 7/ L q s t ' bn f ( n f a ( n f M SymApMq AltApMq commutes. r r r Number theory 19

Proof. By the universal property of tensor product, there exists a unique A-linear map f˘: M bn Ñ L such ˘ bn ˘ that f “ f ˝ t. By definition, f is symmetric (resp. alternating) if and only if IspMq X M Ă Kerpfq (resp. bn ˘ ˘ n IapMqXM Ă Kerpfq), i.e. if and only if the map f factorizes through an A-linear map f : SymApMq Ñ L n (resp. f : AltApMq Ñ L).  r Proposition 6.25. (Universal property of the symmetric algebra). Let f : A Ñ B be a commu- r tative A-algebra. The map

HomA -algpSymApMq,Bq Ñ HomA -modpM,Bq is bijective. In other words, any A-linear map ψ : M Ñ B extends uniquely into a morphism of A-algebras

ψ : SymApMq Ñ B. ψ M / B p 7 ' ψ SymApMq p Proof. If h: SymApMq Ñ B is a morphism of A-algebras, and ψ “ h|M , then ψ is A-linear, and for n P Z, we have hpx1 ¨ x2 ¨ ¨ ¨ xnq “ ψpx1qψpx2q ¨ ¨ ¨ ψpxnq for all x1, . . . , xn P M, which implies that h is entirely determined by ψ (we are just using the fact that M generates SymApMq as an A-algebra). This shows that the map HomA -algpSymApMq,Bq Ñ HomA -modpM,Bq is well defined and injective. n Let ψ P HomA -modpM,Bq. If n P Zě0, the map M Ñ B; px1, . . . , xnq ÞÑ ψpx1qψpx2q ¨ ¨ ¨ ψpxnq is n-linear, so 8 bn factors through a map hn : M Ñ B. The map h “ hn : TpMq Ñ B is a morphism of A-algebras. As B n“0 is commutative, we havep IspMq Ă Kerphq, so h factorsp À throughp a morphism of A-algebras h: SymApMq Ñ B such that h|M “ ψ, which shows the surjectivity of HomA -algpSymApMq,Bq Ñ HomA -modpM,Bq.  p p Similarly: Proposition 6.26. (Universal property of the exterior algebra). Let f : A Ñ B be an anticom- mutative A-algebra. The map

HomA -algpAltApMq,Bq Ñ HomA -modpM,Bq is bijective. In other words, any A-linear map ψ : M Ñ B extends uniquely into a morphism of A-algebras ψ : AltApMq Ñ B. ψ M / B p 8 & ψ AltApMq p Corollary 6.27. Let M1 and M2 be A-modules. There are natural isomorphisms

SymApM1q bA SymApM2q » SymApM1 ‘ M2q

AltApM1q bA AltApM2q » AltApM1 ‘ M2q Proof. Let f : A Ñ B be an A-algebra. Assume B is commutative: we have natural bijections

HomA -algpSymApM1 ‘ M2q,Bq » HomA -modpM1 ‘ M2,Bq

» HomA -modpM1,Bq ˆ HomA -modpM2,Bq

» HomA -algpSymApM1q,Bq ˆ HomA -algpSymApM2q,Bq

» HomA -algpSymApM1q bA SymApM2q,Bq by the universal property of symmetric algebras and tensor product of A-algebras. Since this holds for any commutative A-algebra B, we get an isomorphism SymApM1 ‘ M2q » SymApM1q bA SymApM2q. The case of the exterior algebra is similar. 

Remark 6.28. If n, k P Zě0 and x1, . . . , xk P M1, y1, . . . , yn´k P M2, then bn x1 b ¨ ¨ ¨ xk b y1 b ¨ ¨ ¨ b yn´k P pM1 ‘ M2q so we get a map k bk bn´k bn M1 bA M2 Ñ pM1 ‘ M2q k“0 à 20 Number theory

This map is not an isomorphism in general. For instance, using proposition 5.5 (2) we have b2 b2 b2 pM1 ‘ M2q “ M1 ‘ M1 bA M2 ‘ M2 bA M1 ‘ M2 and the factor M2 bA M1 is not included in the image. If we add all those maps, we get a graded morphism of A-algebras

TApM1q bA TApM2q Ñ TApM1 ‘ M2q (which is not an isomorphism in general). It induces graded morphisms of A-algebras „ „ SymApM1q bA SymApM2q Ñ SymApM1 ‘ M2q AltApM1q bA AltApM2q Ñ AltApM1 ‘ M2q which are nothing but those provided by corollary 6.27. Considering the graded pieces of the graded isomorphisms of corollary 6.27, we get A-linear isomorphisms: n k n´k „ n SymApM1q bA SymA pM2q Ñ SymApM1 ‘ M2q k“0 à n k n´k „ n AltApM1q bA AltA pM2q Ñ AltApM1 ‘ M2q. k“0 à d Corollary 6.29. Assume M “ Axk is free of rank d. k“1 (1) We have SymApMq » ArX1À,...,Xds (where Xk corresponds to the image of p0, xk, 0,...q P TpMq), so n n`d´1 in particular SymApMq is a free module of rank n (a basis being given by homogeneous monomials of degree n). n d ` ˘ (2) The A-module AltApMq is free of rank n with basis pxi1 ^ ¨ ¨ ¨ ^ xin q0ăi1㨨¨ăidďn, so AltApMq is free of rank 2d. ` ˘ Proof. The case d “ 1 is nothing but example 6.21. The general case follows by induction, using corollary 6.27 for the symmetric algebra, and the second isomorphism above for the exterior power. 

Definition 6.30. Let M be a free A-module of rank d and f P EndApMq. By functoriality, f induces an d d d A-linear endomorphism AltApfq: AltApMq Ñ AltApMq, which is the multiplication by a scalar detpfq P A d since AltApMq is free of rank 1 over A by corollary 6.29 (2). This scalar is called the determinant of f.

Remark 6.31. This definition matches the “usual” one: let B “ pe1, . . . , edq be a basis of M and pαi,jq1ďi,jďd P d d MdpAq the matrix of f in B, so that fpeiq “ αi,jej. We have AltApMq “ Ae where e “ e1 ^ ¨ ¨ ¨ ^ ed, so j“1 that: ř d d d AltApfqpeq “ α1,jej ^ ¨ ¨ ¨ ^ αn,jej j“1 j“1 ´ ÿ ¯ ´ ÿ ¯

“ α1,j1 α2,j2 ¨ ¨ ¨ αd,jd ej1 ^ ej2 ^ ¨ ¨ ¨ ^ ejn

1ďj1,...,jdďd ÿ “0 if jk“j` with k‰`

“ α1,σp1q ¨ ¨ ¨ αd,σpdq eσp1q ^looooooooooomooooooooooon ¨ ¨ ¨ ^ eσpdq σPSd ÿ εpσqe loooooooooomoooooooooon “ εpσqα1,σp1q ¨ ¨ ¨ αd,σpdq e σPS ´ ÿd ¯ ˆ 6.32. Symmetric and anti-symmetric tensors. Assume from now on that n P Zě2 and that n! P A . n n bn If x1, . . . , xn P M , put fspx1, . . . , xnq “ xσp1q b ¨ ¨ ¨ b xσpnq. This defines a map fs : M Ñ M σPSn n bn which is n-linear and symmetric: it factors uniquelyř through an A-linear map ιs : SymApMq Ñ M (the symmetrization operator). Likewise, put fapm1, . . . , mnq “ εpσqmσp1q b¨ ¨ ¨bmσpnq: this defines a map σPSn n bn fa : M Ñ M which is n-linear and antisymmetrical (whenceř alternating given the hypothesis): it factors n bn uniquely through an A-linear map ιa : AltApMq Ñ M (the anti-symmetrization operator). bn 1 Endow M with the action of Sn given by σpm1 b ¨ ¨ ¨ b mnq “ mσp1q b ¨ ¨ ¨ b mσpnq. Then n! ιs ˝ πs (where bn n bn Sn πs : M Ñ SymApMq is the canonical map) is a projector onto the subspace M (of invariants under 1 bn n the action of Sn). Similarly, n! ιa ˝ πa (where πa : M Ñ AltApMq is the canonical map) is a projector bn ` ˘ onto the subspace of anti-invariants, i.e. elements x P M such that σpxq “ εpσqx for all σ P Sn. Number theory 21

b2 2 2 Remark 6.33. When n “ 2, the previous projectors provide a decomposition M “ SymApMq ‘ AltApMq. Indeed, as 2 P Aˆ we have pSpMq X M b2q ‘ pApMq X M b2q “ M b2 and they provide identifications 2 b2 2 b2 SymApMq “ ApMq X M and AltApMq “ SpMq X M .

7. Flatness Definition 7.1. A complex of A-modules is a sequence of A-linear maps f : M M (where I Z ‚ i i Ñ i`1 iPI Ă is an interval) such that f ˝ f “ 0 for all i P I. It is exact when Kerpf q “ Impf q for all i P I. i`1 i `i`1 i ˘ ‚ A short exact sequence of A-modules is an exact complex of the form f g 0 Ñ M 1 ÝÑ M ÝÑ M 2 Ñ 0 f g Remark 7.2. If 0 Ñ M 1 ÝÑ M ÝÑ M 2 Ñ 0 is exact, then M 1 » Kerpgq and M 2 » Cokerpgq. Proposition 7.3. Let f g (♣) M 1 ÝÑ M ÝÑ M 2 Ñ 0 be a diagram of A-modules. Then (♣) is an exact sequence if and only if for any A-module N, the sequence

2 ˝g ˝f 1 (♠) 0 Ñ HomApM ,Nq ÝÑ HomApM,Nq ÝÑ HomApM ,Nq is exact. Proof. The exactness of the sequence (♠) for all A-module N means that for any A-linear map v : M Ñ N, the composite v ˝ f is zero if and only if v factors through g, i.e. if and only if there exists a (unique) A-linear map u: M 2 Ñ N such that v “ u ˝ g, which precisely means that g : M Ñ M 2 has the universal property of the cokernel of f. This is thus equivalent to the exactness of (♣).  Remark 7.4. (1) Proposition 7.3 implies in particular that if N is an A-module, the functor

HomAp., Nq: ModpAq Ñ ModpAq is left exact. f g (2) Similarly, a diagram of A-modules 0 Ñ M 1 ÝÑ M ÝÑ M 2 is an exact sequence if and only if for any 1 f˝ g˝ 2 A-module N, the sequence 0 Ñ HomApN,M q ÝÑ HomApN,Mq ÝÑ HomApN,M q is exact. this implies in particular that for any A-module N, the functor HomApN,.q: ModpAq Ñ ModpAq is left exact.

Proposition 7.5. Let N be an A-module. The functor ModpAq Ñ ModpAq; M ÞÑ M bA N is right exact. f g This means that if 0 Ñ M 1 ÝÑ M ÝÑ M 2 Ñ 0 is an exact sequence of A-modules, then the complex

1 fbIdN gbIdN 2 M bA N ÝÝÝÝÑ M bA N ÝÝÝÝÑ M bA N Ñ 0 is exact. Proof. By proposition 7.3, it is enough to check the exactness of the sequence 2 1 0 Ñ HomApM bA N,Lq Ñ HomApM bA N,Lq Ñ HomApM bA N,Lq i.e. that of the sequence 2 1 0 Ñ BilApM ,N; Lq Ñ BilApM,N; Lq Ñ BilApM ,N; Lq 1 for any A-module L. This is trivial: an element ϕ lies in the kernel of BilApM,N; Lq Ñ BilApM ,N; Lq if 1 2 and only if ϕp., yq vanishes on M hence factors through M for all y P N, i.e. if and only if ϕ “ ψ ˝pg bIdN q 2 for some unique ψ P BilApM ,N; Lq.  2 Example 7.6. The sequence 0 Ñ Z ÝÑ Z Ñ Z {2 Z Ñ 0 is exact. After tensoring by Z {2 Z, we get the sequence 2 0 0 Ñ Z {2 Z ÝÝÑ“ Z {2 Z Ñ Z {2 Z Ñ 0.

Definition 7.7. An A-module M is called flat if the functor ModpAq Ñ ModpAq; M ÞÑ M bA N is exact, f g that is if for all exact sequence 0 Ñ M 1 ÝÑ M ÝÑ M 2 Ñ 0, the complex

1 fbIdN gbIdN 2 0 Ñ M bA N ÝÝÝÝÑ M bA N ÝÝÝÝÑ M bA N Ñ 0 is a short exact sequence.

1 fbIdN Remark 7.8. By proposition 7.5, N is flat if and only if M bA N ÝÝÝÝÑ M bA N is injective whenever M 1 Ñ M is injective. 22 Number theory

Proposition 7.9. An A-module N is flat over A if and only if for all ideal I Ă A of finite type, the natural map I bA N Ñ IN is injective.

Proof. ‚ Assume N is flat over A. As I Ñ A is injective, so is I bA N Ñ N. ‚ Conversely, assume that the natural map I bA N Ñ IN is injective for every ideal of finite type I Ă A. Let r I Ă A be any ideal. An element ξ P KerpI bA N Ñ INq can be written ξ “ αi b xi with α1, . . . , αr P I k“1 and x1, . . . , xr P N. Let J Ă A be the ideal generated by α1, . . . , αr, so thatřξ P KerpJ bA N Ñ JNq. As J is of finite type, the map J bA N Ñ JN is injective, hence ξ “ 0 in J bA N, so ξ “ 0 in I bA N. This shows that the natural map I bA N Ñ IN is injective for any ideal I Ă A. 1 1 Let M Ă M be a submodule: we want to show that M bA N Ñ M bA N is injective. As above, 1 we can reduce to the case where M{M is of finite type, so that there exist m1, . . . , mr P M such that 1 1 1 M “ M ` Am1 ` ¨ ¨ ¨ ` Amr. For k P t0, . . . , ru, put Mk “ M ` Am1 ` ¨ ¨ ¨ ` Amk, so that M “ M0 Ă 1 M1 Ă ¨ ¨ ¨ Ă Mr´1 Ă Mr “ M. The map M bA N Ñ M bA N is the composite

M0 bA N Ñ M1 bA N Ñ ¨ ¨ ¨ Ñ Mr´1 bA N Ñ Mr bA N so it is enough to show the injectivity of each map Mk´1 bA N Ñ Mk bA N: we can reduce to the case where M “ M 1 ` Am. Put I “ ta P A ; am P M 1u: this is an ideal in A. The map π : M 1 ‘ A Ñ M; px, aq ÞÑ x ` am is surjective. If px, aq P Kerpπq, then am “ ´x P M 1, so a P I. This implies that the map ι: I Ñ Kerpπq; λ ÞÑ p´λm, λq ι π is an isomorphism. Form the exact sequence 0 Ñ I ÝÑ M 1 ‘ A ÝÑ M Ñ 0 we get the exact sequence

ιbIdN 1 πbIdN I bA N ÝÝÝÝÑpM bA Nq ‘ N ÝÝÝÝÑ M bA N Ñ 0

1 Let ξ P KerpM bA N Ñ M bA Nq. Then pξ, 0q P Kerpπ b IdN q, so there exists η P I bA N such that pξ, 0q “ pι b IdN qpηq. Projecting on the second factor, the image of η P I bA N in N is zero. As the map I bA N Ñ N in injective, we have η “ 0, whence ξ “ 0, as required.  Remark 7.10. There is a more natural proof of this result using derived functors of the tensor product. Proposition 7.11. (1) If N is projective over A (i.e. a direct summand in a free A-module), then M is flat. In particular, flatness is automatic when A is a field. (2) If A is principal, that N is flat if and only if it is torsion-free.

Proof. (1) This is true when N is free by example 5.8 (2). In general, write N ‘ S “ L with L a free A-module. Let f : M 1 Ñ M be an injective map of A-modules. By example 5.8 (2) again, the injective map f b IdL identifies with pf b IdN q ‘ pf b IdSq, so f b IdN is injective as well. (2) Assume N is flat and let α P Azt0u. The multiplication map α: A Ñ A is injective: so is α b IdN : A bA N Ñ A bA N. The latter identifies with the multiplication map α: N Ñ N, so N has no α-torsion. Assume N is torsion-free. If I Ă A is a nonzero ideal, then I “ αA with α P Azt0u. The map I bA N Ñ IN identifies to the multiplication by α on N : it is injective since N is torsion-free. This implies that N is flat over A by proposition 7.9.  Remark 7.12. There are flat modules that are not projective. For instance Q is flat over Z (since it is torsion-free), but it is not projective (because it is divisible).

8. Localization Definition 8.1. A subset S Ă A is called multiplicative if 0 R S, 1 P S and if S is stable under multiplication. Example 8.2. (1) Aˆ. n (2) tf unPZě0 where f P A is not nilpotent. (3) Azp where p Ă A is a prime ideal. ι Proposition 8.3. Let S Ă A a multiplicative set. There exists an A-algebra A ÝÑ S´1A, unique up to isomorphism, having the following universal property: if f : A Ñ B is a ring homomorphism such that p@s P Sq fpsq P Bˆ, then there exists a unique ring homomorphism f : S´1A Ñ B such that f “ f ˝ ι.

f A / B r r = ι ! f S´1A r Number theory 23

Proof. Endow the set A ˆ S with the binary relation „ defined by

pa1, s1q „ pa2, s2q ô pDt P Sq tpa1s2 ´ a2s1q “ 0 This is an equivalence relation. Denote by S´1A “ pAˆSq{ „ the quotient set. If pa, sq P AˆS, we denote by a ´1 a1 a2 a1s2`a2s1 its image in S A. Let pa1, s1q, pa2, s2q P A ˆ S. One checks easily that the elements ` :“ s s1 s2 s1s2 and a1 . a2 :“ a1a2 only depend one a1 and a2 , and that this defines two internal laws ` and . over S´1A, s1 s2 s1s2 s1 s2 ´1 1 making S A a commutative ring with unit 1 . Moreover, the map ι: A Ñ S´1A a a ÞÑ 1 s ´1 1 is a ring homomorphism. Note that if s P S, then ιpsq “ 1 is invertible in S A, with inverse s . Let f : A Ñ B a ring homomorphism such that p@s P Sq fpsq P Bˆ. The map f : S´1A Ñ B a ÞÑ fpsq´1fpaq r s is a well defined ring homomorphism, and it is the unique one such that f “ f ˝ ι. The unicity of pS´1A, ιq follows from the universal property.  r Definition 8.4. The A-algebra S´1A is the localization of A with respect to the multiplicative set S. Remark 8.5. (1) As usual, if a P A, we will write a instead of ιpaq its image in S´1A. (2) In some sense, S´1A is the “minimal” A-algebra in which elements in S are invertible. (3) When A is an integral domain, „ is nothing but the "usual" relation pa1, s1q „ pa2, s2q ô a1s2 “ a2s1. When A is not a domain, the latter is not an equivalence relation (why?), and the "t" is necessary. (4) Kerpιq “ ta P A ; pDs P Sq sa “ 0u, so ι is injective when A is an integral domain. (5) Unless A is a factorial domain, there is no notion of “irreducible fraction”. Example 8.6. (1) Assume A is an integral domain. Then Azt0u is multiplicative (t0u is prime), and pAzt0uq´1A “ FracpAq is the fraction field of A. For instance, FracpZq “ Q, and FracpKrXsq “ KpXq when K is a field. If moreover S Ă A is a multiplicative set, the universal property provides an injective ring homomorphism S´1A Ñ FracpAq: localizations of A identify with subrings of FracpAq. (2) More generally, if we do not assume integrity of A, the set S “ tf P A ; f is not a zero-divisor in Au Ă A is multiplicative. In this case the localization QpAq :“ S´1A is called the total ring of fractions of A. n (3) Let f P A. We denote by Apfq the localization of A with respect to the multiplicative set tf unPZě0 . One can easily show that Apfq » ArXs{xfX ´ 1y. For instance, Zp10q is nothing but the ring of decimal numbers. (4) If p Ă A is a prime ideal, we denote by Ap the localization of A with respect to the multiplicative set Azp. When A is an integral domain and p “ t0u, one recovers FracpAq. (5) Exercise: find multiplicative sets S Ă Z other than Z zt0u such that S´1 Z “ Q. Definition 8.7. Let S Ă A be a multiplicative set and M an A-module. The localization S´1M of M with respect to S is defined similarly as S´1A: it is the quotient of the set M ˆ S by the equivalence relation ´1 given by pm1, s1q „ pm2, s2q ô pDt P Sq tpm1s2 ´ m2s1q “ 0. This is a S A-module with the laws given by m1 m2 m1s2`m2s1 a m am ´1 ` :“ and . 1 :“ 1 . Moreover, an A-linear map f : M Ñ N induces a S A-linear map s1 s2 s1s2 s s ss ´1 ´1 m fpmq fS : S M Ñ S N (such that fS s “ s for all m P M and s P S). It enjoys the following property: for any S´1A-module N, the natural map ` ˘ ´1 HomS´1ApS M,Nq Ñ HomApM,Nq is an isomorphism. In particular, if I Ă A, is an ideal (i.e. a submodule of A), S´1I is an ideal in S´1A.

Proposition 8.8. (1) IdM S “ IdS´1M . (2) If f : M Ñ M 1 and g : M 1 Ñ M 2 are A-linear maps, then pg ˝ fq “ g ˝ f . ` ˘ S S S (3) If M Ă N, then S´1M Ă S´1N and S´1pN{Mq » S´1N{S´1M. ´1 ´1 (4) If f : M Ñ N is A-linear, then KerpfSq “ S Kerpfq and CokerpfSq “ S Cokerpfq. ι Proof. (3) The composite M Ă N ÝÑ S´1N extends into i: S´1M Ñ S´1N (by S´1A-linearity). Let ´1 m x P S M: write x “ s with m P M and s P S. If ipxq “ 0, there exists t P S such that tm “ 0 in 24 Number theory

m ´1 M Ă N, which implies that x “ s “ 0 in S M: the map i is injective. We consider it as an inclusion in S´1M Ă S´1N. π The canonical map π : N Ñ N{M induces a S´1A-linear map S´1N ÝÝÑS S´1pN{Mq. It is surjective: if ´1 n n x P S pN{Mq, there exists n P N{M and s P S such that x “ s . Let n P N lifting n: we have πS s “ x. Of course S´1M Ker π . Conversely, if x n Ker π (with n N and s S), we have πpnq 0 in Ă p Sq “ s P p Sq P P s` “˘ ´1 tn ´1 S pN{Mq: there exists t P S such that tπpnq “ πptnq “ 0 in N{M, i.e. tn P M, thus x “ ts P S M. ´1 ´1 ´1 „ ´1 Hence KerpπSq “ S M and S N{S M Ñ S pN{Mq. (4) Follows from (3). 

´1 „ ´1 Proposition 8.9. Let M be an A-module and S Ă A a multiplicative part. Then S A bA M Ñ S M as S´1A-modules. In particular, the A-algebra S´1A is flat.

´1 ´1 a am Proof. (1) The map S A ˆ M Ñ S M; s , m ÞÑ s is bilinear so factors through an A-linear map ´1 „ ´1 a am ´1 m m1 ´1 u: S A bA M Ñ S M, such that u s b m “ s . It is in fact S A-linear. Assume s “ s1 in S M: ` ˘ 1 there exists t P S such that tps1m ´ sm1q “ 0, so 1 b m “ ts b m “ 1 b pts1mq “ 1 b ptsm1q “ ` ˘ s tss1 tss1 tss1 ts 1 1 1 ´1 ´1 m 1 tss1 b m “ s1 b m . This implies that the map v : S M Ñ S A bA M given by v s “ s b m is well defined, and it is an inverse of u. ` ˘ (2) This is a reformulation of proposition 8.8 (3)  If S, S1 Ă A are multiplicative sets, then SS1 :“ tss1 ; s P S, s1 P S1u is also a multiplicative set of A. Proposition 8.10. Let S be the image of S in S1´1A, then there is an natural isomorphism of rings S ´1pS1´1Aq Ñp„ SS1q´1A. Proof. Let f : A Ñ B be an A-algebra such that fpSS1q Ă Bˆ. As fpS1q Ă Bˆ, the map f extends uniquely into a ring homomorphism f : S1´1A Ñ B. Similarly, fpSq Ă Bˆ, so f extends uniquely into a ring homomorphism f : S ´1pS1´1Aq Ñ B. This implies that S ´1pS1´1Aq has the universal property defining 1 ´1 ´1 1´1 „ 1 ´1 pSS q A: there is an natural isomorphismr of rings S pS Aq Ñpr SS q A. r  p Corollary 8.11. If M is an A-module, there is a natural isomorphism S´1pS1´1Mq Ñp„ SS1q´1M.

´1 1´1 „ 1 ´1 ´1 Proof. Tensored with M, the isomorphism S AbA S A ÑpSS q A provides an isomorphism pS AbA 1´1 „ 1 ´1 1´1 „ 1´1 S AqbA M ÑpSS q AbA M (cf proposition 8.9 (1)). As there are isomorphisms S AbA M Ñ S M 1 ´1 „ 1 ´1 and pSS q A bA M ÑpSS q M (cf proposition 8.9 (1) again), we deduce a chain of isomorphisms

´1 1´1 ´1 1´1 S A bA pS A bA Mq / pS A bA S Aq bA M   ´1 1´1 1 ´1 S A bA pS Mq pSS q A bA M

  ´1 1´1 1 ´1 S pS Mq / pSS q M  Lemma 8.12. Let M be an A-module and N 1 a sub-S´1A-module of S´1M. Then N 1 “ S´1N where N is the inverse image of N 1 under the natural map M Ñ S´1M.

m 1 m ´1 n ´1 Proof. If x “ s P N , then sx “ 1 , i.e. m P N, so x P S N. Conversely, x “ s P S N (with n P N and n 1 1 1 ´1 s P S), then 1 P N , thus x P N since N is a S A-module.  Corollary 8.13. Let S Ă A is a multiplicative set. Ideals in S´1A are localizations of ideals in A. In particular, A is noetherian implies S´1A is noetherian. Notation. We denote by SpecpAq the set of prime ideals in A. It is called the spectrum of A. Proposition 8.14. Let S Ă A be a multiplicative set. The maps ´1 tp P SpecpAq ; p X S “ ∅u Ø SpecpS Aq p ÞÑ S´1p q X A :“ ι´1pqq Ðß q are increasing (for the inclusion) bijections inverse one to the other. Number theory 25

Proof. Let p P SpecpAq such that p X S “ ∅. Then S´1A{S´1p » S´1pA{pq (cf proposition 8.8). Let S be the image of S in A{p: as p X S “ ∅, we have 0 R S, and S is a multiplicative set in A{p. As A{p is an integral domain, so is its localization S´1pA{pq “ S ´1pA{pq Ă FracpA{pq, so that S´1p is prime in S´1A. Conversely, if q P SpecpS´1Aq, then A{ι´1pqq ãÑ S´1A{q is an integral domain: we have q X A P SpecpAq. If s P pq X Aq X S, then s P q. As s is invertible in S´1A, we have q “ S´1A, which is not: we have pq X Aq X S “ ∅. Let p P SpecpAq be such that p X S “ ∅. We have of course p Ă S´1p X A. Conversely, let a P S´1p X A: α write a “ s with α P p and s P S. As sa “ α P p and s R p (because p X S “ ∅), we have a P p, which proves the equality p “ S´1p X A. ´1 ´1 a Let q P SpecpS Aq. We have of course S pq X Aq Ă q. Conversely, let x P q : write x “ s with a P A and a ´1 ´1 s P S. We have sx “ a P q X A, so x “ s P S pq X Aq, which proves the equality q “ S pq X Aq.  Remark 8.15. In particular we have SpecpS´1Aq Ă SpecpAq. The set SpecpAq can be equipped with a topological space structure (and even more...) and the bijection of proposition 8.14 identifies SpecpS´1Aq to an open subset of SpecpAq, which explains the terminology of "localization". Definition 8.16. A local ring is a ring having only one maximal ideal. Exemples 8.17. (1) a field is a local ring. (2) If K is a field, the ring of formal series KrrXss is local, with maximal ideal XKrrXss. (3) Exercise: A is local if and only if AzAˆ is an ideal(7): it is then the maximal ideal of A.

Definition 8.18. Let A and B be local rings with maximal ideals mA and mB respectively. A ring homo- morphism f : A Ñ B is local when fpmAq Ă mB. Example 8.19. Let A be a local ring, m its maximal ideal, k “ A{m its residue field. Then the canonical projection A Ñ k is a local homomorphism. Assume moreover that A is an integral domain, and let K “ FracpAq be its fraction field. Then the inclusion A Ñ K is not local.

Corollary 8.20. If p P SpecpAq, then SpecpApq “ tqAp ; q P SpecpAq, q Ă pu. In particular, Ap is a local ring with maximal ideal pAp. Proof. The equality follows from the equivalence q X pAzpq “ ∅ ô q Ă p and proposition 8.14. Bijections of loc. cit. being increasing (for inclusions), maximal elements correspond. 

Lemma 8.21. Let M be an A-module. Then M “ t0u if and only if Mm “ t0u for all maximal ideal m Ă A.

Proof. Assume Mm “ t0u for all maximal ideal m Ă A. Let m P M. Put I “ ta P A, am “ 0u: this is an m ideal in A. Assume I ‰ A: there exists m Ă A maximal such that I Ă m (theorem 1.7). As m “ 1 is 0 in Mm, there exists t P Azm such that tm “ 0 in M, i.e. t P I. We have thus t P Izm, which is a contradiction: I “ A and m “ 0.  Proposition 8.22. (Local-global principle). Let M be an A-module and M 1, M 2 submodules of M. 1 2 1 2 1 2 1 2 Then M Ă M (resp. M “ M ) if and only if Mm Ă Mm (resp. Mm “ Mm) in Mm for all maximal ideal m of A.

1 2 1 2 Proof. If M Ă M , we already know that Mm Ă Mm for all maximal ideal m in A (proposition 8.8 (3)). 1 2 2 Conversely, assume that Mm Ă Mm for all maximal ideal m in A. Put M “ M{M and π : M Ñ M the 1 1 1 2 canonical map, so πpM q Ă M . By assumption, we have πpM qm “ t0u (because the image of Mm Ă Mm in 2 M m “ Mm{Mm is zero, cf proposition 8.8 (3)) for all maximal ideal m in A. By lemma 8.21, this implies 1 1 2 that πpM q “ t0u in M , i.e. M Ă M .  Remark 8.23. An important special case of last proposition is the following: if I and J are ideals in A, then I Ă J if and only if Im Ă Jm for all maximal ideal m in A. 8.24. Discrete valuation rings. Definition 8.25. A discrete valuation ring (DVR) is a PID having a unique nonzero prime ideal. A generator of this nonzero prime ideal is called a uniformizer of A. Remark 8.26. Assume that A is a DVR. Its unique nonzero prime ideal m is maximal: the ring A is local. Elements is m are not invertible: as m ‰ 0, the ring A is not a field.

(7)If A is local with maximal ideal m, then m Ă AzAˆ, and if a P AzAˆ, the ideal aA is strict: it is contained in a maximal ideal (cf theorem 1.7), hence a P m, which proves the equality m “ AzAˆ. Conversely, if m :“ AzAˆ is an ideal, and if I Ă A is a strict ideal, we have I X Aˆ “ ∅, i.e. I Ă m and m contains all ideals in A. 26 Number theory

Proposition 8.27. Assume that A is a DVR, and denote by m its maximal ideal and π a uniformizer. vpaq ˆ (1) Any element a P Azt0u can be written uniquely a “ uπ with u P A and vpaq P Zě0; i i (2) nonzero ideals in A are of the form m “ π A (with i P Zě0); (3) mi “ t0u; iPZě0 Ş Proof. (1) As A is a PID, it is a UFD. As m “ πA is the only nonzero prime ideal, π is the only irreducible element (up to multiplication by an invertible element). The prime decomposition of a P Azt0u is thus of vpaq ˆ the form a “ uπ where u P A and vpaq “ vπpaq P Zě0 is the π-adic valuation of a. (2) If I Ă A is an ideal, it is principal: we have I “ aA with a P A. If I ‰ t0u, then a ‰ 0, so a “ uπi with ˆ i i u P A and i “ vpaq P Zě0, thus I “ π A “ m . (3) If a P Azt0u, we have a “ uπi with u P Aˆ and i “ vpaq, so a P mizmi`1, and a R mi. Thus iPZě0 i m “ t0u. Ş  iPZě0 Ş 9. Integral extensions In what follows, f : A Ñ B is an A-algebra. Definition 9.1. (1) An element b P B is integral over A if there exists a monic polynomial P P ArXs such that P pbq “ 0. The equality P pbq “ 0 is then called an equation of integral dependence of b over A. (2) We say that B is integral over A (or that A Ñ B is integral) when all its elements are integral over A. ? Example 9.2. 2 C is integral over Z, but ?1 is not. P 2 Proposition 9.3. Let b P B. The following are equivalent: (i) b is integral over A; (ii) Arbs is a finite A-algebra; (iii) there exists a sub-A-module B1 Ă B of finite type such that B1 contains an element which is not a zero divisor, and bB1 Ă B1 (i.e. B1 is stable under multiplication by b). Proof. ‚ Assume (i): let P P ArXs monic and such that P pbq “ 0. If degpP q “ n, the A-module Arbs is generated by t1, b, . . . , bn´1u (euclidean division), hence of finite type. ‚ Assume (ii): the A-module B1 “ Arbs satisfies (iii). 1 1 ‚ Assume (iii): let pβ1, . . . , βnq be a generating family of the A-module B . As bβi P B , there exist n M “ pai,jq1ďi,jďn P MnpAq such that bβi “ ai,jβj for all i P t1, . . . , nu. Put X “ pβiq1ďiďn P Mnˆ1pBq: j“1 we have MX “ bX, i.e. ř

(˚) pb In ´MqX “ 0.

Let P pXq “ detpX In ´Mq: this is a monic polynomial of degree n, with coefficients in A. Multiplying 1 equality (˚) by the transpose of the cofactors matrix of b In ´M, we get P pbqX “ 0, so P pbqB “ 0, whence 1 P pbq “ 0 (since B contains an element which is not a zero divisor by hypothesis). 

Lemma 9.4. Let b1, . . . , bn P B such that bi is integral over Arb1, . . . , bi´1s for all i P t1, . . . , nu. Then the A-algebra Arb1, . . . , bns is finite.

Proof. By induction on n P Zą0, the case n “ 1 following from proposition 9.3. Let n P Zą1 and put 1 1 1 1 A “ Arb1, . . . , bn´1s Ă B. By induction, the A-algebra A is finite. As bn is integral over A , the A -algebra 1 1 A rbns is finite: the A-algebra Arb1, . . . , bns “ A rbns is finite.  Proposition 9.5. The A-algebra B is finite if and only if it is integral and of finite type. Proof. If B is finite over A, it is integral by proposition 9.3 (implication (iii)ñ(i) with B1 “ B). Moreover, if tb1, . . . , bnu generates the A-module B, the morphism of A-algebras ArX1,...,Xns Ñ B sending Xi to bi is surjective, so that B is of finite type (as an algebra) over A. Conversely, assume B is integral and of finite type over A. We can write B “ Arb1, . . . , bns, and as b1, . . . , bn are integral over A, the A-module B is of finite type by lemma 9.4.  Proposition 9.6. If A Ñ B and B Ñ C are integral, so is A Ñ C. n n´1 Proof. Let c P C and P pcq “ 0, with P pXq “ X ` b1X ` ¨ ¨ ¨ ` bn P BrXs, an equation of integral 1 dependence. As A Ñ B is integral, the elements b1, . . . , bn are integral over A: by lemma 9.4, B “ 1 Arb1, . . . , bns is finite over A. As B rcs is finite over B, it is finite over A, which implies that c is integral 1 over A (proposition 9.3, noting that 1 P B rcs).  Number theory 27

Corollary 9.7. Let b, b1 P B be integral over A. Then b ´ b1 and bb1 are integral over A. Proof. By lemma 9.4, the morphism A Ñ Arb, b1s is finite hence integral: as b ´ b1, bb1 P Arb, b1s, they are integral over A.  Remark 9.8. If b P Bˆ is integral over A, the inverse b´1 P B is not integral over A in general. Definition 9.9. (1) By corollary 9.7, the set of elements in B that are integral over A is a sub-A-algebra of B, which is called the integral closure of A in B. (2) Assume A is an integral domain and put K “ FracpAq. The integral closure of A is its integral closure in K. We say that A is integrally closed if it is equal to its integral closure, i.e. when the only element in K that are integral over A are elements in A. Proposition 9.10. UFD are integrally closed. In particular, PID are integrally closed. Proof. Assume that A is a UFD, put K “ FracpAq and let x P K integral over A. Write x “ a{b with n n´1 a P A and b P Azt0u coprime. Let x ` α1x ` ¨ ¨ ¨ ` αn “ 0 be an equation of integral dependence (with n α1, . . . , αn P A). Multiplying by b , we get n n´1 n a ` α1a b ` ¨ ¨ ¨ ` αnb “ 0 n ˆ ´1 so that b divides a . As a and b are coprime, this implies that b P A , whence x “ ab P A.  Example 9.11. Let F be a field, t an indeterminate, and put A “ F rt2, t3s Ă B “ F rts. Then we have FracpAq “ FracpBq “ F ptq. As B is a PID, it is integrally closed by proposition 9.10. The element t is integral over A, but t R A, so that A is not integrally closed (hence not a UFD by proposition 9.10). Proposition 9.12. Assume that A is an integral domain, put K “ FracpAq and let L{K be an algebraic field extension. Denote by B the integral closure of A in L. If x P L, there exists a P Azt0u such that ax P B. In particular(8) L “ FracpBq and B is integrally closed.

d d´1 Proof. Let X `α1X `¨ ¨ ¨`αd P KrXs be the minimal polynomial of x over K, and a P Azt0u such that d d´1 d aαi P A for all i P t1, . . . , du. The minimal polynomial of ax over K is then X `aα1X `¨ ¨ ¨`a αn P ArXs, so ax P B. This implies that FracpBq “ L. If x P L is integral over B, then it is integral over A (proposition 9.6), i.e. x P B, and B is integrally closed.  Proposition 9.13. (Integral closure commutes to localization). Under the hypothesis of propo- sition 9.12, let S Ă A be a multiplicative part. The integral closure of S´1A Ă K in L is S´1B.

n n´1 Proof. Let b P B and b ` a1b ` ¨ ¨ ¨ ` an “ 0 an equation of integral dependence over A. If s P S and b ´1 n a1 n´1 an ´1 x “ s P S B, then x ` s x ` ¨ ¨ ¨ ` sn “ 0, which shows that x is integral over S A. Conversely, let ´1 n n´1 ´1 x P L integral over S A and x ` α1x ` ¨ ¨ ¨ ` αn “ 0 an equation of integral dependence over S A. There exists s P S such that ai :“ sαi P A for all i P t1, . . . , nu (take a common denominator to the αi). n n´1 n´2 n´2 n´1 Put b “ sx P L: we have b ` a1b ` sa2b ` ¨ ¨ ¨ ` s an´1b ` s an “ 0, so that b is integral over ´1 A. We thus have b P B, and x P S B.  Definition 9.14. Recall that a number field is a finite extension of Q (usually seen as a subfield of C). If K is a number field, its ring of integers is the integral closure OK of Z in K. On la note OK . By last ´1 proposition, it is an integrally closed ring and K “ pZ zt0uq OK . Proposition 9.15. Assume A is integrally closed, let K “ FracpAq and L{K be an algebraic extension. An element in L is integral over A if and only if its minimal polynomial over K has coefficients in A. Proof. Let x P L and P P KrXs its minimal polynomial over K. If P P ArXs, the equality P pxq “ 0 is an equation of integral dependence, and x is integral over A. Conversely, if x P L is integral over A, fix an algebraic closure L of L, and let x1, . . . , xn be the roots of P in L (i.e. the conjugates of x, counted with multiplicities). If i P t1, . . . , nu, there exists a K-isomorphism of fields f : Kpxq Ñ Kpxiq mapping x to xi (isomorphism extension theorem). If Qpxq “ 0 is an equation of integral dependence (with Q P ArXs), then Qpxiq “ Qpfpxqq “ fpQpxqq “ 0, so that xi is integral over A for all i P t1, . . . , nu. From corollary 9.7, so are the coefficients of P (which are, up to a sign, symmetric polynomials in x1, . . . , xn). As those coefficients belong to K and A is integrally closed in K by hypothesis, we have P P ArXs.  ? 2 2 1 Example 9.16. 2 is not integral over Z (its minimal polynomial over Q is X ´ 2 R ZrXs).

(8)As the proof shows, we have in fact L “ pAzt0uq´1B. 28 Number theory ? Exercise 9.17. Let d P Z zt0, 1u without square factor and K “ Qp dq. Then ? 1` d Z 2 if d ” 1 mod 4 Z OK “ ? Z d if d 1 mod 4 Z # r“ s ‰ ı Proposition 9.18. Assume A Ñ B is injective and that B is an integral domain(9) and integral over A. then A is a field if and only if B is a field.

Proof. ‚ Assume A is a field, and let b P Bzt0u. As B is integral over A, there is an equation of integral n n´1 dependence b ` a1b ` ¨ ¨ ¨ ` an “ 0 with a1, . . . , an P A. As B is an integral domain, we can assume that an ‰ 0 (otherwise we can divide the equation by b): we have bc “ 1 with

´1 n´1 n´2 c “ ´an pb ` a1b ` ¨ ¨ ¨ ` an´1q P B so that b is invertible in B, and B is a field. ‚ Conversely, assume that B is a field. If a P Azt0u, then a has a nonzero (by injectivity of A Ñ B) hence invertible image in B: let a´1 P B be its inverse. As B is integral over A, there is a equation of integral ´1 n ´1 n´1 dependence pa q ` α1pa q ` ¨ ¨ ¨ ` αn “ 0 with α1, . . . , αn P A and

´1 n´1 a “ ´α1 ´ α2a ´ ¨ ¨ ¨ ´ αna P A so that A is a field. 

Proposition 9.19. Assume f : A Ñ B is integral. (1) If M Ă B is a maximal ideal, then M X A is a maximal ideal in A. (2) If f is injective and m Ă A is a maximal ideal, there exists a prime ideal M Ă B such that m “ M X A, and any such M is maximal in B.

Proof. (1) Assume M Ă B is maximal, and put m “ M X A. the morphism A{m Ñ B{M is injective. The A{m-algebra B{M is integral because B is over A (if b P B and P pbq “ 0 is an equation of integral dependence with P P ArXs, we have P pbq “ 0 where P P pA{mqrXs and b P B{M denote the reductions of P modulo mArXs and of b modulo M respectively). As B{M is a field, so is A{m by proposition 9.18, and m is maximal in A. (2) Let m Ă A be a maximal ideal. Assume that mB “ B, i.e. 1 P mB: we can write

r (˚) 1 “ αibi i“1 ÿ 1 1 with α1, . . . , αn P m and b1, . . . , bn P B. As B is integral over A, so is B “ Arb1, . . . , bns. As B is of finite 1 1 type over A, the A-algebra B is in fact finite (cf proposition 9.5): we can write B “ Aβ1 ` ¨ ¨ ¨ ` Aβn. On 1 1 the other hand, equality (˚) implies that mB “ B : for all i P t1, . . . , nu, there exists λi,1, . . . , λi,n P m such that n βi “ λi,jβj. j“1 ÿ 1 If M “ pλi,jq1ďi,jďn P MnpAq and X “ pβiq1ďiďn P Mnˆ1pB q, we have MX “ X, thus pIn ´MqX “ 0: mul- 1 tiplying by the transpose of the cofactor matrix of In ´M, we get detpIn ´MqX “ 0, i.e. detpIn ´MqB “ 0, 1 thus detpIn ´Mq “ 0 in B since 1 P B . Because f is injective, we have detpIn ´Mq “ 0 in A: as detpIn ´Mq ” 1 mod m, we deduce that 1 P m which is absurd, so we necessarily have mB ‰ B. As the ideal mB Ă B is strict, there exists a maximal ideal M Ă B such that mB Ă M (cf theorem 1.7). We of course m Ă M X A, whence m “ M X A since m is maximal in A. If P Ă B is a prime ideal such that m “ P X A, the morphism A{m Ñ B{P is injective. It makes B{P an integral A{m-algebra since B is over A, and B{P is an integral domain: as A{m is a field, so is B{P (cf proposition 9.18), i.e. P is maximal in B. 

10. Discriminants Let A be a ring.

(9)This implies that A is an integral domain. Number theory 29

10.1. Traces and norms.

(10) Definition 10.2. (1) Let M be a free A-module of finite rank and f P EndApMq. If B is an A-basis of M, we can describe f by its matrix pai,jq1ďi,jďn in B (where n “ rkApMq). The trace, the determinant and the characteristic polynomial of f are n Trpfq “ ai,i P A, detpfq “ detpai,jq1ďi,jďn P A, i“1 ÿ

and χf pXq “ det XIn ´ pai,jq1ďi,jďn P ArXs respectively. They depend on f and not on the choice` of the basis B. Recall˘ that Trpf`αgq “ Trpfq`α Trpgq, n detpfgq “ detpfq detpgq and detpαfq “ α detpfq for α P A and f, g P EndApMq. (11) (2) Let B be a free A-algebra of finite rank over A. If x P B, let mx P EndApBq be the map defined by mxpbq “ xb for all b P B. Put

TrB{Apxq “ Trpmxq P A, NB{Apxq “ detpmxq P A and χx,B{A “ χmx P ArXs that we call the trace, the norm and the characteristic polynomial of x respectively (note that χmx is monic). Proposition 10.3. Let B be a free A-algebra of rank n, x, y P B and a P A. Then

(1) TrB{Apx ` yq “ TrB{Apxq ` TrB{Apyq ; (2) TrB{Apaq “ na ; (3) NB{Apxyq “ NB{Apxq NB{Apyq ; n (4) NB{Apaq “ a .

Proposition 10.4. Let L{K be a finite field extension, x P L, and x1, . . . , xn the roots (in some algebraic closure K of K, counted with multiplicities) of the minimal polynomial P of x over K. Then

n n rL:Kpxqs rL:Kpxqs TrL{K pxq “ rL : Kpxqs xi, NL{K pxq “ xi and χx,L{K “ P i“1 i“1 ÿ ´ ź ¯ Proof. Assume first that L “ Kpxq. Let B “ p1, x, . . . , xn´1q: this is a basis of L over K. Let P P KrXs be n n´1 the minimal polynomial of x over K: write P pXq “ X ´ λ1X ´ ¨ ¨ ¨ ´ λn. The matrix of multiplication by x in B is the companion matrix:

0 .¨¨¨ ¨¨¨ 0. λn 1 . . . λ . . . n.´1 C “ Cpλ1, . . . , λnq “ 0 ...... P MnpKq . . . ¨ . . 1 0 λ2 ˛ 0 ¨¨¨ 0 1 λ1 n n´1˝ ‚ We have χC pXq “ detpXIn ´ Cq “ X ´ λ1X ´ ¨ ¨ ¨ ´ λn, so that χx,L{K “ P . In particular, we have d d n´1 TrL{K pxq “ λ1 “ xi and NL{K pxq “ p´1q λn “ xi. i“1 i“1 In general, let d “ř rL : Kpxqs and py1, . . . , ydq be basisś of L over Kpxq, so that L “ Kpxqy1 ‘ ¨ ¨ ¨ ‘ Kpxqyd. d As the multiplication by x preserves each factor Kpxqyi, we have TrL{K pxq “ d TrKpxq{K pxq “ d xi, i“1 d d d d d ř NL{K pxq “ NKpxq{K pxq “ xi and χx,L{K “ χx,Kpxq{K “ P .  i“1 ´ ś ¯ Corollary 10.5. Assume L{K is not separable. Then TrL{K “ 0.

Proof. We have charpKq “ p ą 0. Let x P L, and x1, . . . , xn the roots (in some algebraic closure K of K, counted with multiplicities) of its minimal polynomial P over K. If x is separable over K, then L{Kpxq is n not separable, hence p | rL : Kpxqs, thus TrL{K pxq “ rL : Kpxqs xi “ 0. If x is not separable over K, i“1 pe e we have P pXq “ QpX q with e P Zą0 and Q P KrXs separable:ř each root of P has multiplicity p . This n n implies that xi “ 0, hence TrL{K pxq “ rL : Kpxqs xi “ 0.  i“1 i“1 ř ř (10)It is possible to extend the following definitions to the case where M is a projective module of finite rank. This generalization is useful when working with extension of number fields whose ring of integers is not a PID for instance. (11)I.e. such that B is free seen as an A-module. 30 Number theory

n Example 10.6. (1) Let K be a field, x algebraic over K and P pXq “ X ` a1X ` ¨ ¨ ¨ ` an P KrXs its n minimal polynomial. We have TrKpxq{K pxq “ ´a1, NKpxq{K pxq “ p´1q an and χx,L{K “ P . (2) If L{K is a separable finite extension, K an algebraic closure of K and HomK-algpL, Kq “ tσ1, . . . , σdu, we have d “ rL : Ks, and

d d

TrL{K pxq “ σipxq and NL{K pxq “ σipxq i“1 i“1 ÿ ź ? ? (3) Let d P Z zt0, ?1u be a squarefree? integer and? K “ Qp dq. We have K “ Q ‘ Q d and GalpK{ Qq “ tId , σu where σp dq “ ´ d. If z “ x ` y d P K (with x, y P Q), we thus have Tr pzq “ 2x and K ? ? K{ Q 2 2 NK{ Qpzq “ px ` y dqpx ´ y dq “ x ´ dy .

Corollary 10.7. Let A be an integrally closed domain, K “ FracpAq, L{K a finite extension and B the integral closure of A in L. If b P B, then TrL{K pbq, NL{K pbq P A and χb,L{K P ArXs. Moreover, we have ˆ ˆ b P B ô NL{K pbq P A .

Proof. As the conjugates of b are also integral over A (because its minimal polynomial has coefficients in A, cf proposition 9.15), so are their sum, their product, and more generally any symmetric polynomial evaluated on these conjugates. This implies that TrL{K pbq, NL{K pbq P A and χb,L{K P ArXs. Let b P Bzt0u and P its minimal polynomial over K. By proposition 9.15, we have P P ArXs. Write d d´1 ´1 d ad´1 d´1 a1 1 P pXq “ X `a1X `¨ ¨ ¨`ad: the minimal polynomial of b over K is then X ` X `¨ ¨ ¨` X` . ad ad ad ˆ ˆ d rL:Kpbqs By proposition 9.15, we have thus b P B ô ad P A . We conclude since NL{K pbq “ p´1q ad .  ` ˘ ? Exemples?10.8. (1) Let d?P Z zt0,?1u be a squarefree integer and K “ Qp dq. If d ı 1 mod 4 Z, we have O “ Zr ds. If z “ x ` y d P Zr ds, then N pzq “ x2 ´ dy2 (cf example 10.6 (3)). As Zˆ “ t˘1u, we K ? K{ Q ˆ 2 2 2 2 thus have? z P Zr ds ô x ´ dy P t˘1u. When d ă 0, this is equivalent to x ´ dy “ 1: if d ď ´2, we have Zr dsˆ “ t˘1u and when d “ ´1, we have Zrisˆ “ t˘1, ˘iu. (2) Let p be an odd prime number, ζ P C a primitive p-th root of unity and K “ Qpζq. The minimal p´1 p´2 polynomial of ζ over Q is P pXq “ X ` X ` ¨ ¨ ¨ ` X ` 1. We thus have TrK{ Qpζq “ ´1 and NK{ Qpζq “ 1, so TrK{ Qpζ ´ 1q “ TrK{ Qpζq ´ TrK{ Qp1q “ ´p. The minimal polynomial of ζ ´ 1 over Q is P pX ` 1q, whence NK{ Qpζ ´ 1q “ P p1q “ p. Similarly, the minimal polynomial of ζ ` 1 over Q is P pX ´ 1q, 1 thus NK{ Qpζ ` 1q “ P p´1q “ 1 (which shows that ζ`1 is integral over Z by the preceding corollary).

Proposition 10.9. (Transitivity). If L{K and K{F are finite field extensions, we have

TrL{F “ TrK{F ˝ TrL{K and NL{F “ NK{F ˝ NL{K

Lemma 10.10. Let L{K and K{F be algebraic extension, and F an algebraic extension of F . There exists a bijection

„ HomF -algpL, F q Ñ HomK-algpL, F q ˆ HomF -algpK, F q.

Proof. For each ρ P HomF -algpK, Lq, fix an embedding ρ P HomF -algpF, F q (use Steinitz’ theorem). If K ´1 σ P HomF -algpL, F q let σK denote its restriction to K and put σ “ σ|K ˝ σ. By construction, the field K K K is invariant under σ : we have σ P HomK-algpL, F q. Wep thus have a map y

HomF -algpL, F q Ñ HomK-algpL, F q ˆ HomF -algpK, F q K σ ÞÑ pσ , σK q

K It is injective because σ “ σK ˝ σ . It is surjective since pρ, τq P HomK-algpL, F q ˆ HomF -algpK, F q, and if K σ “ ρ ˝ τ, then we have σK “ ρ and σ “ τ.  x p Number theory 31

Proof of proposition 10.9. ‚ Case where L{F is separable. Keep notations from lemma 10.10. Let x P L: by example 10.6, we have

TrL{F pxq “ σpxq (because L{F is separable, cf example 10.6 (2)) σPHomFÿ-algpL,F q “ ρ τpxq (by lemma 10.10) τPHom pL,F q Kÿ-alg ` ˘ ρPHomF -algpK,F q p “ ρ τpxq ρPHomFÿ-algpK,F q ´ τPHomKÿ-algpL,F q ¯ p “ ρpTrL{K pxqq (because L{K is separable, cf example 10.6 (2)) ρPHomFÿ-algpK,F q p As TrL{K pxq P K, we have ρpTrL{K pxqq “ ρpTrL{K pxqq for all ρ P HomF -algpK, F q, which implies that TrL{F pxq “ TrK{F pTrL{K pxqq (cf example 10.6 (2)). The proof is the same for the norm, replacing sums by products. p ‚ Case where L{F is not separable. By corollary 10.5, we have TrL{F “ 0. Also, one among L{K and K{F is not separable, so TrL{K “ 0 or TrK{F “ 0 (cf corollary 10.5), so the statement on traces is clear. Let rL:F pxqs rKpxq:F pxqs rL:Kpxqs x P L. By proposition 10.4, we have NL{F pxq “ NF pxq{F pxq “ NF pxq{F pxq and N pxq “ N pxqrL:Kpxqs: the statement on norms is equivalent to the equality L{K Kpxq{K ` ˘ rKpxq:F pxqs (˚) NF pxq{F pxq “ NK{F pNKpxq{K pxqq. When x P K, we have Kpxq “ K so the equality follows from proposition 10.4 in that case. In general, n n´1 let P pXq “ X ` an´1X ` ¨ ¨ ¨ ` a1X ` a0 P KrXs be the minimal polynomial of x over K, so that n nd NKpxq{K pxq “ p´1q a0, whence NK{F pNKpxq{K pxqq “ p´1q NK{F pa0q where d “ rK : F s. Fix a basis j n´1 n´1 B “ pe1, . . . , edq of K over F . Then B “ peix q1ďiďd “ pe1, . . . , ed, xe1, . . . , xed, . . . , x e1, . . . , x edq is 0ďjăn a basis of Kpxq over F . As xn “ ´a ´ a x ´ ¨ ¨ ¨ ´ a xn´1, the matrix of the multiplication by x in the 0r 1 n´1 basis B is 0 ¨¨¨ ¨¨¨ 0 ´M0 . . . In . . ´M1 ...... r M “ 0...... P MndpF q ¨ . . . . . 0 . ˛ 0 ¨¨¨ 0 In ´Mn´1 ˝ ‚ (12) (companion matrix by blocks), where Mi is the matrix of the multiplication by ai in the basis B. Then pnd´dqd pn´1qd2 d nd we have TrKpxq{F pxq “ detpMq “ p´1q detp´M0q “ p´1q p´1q NK{F pa0q “ p´1q NK{F pa0q d2 d (because detpM0q “ NK{F pa0q and p´1q “ p´1q ), proving equality (˚).  10.11. Discriminant.

Definition 10.12. Let B be a free A-algebra of rank n and x1, . . . , xn P B. The discriminant of px1, . . . , xnq is D x , . . . , x det Tr x x A p 1 nq “ B{Ap i jq 1ďi,jďn P ´ ¯ Proposition 10.13. Under the hypothesis of definition` 10.12,˘ let M “ pai,jq1ďi,jďn P MnpAq and yi “ n ai,jxj P B for i P t1, . . . , nu. Then j“1 ř 2 Dpy1, . . . , ynq “ detpMq Dpx1, . . . , xnq Proof. Put X Tr x x and Y Tr y y . For all i, j 1, . . . , n , we have “ B{Ap i jq 1ďi,jďn “ B{Ap i jq 1ďi,jďn P t u ` ˘ n `n ˘ n n yiyj “ ai,kxk aj,lxl “ ai,kxkxlaj,l k“1 l“1 k“1 l“1 ´ ÿ ¯´ ÿ ¯ ÿ ÿ hence n n

TrB{Apyiyjq “ ai,k TrB{Apxkxlqaj,l k“1 l“1 T 2 ÿ ÿ 2 thus Y “ MX M, hence detpY q “ detpMq detpXq i.e. Dpy1, . . . , ynq “ detpMq Dpx1, . . . , xnq. 

(12)This follows from the equality det 0 X 1 rs det X whenever X M F , an equality which follows from a Ir Y “ p´ q p q P sp q straightforward induction on r (developing the determinant along the first column). ` ˘ 32 Number theory

Corollary 10.14. Under the hypothesis of definition 10.12, let px1, . . . , xnq and py1, . . . , ynq be bases of B over A. Then Dpy1, . . . , ynqA “ Dpx1, . . . , xnqA n Proof. There exists M “ pai,jq1ďi,jďn P GLnpAq such that yi “ ai,jxj P B for i P t1, . . . , nu. We j“1 2 ˆ have then Dpy1, . . . , ynq “ detpMq Dpx1, . . . , xnq (10.13): as detpMřq P A , we have Dpy1, . . . , ynqA “ Dpx1, . . . , xnqA. 

Remark 10.15. When B “ px1, . . . , xnq is a basis of B over A, the element Dpx1, . . . , xnq is the discriminant of the bilinear form B ˆ B Ñ A; px, yq ÞÑ TrB{Apxyq in the basis B.

Definition 10.16. By corollary 10.14, under the hypothesis of definition 10.12, the ideal Dpx1, . . . , xnqA does not depend of basis px1, . . . , xnq of B over A. This principal ideal is called the discriminant of B over A and is denoted dB{A. Proposition 10.17. Under the hypothesis of definition 10.12, let S Ă A be a multiplicative part. then S´1B is bree over S´1A and ´1 dS´1B{S´1A “ S dB{A. ´1 ´1 Proof. This is obvious since a basis of B over A provides a basis of S B over S A. 

Remark 10.18. The previous proposition shows that the definition of the ideal dB{A sheafifies: one can define it for locally free sheaves on a scheme. This shows in particular that it generalizes to the case where B is projective over A.

Proposition 10.19. Under the hypothesis of definition 10.12, if dB{A contains an element which is not a zero divisor, and if x1, . . . , xn P B, the following conditions are equivalent:

(i) px1, . . . , xnq is a basis of B over A ; (ii) Dpx1, . . . , xnq generates dB{A.

Proof. Implication (i)ñ(ii) follows from definition of the ideal dB{A. Conversely, assume that Dpx1, . . . , xnq generates dB{A. Let pb1, . . . , bnq be a basis of B over A and d “ Dpb1, . . . , bnq so that dB{A “ dA. There n exists M “ pai,jq1ďi,jďn P MnpAq such that xi “ ai,jbj for all i P t1, . . . , nu. By proposition 10.13, we j“1 2 ˆ have Dpx1, . . . , xnq “ detpMq d. As Dpx1, . . . , xnřq generates dB{A “ dA, there exists u P A such that 2 Dpx1, . . . , xnq “ ud, so that dpu ´ detpMq q “ 0. As d is not a zero divisor (otherwise dB{A would only contain zero didisors, which is excluded by the hypothesis), we have detpMq2 “ u thus detpMq P Aˆ, so that M P GLnpAq, which implies that px1, . . . , xnq is a basis of B over A. 

Corollary 10.20. Under the hypothesis of definition 10.12, assume moreover that A is a UFD. Let x1, . . . , xn P B be such that d “ Dpx1, . . . , xnq P Azt0u is squarefree. Then px1, . . . , xnq is a basis of B over A, and dB{A “ dA.

Proof. Let pe1, . . . , enq be a basis of B over A: there exists M “ pai,jq1ďi,jďn P MnpAq such that for all n 2 i P t1, . . . , nu, we have xi “ ai,jej. We have Dpx1, . . . , xnq “ detpMq Dpe1, . . . , enq (proposition 10.13), j“1 2 ˆ i.e. dA “ detpMq dB{A. As řd is squarefree by hypothesis, we have detpMq P A , so that dB{A “ dA. By proposition 10.19, the family px1, . . . , xnq is thus a basis of B over A. 

Theorem 10.21. (Dedekind). Let K{F and L{F be extensions. Then elements in HomF -algpK,Lq are linearly independent in the L-vector space HomF - linpK,Lq. r Proof. Assume the contrary. Let λiσi “ 0 with λi P L and σi P HomF -algpK,Lq for i P t1, . . . , ru be i“1 a non trivial linear dependence relationř such that r is minimal. By minimality, we have λi ‰ 0 for all i P t1, . . . , ru, and the σi are pairwise distinct. After dividing the relation by λr, we may assume that λr “ 1. For all x P K, we have thus r´1 (˚) λiσipxq ` σrpxq “ 0. i“1 ÿ Equality (˚) applied to the product of x, y P K gives r´1 λiσipxqσipyq ` σrpxqσrpyq “ 0 i“1 ÿ Number theory 33

Subtracting σrpyq times (˚) to the preceding equality gives r´1 λiσipxqpσipyq ´ σrpyqq “ 0 i“1 ÿ for all x, y P K. In particular, y being fixed, we have r´1 λipσipyq ´ σrpyqqσi “ 0. i“1 ÿ By minimality of r, the coefficients of this linear combination are all zero: we have σipyq “ σrpyq for all y P K. The σi being pairwise distinct, this implies r “ 1, which is impossible. 

Proposition 10.22. Let L{K be a finite separable field extension, K an algebraic closure of K, and x1, . . . , xn a basis of L over K. Write HomK-algpL, Kq “ tσ1, . . . , σnu. Then 2 Dpx1, . . . , xnq “ det pσipxjqq1ďi,jďn ‰ 0. n ` ˘ Proof. Recall that TrL{K pxq “ σkpxq for all x P L (exemple 10.6 (2)). We have k“1 ř n n TrL{K pxixjq “ σkpxixjq “ σkpxiqσkpxjq k“1 k“1 ÿ ÿ so that Tr x x tMM where M σ x M K . We have thus L{K p i jq 1ďi,jďn “ “ p ip jqq1ďi,jďn P np q t 2 2 ` D˘ px1, . . . , xnq “ det MM “ detpMq “ det pσipxjqq1ďi,jďn .

It remains to check that detpMq ‰ 0. Let` X “˘ pλiq1ďiďn P M1ˆnp`Kq such that XM˘ “ 0. We have then n n λiσipxjq “ 0 for all j P t1, . . . , nu. By K-linearity, this implies λiσi “ 0 in HomK- linpL, Kq. As i“1 i“1 Lř{K is separable, Dedekind’s theorem (theorem 10.21) implies that X ř“ 0: the matrix M is invertible, and detpMq ‰ 0.  n Corollary 10.23. Let L{K be a separable field extension of degree n. A family px1, . . . , xnq P L is a K-basis of L if and only if Dpx1, . . . , xnq ‰ 0. Proposition 10.24. (Transitivity of discriminant). Let K{F and L{K be two finite separable field extensions, x1, . . . , xn a basis of K over F and py1, . . . , ymq a basis of L over K. Then rL:Ks Dpxiyjq 1ďiďn “ Dpx1, . . . , xnq NK{F pDpy1, . . . , ymqq. 1ďjďm

Proof. Write HomF -algpK, F q “ tρ1, . . . , ρnu and HomK-algpL, F q “ tτ1, . . . , τnu (where F is an algebraic closure of F ). Fix liftings ρ1,..., ρn P HomF -algpF, F q of ρ1, . . . , ρn: we have HomF -algpL, F q “ ρiτj 1ďiďn 1ďjďm 2 (cf lemma 10.10). On the other hand, we have Dpxiyjq 1ďiďn “ detpMq where M P Mmn pF q( is the p p 1ďjďm p 2 matrix with entries ρiτjpxky`q “ ρipxkqρiτjpy`q for pi, jq, pk, `q P t1, . . . , nu ˆ t1, . . . , mu (cf proposition ρ1px1qρ1pY q ¨¨¨ ρ1pxnqρ1pY q . . 10.22). Put Y “ pτjpy`qq1 j,` m P MmpF q: we have M “ ` . . ˘ “ M1M2 (block p ď ď p . . ˆ ρnpx1qρpnpY q ¨¨¨ ρnpxnqρpnpY q ˙ ρ1px1q Im ¨¨¨ ρ1pxnq Im . . matrix) where M1 “ diag ρ1pY q,..., ρnpY q P MmnpF q and M2 “ p . p . P MmnpF q. We ρnpx1q Im ¨¨¨ ρnpxnq Im 2 2 ˆ ˙ have detpM1q “ ρ Hom` F K,F ρpdetpY q˘ q “ NK{F pDpy1, . . . , ymqq. On the other hand, there exists a P -algp p q p permutation matrix P GL Z such that P ´1M P diag X,...,X with X ρ x M F . ś P mnp q 2 “ p q “ p ip kqq1ďi,kďn P np q m 2 rL:Ks We thus have detpM2q “ detpXq ,p whence detpM2q “ Dpx1, . . . , xnq (because rL : Ks “ m). At the 2 2 2 rL:Ks end, we have Dpxiyjq 1ďiďn “ detpMq “ detpM1q detpM2q “ Dpx1, . . . , xnq NK{F pDpy1, . . . , ymqq. 1ďjďm  Corollary 10.25. (Transitivity of discriminant). Let A be an integral domain, F “ FracpAq and K{F and L{K finite separable fields extensions. Let B (resp. C) be the integral closure of A in K rkB pCq (13) (resp. L). Assume B is free over A and C is free over B. Then dC{A “ dB{A NB{ApdC{Bq (where NB{ApdBq “ NB{ApdqA).

(13)This does not depend on the choice of the generator d. 34 Number theory

10.26. Discriminant of polynomials.

Definition 10.27. Let K a field, P P KrXs monic and α1, . . . , αn P K the roots of P in an algebraic closure K of K (counted with multiplicities). The discriminant of P is

2 npn´1q discpP q “ pαi ´ αjq “ p´1q 2 pαi ´ αjq 1ďiăjďn 1ďi‰jďn ź ź It is a symmetric polynomial in the roots of P , hence a polynomial in the coefficients of P , and discpP q P K. By definition, P is separable if and only if discpP q ‰ 0. Lemma 10.28. With notations of definition 10.27, we have

n npn´1q 1 discpP q “ p´1q 2 P pαiq i“1 ź n 1 1 Proof. We have P pXq “ pX ´ αjq, hence P pαiq “ pαi ´ αjq which implies that i“1 1ďj‰iďn 1ďj‰iďn n 1 ř ś npn´1q ś P pαiq “ pαi ´ αjq “ p´1q 2 discpP q.  i“1 1ďi‰jďn ś ś Example 10.29. (1) The discriminant of X2 ` aX ` b is a2 ´ 4b. That of X3 ` pX ` q is ´4p3 ´ 27q2 (exercise). n n (2) Let n P Zą0 and P pXq “ X ´ 1 P QrXs. Put µn “ tz P C ; z “ 1u: we have P pXq “ pX ´ ζq. ζPµn 1 n´1 n`1 1 n n2´1 For ζ P µn, we have P pζq “ nζ : as ζ “ p´1q , we have P pζq “ n p´1q , andś thus ζPµn ζPµn ś ś 2 npn´1q 1 n `n´2 n discpP q “ p´1q 2 P pζq “ p´1q 2 n ζPµ źn Remark 10.30. Up to a normalization, the discriminant is nothing but the resultant of P and P 1.

Proposition 10.31. Let L{K a separable field extension of degree d, α P L such that L “ Krαs and P P KrXs the minimal polynomial of α over K. Then p1, α, α2, . . . , αn´1q is a basis of L over K and

2 n´1 npn´1q 1 Dp1, α, α , . . . , α q “ discpP q “ p´1q 2 NL{K pP pαqq

Proof. Let K be an algebraic closure of K and HomK-algpL, Kq “ tσ1, . . . , σnu. the conjugates of α are the αi :“ σipαq for i P t1, . . . , nu. The extension L{K is separable: by proposition 10.22, we have

n´1 j´1 2 j´1 2 Dp1, α, . . . , α q “ det pσipα qq1ďi,jďn “ det pαi q1ďi,jďn j´1 ` ˘ ` ˘ As det pαi q1ďi,jďn “ pαi ´ αjq (Vandermonde determinant), this proves the first equality. 1ďiăjďn n ` ˘ ś npn´1q 1 By lemma 10.28, we have discpP q “ p´1q 2 P pαiq. For i P t1, . . . , nu, we have αi “ σipαq, hence i“1 n n 1 1 1 ś P pαiq “ σpP pαqq “ NL{K pP pαqq, proving the second equality.  i“1 i“1 ś ś Example 10.32. Let K be a field and P pXq “ Xn ` aX ` b P KrXs, that we assume irreducible and separable. If α is a root of P in an algebraic closure of K, we have(14)

2 n´1 npn´1q 1 Dp1, α, α , . . . , α q “ discpP q “ p´1q 2 NKpαq{K pP pαqq npn´1q n n´1 n´1 n´1 n “ p´1q 2 n b ` p´1q pn ´ 1q a

For n P t2, 3u, we recover formulas of example 10.29 (1). ` ˘

(14) 1 n´1 ´aα´b nb ´1 n a n´1 1 We have P pαq “ nα ` a “ n α ` a “ ´ α ´ pn ´ 1qa. The minimal polynomial of α being X ` b X ` b , nb n n´1 n n´1 1 1 that of ´ α is QpXq “ X ´ naX ` p´nq b and that of P pαq is thus QpX ` pn ´ 1qaq: we have NKpαq{K pP pαqq “ p´1qnQppn ´ 1qaq “ nnbn´1 ` p´1qn´1pn ´ 1qn´1an. Number theory 35

10.33. Integral closure in a separable extension. Proposition 10.34. Let L{K be a finite separable field extension. L ˆ L Ñ K

px, yq ÞÑ TrL{K pxyq is a non degenerate pairing.

Proof. Bilinearity follows from la proposition 10.3. Let x P L be such that TrL{K pxyq “ 0 for all y P L. n Let K an algebraic closure de K and HomK-algpL, Kq “ tσ1, . . . , σnu, we have TrL{K pxyq “ σipxqσipyq, i“1 n ř so that σipxqσi. As tσ1, . . . , σnu is linearly independent in HomK-linpL, Kq (Dedekind’s theorem, cf i“1 theorem ř10.21), this implies σipxq “ 0 for all i P t1, . . . , nu, thus x “ 0. The kernel of the bilinear map is zero: it is non degenerate.  Remark 10.35. By corollary 10.5, the preceding proposition is an equivalence. Corollary 10.36. Let L{K be a finite separable field extension. The map

L Ñ HomK- linpL, Kq

x ÞÑ y ÞÑ TrL{K pxyq is an isomorphism of K-vector spaces. If px1, .` . . , xnq is a basis˘ de L over K, there exists a unique basis py1, . . . , ynq of L over K such that TrL{K pxiyjq “ δi,j for all i, j P t1, . . . , nu: it is called the dual basis of px1, . . . , xnq.

Proof. The map f : L Ñ HomK- linpL, Kq is is the linear map associated to the symmetric bilinear map px, yq ÞÑ TrL{K pxyq. As the latter is not degenerate, the map f is injective: it is an isomorphism since dimK HomK- linpL, Kq “ dimK pLq. If px1, . . . , xnq is a basis of L over K, the family pfpx1q, . . . , fpxnqq is a basis of HomK- linpL, Kq over K. The family py1, . . . , ynq satisfies Tr pxiyjq “ fpxiqpyjq “ δi,j for all ` ˘ L{K i, j P t1, . . . , nu if and only if it is the dual basis of pfpx1q, . . . , fpxnqq in L: it exists and is unique.  Proposition 10.37. Let A be an integrally closed domain, K its fraction field and L{K a finite separable field extension. Let B be the integral closure of A in L. Then B contains a basis of L over K, and it is a sub-A-module of a free A-module of rank rL : Ks contained in L.

Proof. If pe1, . . . , enq is a basis of L over K, there exists a P Azt0u such that xi :“ aei P B for all i P t1, . . . , nu (cf proposition 9.12). The family px1, . . . , xnq is still a basis of L over K, made of elements in B. 1 Let py1, . . . , ynq be the dual basis of px1, . . . , xnq for the trace form, and B the sub-A-module of L generated 1 by ty1, . . . , ynu. As py1, . . . , ynq is a basis of L over K, the A-module B is free of rank n “ rL : Ks. If n n x P B, write x “ λjyj with λ1, . . . , λn P K: as xix P B thus TrL{K pxixq “ λj TrL{K pxiyjq “ λi P A j“1 j“1 1 for all i P t1, . . . , nřu (corollary 10.7), we have x P B . ř  Proposition 10.38. Under the hypothesis of proposition 10.37, we have in fact the following more explicit statement. If px1, . . . , xnq is a basis of L over K made of elements in B, we have 1 B Ă Ax ‘ ¨ ¨ ¨ ‘ Ax d 1 n hence d “ Dpx1, . . . , xdq. ` ˘

Proof. By the proof of proposition 10.37, if py1, . . . , ynq is the dual basis of px1, . . . , xnq, we have 1 B Ă B “ Ay1 ‘ ¨ ¨ ¨ ‘ Ayn n Write yi “ αi,jxj with αi,j P K for all i, j P t1, . . . , nu. We have j“1 ř n

δi,j “ TrL{K pxiyjq “ αj,k TrL{K pxixkq k“1 ÿ so that if M Tr x x M A and N α M K , we have M tN I , i.e. “ L{K p i jq 1ďi,jďn P np q “ p i,jq1ďi,jďn P np q “ n tN M ´1 1 M A by Cramer’s formulas: we have α 1 A for all i, j 1, . . . , n . “ P d `np q ˘ i,j P d P t u  Corollary 10.39. Under the hypothesis of proposition 10.37, we have: 36 Number theory

(1) if A is noetherian, then B is a finite A-algebra (in particular, B is noetherian); (2) if A is a PID, then B is a free A-module of rank rL : Ks. Proof. By proposition 10.37, there exists B1 a sub-A-module of L which is free of rank rL : Ks and such that B Ă B1. (1) If A is noetherian, so is B1 (proposition 3.4): the A-module B is of finite type (thus noetherian by proposition 3.4). In particular, B is finite over A (proposition 9.5). (2) If A is a PID, B is free of finite rank as a sub-A-module of the free A-module of finite rank B1 (theorem 4.11). As it contains a basis de L over K (proposition 10.37), its rank is rL : Ks.  Remark 10.40. Under the hypothesis of proposition 10.37, assume moreover that A is a PID. By corollary 10.20, if x1, . . . , xn P B are such that Dpx1, . . . , xnq is squarefree in A (which is a PID hence a UFD), then px1, . . . , xnq is a basis of B over A.

11. Inverse limits 11.1. Generalities. Let C be a category and pI, ďq a directed set(15) (i.e. a preordered set in which every pair of elements has an upper bound: p@i, j P Iq pDk P Iq i ď k, j ď kq).

Definition 11.2. ‚ A inverse system in C indexed by I is a pair tXiuiPI , tui,jui,jPI where tXiuiPI is a iďj ui,j family of objects of C , and tui,jui,jPI a family of morphisms Xj ÝÝÑ` Xi (called transition˘ morphisms) such iďj that ui,k “ ui,j ˝ uj,k whenever i ď j ď k in I. As often, it will be denoted by pXiqiPI alone. (16) ‚ Let tXiuiPI , tui,jui,jPI be a inverse system in C indexed by I. Its inverse limit (or simply limit) iďj is an object` X P C with˘ morphisms πi : X Ñ Xi for all i P I such that p@i ď j P Iq πi “ ui,j ˝ πj, having the following universal property: whenever Y P C and ψi : Y Ñ Xi are morphisms such that p@i ď j P Iq ψi “ ui,j ˝ ψj, then there exists a unique morphism ψ : Y Ñ X such that p@i P Iq ψi “ πi ˝ ψ.

ψj , X r r πj 7 j ψ Y / X ui,j r πi (  2 Xi ψj

Being the solution of a universal problem, the inverse limits of tXiuiPI , tui,jui,jPI , if it exists, is unique iďj up to isomorphism: it is denoted lim X . ÐÝ i ` ˘ I ‚ A direct system in C indexed by I is an inverse system in C op indexed by I. Its inductive limit (or colimit) is the corresponding inverse limit. Remark 11.3. An inverse system in C indexed by I is nothing but a contravariant functor I Ñ C . There op is the obvious inclusion functor i: C Ñ C I that maps an object to the corresponding constant inverse system. If tXiuiPI , tui,jui,jPI is an inverse system in C indexed by I, its inverse limits, if it exists, is iďj characterized` by ˘ „ Hom Iop i Y , X Hom Y, lim X C p q p iqiPI Ñ C ÐÝ i I ` ˘ ` ˘ for all Y P C , i.e. is a final object in the category of pairs pY, ψq where Y P C and ψ : ipY q Ñ pXiqiPI (one op can also say that it represents the contravariant functor Y ÞÑ HomC I ipY q, pXiqiPI ).

Example 11.4. (1) When I is trivial (i.e. i ď j ô i “ j), the inverse limit` is the product˘ Xi. I (17) u (2) If C is preabelian and u P HomC pX,Y q, the kernel of u is the inverse limit of X ÝÑśY Ð 0. (3) Assume that C is a subcategory of Set that admits products indexed by I. Then lim X x X ; i, j I i j u x x X . ÐÝ i – p iqiPI P i p@ P q ď ñ i,jp jq “ i Ă i I iPI iPI ! ś ) ś (15)Which can be seen as a category whose objects are elements of I and there is exactly one arrow i Ñ j if i ď j, and no arrow otherwise. (16)”Limite projective” in French. (17)Which means that C is additive and has kernels and cokernels. Number theory 37

The maps π : lim X X is the restriction of the projection on the factor of index k. In particular, k ÐÝ i Ñ k I inverse limits exist in Set, Gr, ModR (where R is a commutative ring) and Top. (4) An inverse limit lim X in Gr (resp. Mod , resp. Top) coincide with the inverse limit in Set, ÐÝ i R I endowed with the structure of group (resp. R-module, resp. topological space) induced by the inclusion lim X X . ÐÝ i Ă i I iPI ś Remark 11.5. Assume I “ Zě0 (endowed with the natural order). The data of an inverse system is equivalent to that of a sequence of sets pXnqnPZě0 , and for each n P Zě0, a map ρn : Xn`1 Ñ Xn. The inverse limits is then simply: 8 8 lim X :“ px q P X ; p@n P Z q ρ px q “ x Ă X . ÐÝ n n nPZě0 n ě0 n n`1 n n n n“0 n“0 ! ś ) ś Definition 11.6. A morphism of inverse systems tXiuiPI , tui,jui,jPI Ñ tYiuiPI , tvi,jui,jPI is a family of iďj iďj morphisms pfi : Xi Ñ YiqiPI such that fi ˝ ui,j “ u`i,j ˝ fj whenever i ˘ď j.` ˘

Proposition 11.7. (Functoriality of inverse limits). Let pfi : Xi Ñ YiqiPI be a morphism of inverse systems in a category . Assume that the inverse limits X lim X and Y lim Y exist in . Then there C “ ÐÝ i “ ÐÝ i C iPI iPI exists a unique map f : X Ñ Y such that fi ˝ πX,i “ πY,i ˝ f (where πX,i : X Ñ Xi and πY,i : Y Ñ Yi are the projections). Proof. This follows from the universal property of Y :

fj Xj / Yj π 9 9 X,j πY,j ui,j X / Y vi,j f π πY,i X,i %  %  Xi / Yj fi

 11.8. Exactness properties. References for this section are [1, §1.12] and [4, Section 0594]. Here, we assume that C is a subcategory of Gr that is stable under inverse limits (hence under kernels) and cokernels (hence under images).

Definition 11.9. An exact sequence in C is a sequence of morphisms pfn : Xn Ñ Xn`1qnPJ (where J Ă Z is an interval)

fn fn`1 ¨ ¨ ¨ Ñ Xn ÝÑ Xn`1 ÝÝÝÑ Xn`2 Ñ ¨ ¨ ¨ such that Impfnq “ Kerpfn`1q for all n P J. A short exact sequence is an exact sequence of the form 0 Ñ X1 Ñ X Ñ X2 Ñ 0.

op Proposition 11.10. The inverse limit functor lim: I is left exact. ÐÝ C Ñ C I

1 1 pfiqiPI pgiqi,jPI 2 2 Proof. Let 0 Ñ tXiuiPI , tui,jui,jPI ÝÝÝÝÑ tXiuiPI , tui,jui,jPI ÝÝÝÝÝÑ tXi uiPI , tui,jui,jPI Ñ 0 be an iďj iďj iďj exact sequence of` inverse systems of˘ groups.` The first row in ˘ ` ˘ f g 1 2 0 / Xi / Xn / Xi / 0 iPI iPI iPI ś ? f ś ? g ś ? lim X1 lim X lim X2 ÐÝ i / ÐÝ i / ÐÝ i iPI iPI iPI is exact. This implies the injectivity of f : lim X1 lim X . Let x x lim X be such that g x e ÐÝ i Ñ ÐÝ i “ p iqiPI P ÐÝ i p q “ iPI iPI iPI (the unit in lim X2). By the exactness of the first row, we have x f x1 for a unique x1 x1 X1. ÐÝ i “ p q “ p iqiPI P i iPI iPI 1 1 1 1 1 1 1 If i ď j in I, we have xi “ ui,jpxjq i.e. fipxiq “ ui,jpfjpxjqq “ fipui,jpxjqq, thus xi “ ui,jpxjq by injectivityś of f . Since this holds for all i j in I, we get x1 lim X1, and the proposition follows. i ď P ÐÝ i  iPI 38 Number theory

Remark 11.11. The inverse limit functor is not exact in general. For instance, passing to the inverse limit n n on the exact sequences 0 Ñ p Z Ñ Z Ñ Z {p Z Ñ 0 gives the exact sequence 0 Ñ 0 Ñ Z Ñ Zp, and Z Ñ Zp is not surjective.s

Definition 11.12. Let tXiuiPI , tui,jui,jPI be an inverse system in Set. If i P I, the family pui,jpXjqqjPI iďj of subsets of Xi is decreasing,` in the sense˘ that i ď j1 ď j2 ñ ui,j2 pXj2 q Ă ui,j1 pXi1 q Ă Xi. We say that tXiuiPI , tui,jui,jPI satisfies the Mittag-Leffler condition if for any i P I, the family pui,jpXjqqjPI stabilizes, iďj `i.e. there exists npi˘q ě i such that

p@j ě npiqq ui,jpXjq “ ui,npiqpXnpiqq Ă Xi.

Remark 11.13. If the maps ui,j are all surjective, then tXiuiPI , tui,jui,jPI satisfies the Mittag-Leffler iďj condition. Conversely, assume that tXiuiPI , tui,jui,jPI satisfies` the Mittag-Leffler˘ condition. If i P I and iďj npiq ě i is such that j ě npiq ñ ui,jp`Xjq “ ui,npiqpXnpiq˘q “: Xi Ă Xi. If i ď j in I and x P Xj, let k P I be such that k ě npiq and k ě npjq: we can write x “ uj,kpyq with y P Xk, and ui,jpxq “ ui,kpyq P Xi (since r r k ě npiq). Moreover, if z P Xi, there exists z P Xk such that z “ ui,kpzq “ ui,jpuj,kpzqq P ui,jpXjq, which shows that the maps ui,j : Xj Ñ Xj induce surjective maps ui,j : Xj Ñ Xi. By functoriality, the inclusionsr X X induce an injective mapr lim X lim X . The latter is in fact an equality: if x limr X , then i Ă i ÐÝ i Ñ ÐÝp i p p iqiPI P ÐÝ i iPI iPI r r iPI xri “ ui,jpxjq P ui,jpXjq for all j ě i, hencer xi P Xi for all i P I.

Lemma 11.14. Assume that I is countable. Let tXiuiPI , tui,jui,jPI be an inverse system of nonempty r iďj sets satisfying the Mittag-Leffler condition. Then lim X . ÐÝ` i ‰ ∅ ˘ iPI

Proof. Write I “ tinunPZě0 . Then one can construct inductively a strictly increasing map ϕ: Zě0 Ñ Zě0 such that ϕp0q “ 0 and iϕpnq ě in and iϕpnq ě iϕpn´1q for all n P Zą0. Using notations of remark 11.13, we have Xin “ uin,iϕpmq pXiϕpmq q for some m " n. As the sets Xiϕpmq are nonempty, so are the sets Xin . As the transition maps of the inverse system tXiuiPI , tui,jui,jPI are surjective, we can find inductively a sequence r iďj r pξ q P lim tX u , tu ` u : choose˘ any ξ P X , and ξ , . . . , ξ being constructed, n nPZě0 ÐÝ iϕpnq nPZě0 iϕpnq,irϕpmq měn 0 0 0 n nPZě0 ` ˘ choose ξn`1 P Xiϕpn`r1q such that uiϕpnq,iϕpn`1q pξn`1q “ ξn. If i P I, letr xi “ ui,iϕpnq pξnq for n P Zě0 large enough so that i i . Then x lim X lim X , so the latter is nonempty ď ϕpnq p iqiPI P ÐÝ i “ ÐÝ i  r iPI iPI Remark 11.15. Some examples that show thatr the hypothesis are really necessary in the previous lemma. (1) Put I Z , X Z , and u : Z Z ; x x m n if n m. An element in X lim X “ ě0 n “ ě0 n,m ě0 Ñ ě0 ÞÑ ` ´ ď “ ÐÝ n n is thus a sequence pxnqnPZě0 such that xn “ xn`1 ` 1, so that xn “ x0 ´ n for all n P Zě0. Such sequences do not exist, so X “ ∅. x 1 (2) Put I “ Zě0, Xn “s0, 1r, and un,m : Xm Ñ Xn; x ÞÑ 2m´n . Then u0,npXnq “ 0, 2n , so that if 8 1 pxnqn Z P X “ lim Xn, we have x0 P 0, n “ . ‰ “ P ě0 ÐÝ 2 ∅ n n“0 (3) For each finite subset A Ă R, let XAŞbe‰ the set“ of injections A Ñ N. If A Ă B, the restriction provides a surjective map XB Ñ XA, so we get an inverse system (indexed by the finite subsets of R, partially ordered by the inclusion) with surjective transition maps. However, the inverse limit is the set of injections from R to N: it is empty (this example is due to Waterhouse). Proposition 11.16. Let

1 1 pfiqiPI pgiqi,jPI 2 2 0 Ñ tXiuiPI , tui,jui,jPI ÝÝÝÝÑ tXiuiPI , tui,jui,jPI ÝÝÝÝÝÑ tXi uiPI , tui,jui,jPI Ñ 0 iďj iďj iďj ` ˘ ` ˘ ` ˘ be an exact sequence of inverse systems indexed by I in ModR (where R is a commutative ring). Assume 1 1 that I is countable and that tXiuiPI , tui,jui,jPI has the Mittag-Leffler property. Then the sequence iďj ` f˘ g 0 lim X1 lim X lim X2 0 Ñ ÐÝ i ÝÑ ÐÝ i ÝÑ ÐÝ i Ñ iPI iPI iPI is exact. Number theory 39

Proof. By proposition 11.10, it is enough to show the surjectivity of g. Let x2 “ px2 q P lim X2. For n nPZě0 ÐÝ n n ´1 2 i P I, put Ei “ gi ptxi uq Ă Xi: the set Ei is nonempty since gi is surjective. If j ě i in I and ξ P Ej, 2 2 2 2 then gipui,jpξqq “ ui,jpgjpξqq “ ui,jpxj q “ xi so that ui,jpξq P Ei. This implies that pEi, ui,j|Ej qi,jPI is a sub-inverse system of X , u : we have an inclusion E : lim E lim X , and g x x2 for any p i i,jqi,jPI “ ÐÝ i Ă ÐÝ i p q “ iPI iPI x lim E . We have thus to show that E is nonempty. As I is countable, it is enough to check that the P ÐÝ i iPI inverse system tEiuiPI , tui,j|Ej ui,jPI satisfies the Mittag-Leffler condition (cf lemma 11.14). iďj 1 1 As tXiuiPI , tu`i,jui,jPI has the Mittag-Leffler˘ property, for each i P I, there exists npiq ě i in I such that iďj 2 2 ui,j`pXj q “ ui,npiqpXnpi˘qq for all j ě npiq. Let j ě npiq. We have ui,jpEjq Ă ui,npiqpEnpiqq. Conversely, let 2 2 2 2 ξ P Enpiq. If η is any element in Ej, we have gnpiqpunpiq,jpηqq “ unpiq,jpgjpηqq “ unpiq,jpxj q “ xnpiq “ gnpiqpξq, 1 so that ξ ´ unpiq,jpηq P Kerpgnpiqq “ Impfnpiqq: we can write ξ ´ unpiq,jpηq “ fnpiqpλq with λ P Xnpiq. We have 1 1 1 1 1 1 ui,npiqpξq “ ui,jpηq`ui,npiqpfnpiqpλqq “ ui,jpηq`fipui,npiqpλqq. As ui,npiqpλq P ui,npiqpXnpiqq “ ui,jpXjq, there 1 1 1 1 exists µ P Xj such that ui,npiqpλq “ ui,jpµq, hence ui,npiqpξq “ ui,jpηq ` fipui,jpµqq “ ui,jpη ` fjpµqq. As

η ` fjpµq P Ej, this shows that ui,npiqpξq P ui,jpEjq, showing that the inverse system tEiuiPI , tui,j|Ej ui,jPI iďj satisfies the Mittag-Leffler condition indeed. ` ˘ 11.17. Profinite groups. Definition 11.18. A inverse limit of finite sets (resp. groups) is called a profinite set (resp. a profinite group). We endow these finite sets with the discrete topology, their product with the product topology and their inverse limit with the induced topology. Let p be a prime integer. A pro-p-group is an inverse limit of p-groups. Proposition 11.19. Profinite sets are compact.

Proof. Let tXiuiPI , tui,jui,jPI be a inverse system of finite sets. Being finite, each Xi is compact: by iďj Tychonoff’s` theorem, the product˘ Xi is compact as well. If J Ă I is finite, let πJ : Xi Ñ Xi iPI iPI iPJ be the projection on factors of index J, and lim X the inverse limit of X , u . Then ś P ÐÝ j t jujPJ t i,jśui,jPJ ś J iďj π lim X lim X , and lim X π´1 lim X . Since X is finite, lim`X is closed, so π´1˘ lim X J pÐÝ iq Ă ÐÝ j ÐÝ i “ J pÐÝ jq j ÐÝ j J pÐÝ jq I J I JĂI J J J J J finite is closed in X (by definition of theŞ product topology). Beingś an intersection of closed subsets, lim X is i ÐÝ i iPI I (18) closed in śXj, hence compact .  J ś Remark 11.20. If G is a profinite group and H Ă G an open subgroup, then H is closed as well: GzH “ gH is a union of open subsets, so it is open. gRH Ş Proposition 11.21. Let G be a topological group. Then G is profinite if and only if it is Hausdorff, compact, and admits a basis of neighborhoods of 1 consisting of normal subgroups. Proof. Assume G is profinite: G lim G . Since G is separated (each G is), so is G. Moreover G is “ ÐÝ i i i I I compact thanks to the previous proposition. Finally,ś a basis of neighborhoods of 1 is given by tKerpπiquiPI where πi is the projection to the factor of index i, which consists of normal subgroups. Conversely, assume G is Hausdorff, compact, and admits a basis of neighborhoods of 1 consisting of normal subgroups. Let tNiuiPI be the family of open normal subgroups. As G is compact, the quotient Gi :“ G{Ni is finite for all i P I. Write i ď j if Ni Ą Nj, so that I becomes a directed set (an upper bound of Ni and Nj is given by Ni X Nj). The family tGiuiPI is then a inverse system. The canonical maps πi : G Ñ Gi induce a canonical morphism ψ : G lim G . Its kernel is N 1 (since N is a basis of neighborhoods of Ñ ÐÝ i i “ t u t iuiPI I I 1), so ψ is injective. A sub-basis of neighborhoods ofŞ1 in Gi is given by US “ Gi ˆ t1u, where S I iPIzS iPS ś ś ś (18) ´1 Another way of formulating it: X “ pπi, ui,j ˝ πj q p∆P q where P “ Xk and ∆P “ tpx, xq ; x P P u is the diagonal iďj kPI ´1 of P . As ∆P is closed in P ˆ P , the setsŞ pπi, ui,j ˝ πj q p∆P q are closed inśP for all i ď j, so that X is closed in P . As P is compact (by Tychonof’s theorem), this shows that X is compact as well. 40 Number theory

´1 runs through the finite subsets of I. As ψ pUSq “ Ni is open, the map ψ is continuous. In particular, iPS as G is compact, ψ G is compact hence closed inside lim G . In fact, ψ is surjective, because ψ G is dense p q ŞÐÝ i p q I in lim G . Indeed, let g g lim G and S a finite subset of I; let k I be such that N N , and ÐÝ i “ p iqiPI P ÐÝ i P k “ i I I iPS g G a lift of g G G N . Then g g mod N for all i S, so ψ g g U lim G .Ş As ψ is a P k P k “ { k i “ i P p q P p S X ÐÝ iq I continuous and G is compact, it maps closed subsets to closed subsets: it is open. This shows that ψ is an isomorphism and a homeomorphism.  Remark 11.22. If G is any group, its profinite completion is the natural map G lim G N. In the Ñ ÐÝ { NEG rG:Nsă8 previous proof, we have seen that G is profinite if and only if this is an isomorphism and a homeomorphism. Example 11.23. (1) If p is a prime number, Z lim Z pn Z. p » ÐÝ { nPNą0

(2) If we endow Ną0 with the order given by n ď m ô n | m, then tZ {n ZunPNą0 is an inverse system, whose inverse limit is denoted by Z. This is the profinite completion of Z. Remark 11.24. The maps Z Z and Z Z are injective, but are not isomorphisms: their image is only Ñ p Ñ n n dense (because Z {p Z Ñ Zp {p Z and Z {n Z Ñ Z{nZ are isomorphisms for all n P Ną0). p „ Example 11.25. The natural map Z Ñ Zp is anp isomorphismp and a homeomorphism. This follows from pPP the Chinese remainder theorem. p ś 11.26. Completion of a ring with respect to an ideal. References for this section are [2, §8] and [3,II §5]. Let I Ă A be an ideal. Definition 11.27. Let M be an A-module. n (1) The I-adic topology on M is the topology for which tI MunPZą0 is a basis of neighborhoods of 0. (2) The I-adic completion of M is M lim M InM. The A-module M is I-adically complete when the “ ÐÝ { nPZą0 natural map M Ñ M is bijective. x 8 Remark 11.28. (1)x The I-adic topology on M is separated if and only if InM “ t0u. n“1 (2) The addition M ˆ M Ñ M and the multiplication A ˆ M Ñ M are continuousŞ (19). In particular, the ring operations are continuous on A for the I-adic topology. (3) Each InM is open in M, hence also closed since its complement in M is the open pm ` InMq: mRInM the quotient module M{InM is discrete. Ť (4) M is an A-module. n n n n (5) If f : M1 Ñ M2 is an A-linear map, then fpI M1q Ă I M2, so f induces a map M1{I M1 Ñ M2{I M2 for allx n P Ząp0, hence a map f : M1 Ñ M2 between the I-adic completions. n n (6) In general, M is I-adically separated, but if n P Zą0, the natural map M{I M Ñ M{I M may not be p x x an isomorphism, and the map M Ñ M not an isomorphism, i.e. M may not be complete for the I-adic topology. x x x x x x Example 11.29. Assume m1,..., mr are pairwise distinct maximal ideals in A and e1, . . . , er P Zą0. Put e1 er I “ m1 ¨ ¨ ¨ mr . Denote by Ami the completion of the local ring Ami with respect to the miAmi -topology. The natural map p r A Ñ Ami i“1 à is an isomorphism. Indeed, for all n P Zą0, thep natural mapp r n nei A{I A Ñ A{mi i“1 à

(19)If x, x1, y, y1 P M are such that x ´ x1, y ´ y1 P InM, then px ` yq ´ px1 ` y1q P InM; moreover, if a, a1 P A are such that a ´ a1 P In, then ax ´ a1x1 “ apx ´ x1q ` pa ´ a1qx1 P InM. Number theory 41 is an isomorphism (by the Chinese remainder theorem, cf 1.14). Passing to inverse limits provides an r isomorphism A Ñ lim A{mnei . Now lim A{mnei “ lim A{mn, and the natural map A{mn Ñ A {mnA ÐÝ i ÐÝ i ÐÝ i i mi i mi i“1 n n n is an isomorphism. p À Lemma 11.30. An A-module M is separated and complete if and only if Cauchy sequences in M converge.

Proof. The A-module M is separated and complete if and only if for any sequence pmkqkPZą0 such that k k p@k P Zą0q mk`1 ´ mk P I M, there exists a unique m P M such that p@k P Zą0q mk ” m mod I M. This certainly holds if Cauchy sequences converge. Conversely, assume that M is separated and complete and let pxiqiPZą0 be a Cauchy sequence in M. If k P Zą0, there exists ϕpkq P Zą0 such that i, j ě ϕpkq ñ k xi ´ xj P I M. We can assume that the map ϕ is strictly increasing. Put mk “ xϕpkq P M: we have k k mk`1 ´ mk P I M for all k P Zą0, so there is a m P M such that mk ” m mod I M for all k P Zą0. If k k i ě ϕpkq, we have thus xi ´ mϕpkq, mϕpkq ´ m P I M, whence xi ” m mod I M, showing that pxiqiPZą0 converges to m.  Corollary 11.31. If M is an A-module which is separated and complete for the I-adic topology, then a 8 series mn converges in M if and only if its general term mn tends towards 0. n“0

Theorem 11.32.ř (Hensel’s lemma). Let A be a local ring, m Ă A its maximal ideal and k “ A{m its residue field. Assume that A is m-adically separated and complete, and let F P ArXs be a monic polynomial. Assume there are monic polynomials g, h P krXs such that gcdpg, hq “ 1 and gh “ F , where F is the image of F in krXs. Then there exist monic polynomials F,G P ArXs such that F “ GH, and whose images in krXs are g and h respectively.

Proof. Note that the assumption imply that degpgq`degphq “ d :“ degpP q. Let i P t0, . . . , d´1u. As gcdpg, hq “ 1, there exist ui, vi P krXs such i (20) that gui ´hvi “ X . Replacing ui by its remainder modulo h and vi by its remainder modulo g, we may further assume that degpuiq ă degphq and degpviq ă degpgq. Choose lifts Ui,Vi P ArXs of ui and vi respectively such that degpUiq “ degpuiq and degpViq “ degpviq. Let G1,H1 P ArXs be monic lifts of g and h respectively (so that G1 “ g and H 1 “ h). We construct by induction monic polynomials Gn,Hn P ArXs such that n GnHn ” P mod m rXs n (˚) $Gn`1 ” Gn mod m rXs n &’Hn`1 ” Hn mod m rXs ’ for all n P Zą0. Let n P Zą0 be such that tGiu1ďiďn and% tHiu1ďiďn have been constructed. Conditions (˚) imply that Gn “ g and H n “ h, (21) i and that degpGnq “ degpgq and degpHnq “ degphq. This implies in particular that degpGnUi ´ HnViq ă d and that GnUi ´ HnVi ” X d´1 d´1 i n n`1 mod mrXs. Write P ´ GnHn “ αiX with α0, . . . , αd´1 P m : we have P ´ GnHn ” αipGnUi ´ HnViq mod m rXs. Put i“0 i“0 ř ř d´1 Gn`1 “ Gn ´ αiVi $ i“0 d 1 ’ ř´ &’Hn`1 “ Hn ` αiUi i“0 ’ ř n ’n so that Gn`1 ” Gn mod m rXs and Hn`1 ” Hn mod m% rXs. We have

d´1 2n Gn`1Hn`1 ” GnHn ` αipGnUi ´ HnViq mod m rXs i“0 ÿ ” P mod mn`1rXs

(as n ` 1 ď 2n), which completes the construction of the sequences pG q and pH q . As A is separated and complete for the n nPZą0 n nPZą0 m-adic topology, these sequences converge in ArXs (note that both are given by d sequences of coefficients): denote by G and H their limits. By construction we have F “ GH.  From now on, A is assumed to be noetherian. 8 Notation. ‚ Put A “ In: this is naturally an A-algebra (the product of x in the factor In with y in the n“0 factor Im is xy in the factorÀ In`m). As I is of finite type, so is A as an A-algebra: it is noetherian. ‚ More generally, let M be an A-module endowed with a decreasing filtration, i.e. a decreasing sequence of sub-A-modules pMnqnPZě0 such that IMn Ă Mn`1 for all n P Zě0. The associated graded group is 8 n M “ Mn. It is naturally endowed with an A-module structure (the product of a in the factor I with n“0 m in theÀ factor Mm is am in the factor Mn`m). Lemma 11.33. Assume M is of finite type over A. The following properties are equivalent:

(i) Mn`1 “ IMn for n sufficiently large; n (ii) there exists c P Zě0 such that Mn`c “ I Mc for all n P Zě0; (iii) M is a finitely generated A-module.

(20) i Let indeed ui and vi be these remainders: we have ui “ ui ` hδi with δi P krXs, so that gpui ` hδiq ´ hvi “ X , i.e. i i gui ´ hpvi ´ gδiq “ X . This implies that degphpvi ´ gδiqq “ degpgui ´ X q ă d, thus degpvi ´ gδiq ă degpgq, i.e. i vi ´ gδi “ vri, and grui ´ hvi “ X . r r (21) Ther degree of a monic polynomial is equal to that of its reduction modulor m. r r r 42 Number theory

c Proof. (i)ô(ii) is trivial. If (ii) holds then M is generated by Mi, so that we have (iii). Conversely, i“0 assume (iii): the A-module M can be generated by finitely many elementsř x1, . . . , xr, with xi homogeneous, i.e. belonging to some factor Mni Ă M for i P t1, . . . , ru. Then Mn`1 “ IMn for all n ě c :“ max ni. 1ďiďr 

Theorem 11.34. (Artin-Rees lemma). Let M be an A-module of finite type. If N Ă M is a submodule, n`c n c there exists c P Zě0 such that for every n P Zě0, we have I M X N “ I pI M X Nq.

n Proof. For n P Zě0, put Mn “ I M and Nn “ Mn X N: we have N Ă M . As A is noetherian and M finitely generated as an A-module (by lemma 11.33), so is N : by lemma 11.33 again, there exists c P Zě0 n n`c n c such that Nn`c “ I Nc i.e. I M X N “ I pI M X Nq. 

Remark 11.35. This theorem essentially says that the I-adic topology on N coincides with the topology induced on N by the I-adic topology on M.

Corollary 11.36. Let 0 Ñ M 1 Ñ M Ñ M 2 Ñ 0 be an exact sequence of A-modules of finite type. The sequence 0 Ñ M 1 Ñ M Ñ M 2 Ñ 0 is exact.

Proof. By rightx exactnessx ofx the tensor product (cf proposition 7.5), the sequence M 1{InM 1 Ñ M{InM Ñ M 2{InM 2 Ñ 0

n n is exact (recall that M{I M » M bA pA{I q) for all n P Zě0. On the other hand, there exists c P Zě0 such that InM X M 1 “ In´cpIcM X M 1q for integers n ě c (Artin-Rees lemma, cf theorem 11.34). This implies that for n P Zěc, we have InM 1 Ă InM X M 1 “ In´cpIcM X M 1q Ă In´cM 1 and the sequence 0 Ñ M 1{pIn´cpIcM X M 1qq Ñ M{InM Ñ M 2{InM 2 Ñ 0 1 n´c c 1 is exact. This gives an exact sequence of inverse systems. The inverse system pM {pI pI M X M qqqnPZą0 has the Mittag-Leffler property (the transition maps are surjective): by proposition 11.16, the sequence

0 lim M 1 In´c IcM M 1 M M 2 0 Ñ ÐÝ {p p X qq Ñ Ñ Ñ n is exact. Moreover, the surjective maps x x

M 1{In´cM 1 Ñ I1{pIn´cpIcM X M 1qq Ñ M 1{InM 1 provide surjective maps M 1 lim M 1 In´c IcM M 1 M 1 (here again the surjectivity follows from Ñ ÐÝ {p p X qq Ñ n the Mittag-Leffler condition satisfied by the kernels of these maps), whose composite is the identity: we x x have lim M 1 In´c IcM M 1 „ M 1 hence the result. ÐÝ {p p X qq Ñ  n

x „ Corollary 11.37. Let M be an A-module of finite type. Then A bA M Ñ M.

Proof. This is obvious when M is free. In the general case, letpL1 Ñ L0 Ñx M Ñ 0 be an exact sequence where L0 and L1 are free of finite rank (such a sequence exists since M in of finite type and A noetherian). The exactness of completion on short exact sequences of A-modules of finite type imply that the sequence L1 Ñ L0 Ñ M Ñ 0 is exact. We thus have the following commutative diagram with exact rows p p x A bA L1 / A bA L0 / A bA M / 0 φ φ φ 1  0   p p p L1 / L0 / M / 0

As φ0 and φ1 are isomorphisms, sop is φ. p x  Corollary 11.38. A is flat over A.

Proof. This followsp from corollaries 11.36 and 11.37.  Number theory 43

12. Appendix: Zorn’s lemma The axiom of choice (that we assume) is equivalent to the following: Theorem 12.1. A partially ordered set in which every chain(22) has an upper bound contains at least one maximal element. Remark 12.2. Considering opposite orders, we also have the dual statement: a partially ordered set in which every chain has an lower bound contains at least one minimal element.

13. Exercises ? 1`i 19 The following two exercises show that the ring Z 2 is not euclidean, though a PID. 2 Exercise 13.1. Put A “ Zrζs where ζ ´ ζ ` 5 “ 0.“ We denote‰ by N the norm of the number field Qrζs. (1) Compute Npx ` yζq for x, y P Q an determine Zrζsˆ. (2) Let a, b P Zrζszt0u. Show that there exist q, r P Zrζs such that: (r “ 0 or Nprq ă Npbq) and (a “ bq ` r or 2a “ bq ` r). (3) Show that the ideal 2 Zrζs is maximal in A. (4) Show that A is a PID.

Exercise 13.2. Let A be an integral domain which is not a field. We construct (by induction on n P Zě0) a sequence of subsets of A by: A0 “ t0u and An`1 “ An Y tx P A ; A “ xA ` Anu for all n P Zě0. For 8 x P An, we put φpxq “ inftn P Zě0 ; x P Anu. n“0 Ť 8 (1) Assume that A “ An. Show that A euclidean for the euclidean function φ. n“0 (2) Assume that A is euclideanŤ for a euclidean function ψ : Azt0u Ñ Zą0. Show that: (i) φpxq ď ψpxq for all x P An; nPN 8 Ť (ii) A “ An [Hint: reductio ad absurdum using (i)); n“0 (iii) A is euclideanŤ for the euclidean function φ; (iv) if a divides b in A, then φpaq ď φpbq; (v) there exists x P AzAˆ such that the restriction of the projection A Ñ A{xA to Aˆ Yt0u is surjective. (3) Determine φ in the following cases: A “ Z and A “ krXs (where k is a field). (4) Let A “ Zrζs Ă C where ζ2 ´ ζ ` 5 “ 0. 2 (i) Show that the equation z ´ z ` 5 “ 0 has no solution in F2 nor in F3. (ii) Deduce that A is not euclidean [Hint: reductio ad absurdum using (2-v)]. Exercise 13.3. Let A be a domain. (1) Show that if A is a UFD if and only if non-zero elements can be factored into a product of irreducible elements, and irreducible elements are prime in A. (2) Show that if A is noetherian, non-zero elements can be factored into a product of irreducible elements. (3) Give an example of non noetherian UFD. Exercise 13.4. Let A be a UFD, and S Ă A a multiplicative part. Show that S´1A is a UFD. Exercise 13.5. Let A be a ring, M an A-module of finite type and ϕ: M Ñ An a surjective morphism. Show that M “ N ‘ Kerpϕq, where N is a submodule of M isomorphic to An through ϕ. Show that Kerpϕq is of finite type. Exercise 13.6. Let A be an integral domain and M an A-module. Assume that M can be generated by n elements, and contains a submodule which is free of rank n. Show that M is free of rank n. Exercise 13.7. (1) Let L{K is a finite Galois extension with group G. Show that the natural map

L bK L Ñ L σPG à x b y ÞÑ pxσpyqqσPG is an isomorphism of L-algebras (for the left structure on the LHS, and the diagonal structure on the RHS). (2) More generally, let L{K be a finite separable extension, and F {K be any extension. Show that L bK F is isomorphic, as an F -algebra, to a finite product of separable extensions of F . (3) Is it still true when L{K is not assumed to be separable?

(22)I.e. a totally ordered subset. 44 Number theory

Exercise 13.8. (Nakayama’s lemma). Let A be a ring, I Ă A an ideal and M an A-module of finite type such that IM “ M. (1) Show that there exists an element a P A such that a ” 1 mod I and aM “ t0u. (2) Deduce that if I Ă radpAq, then M “ t0u. (3) Assume that A is local and denote by k its residue field. Show that if k bA M “ t0u, then M “ t0u. (4) Give a counter-example of (3) when M is not assumed to be of finite type. Exercise 13.9. (Nakayama’s lemma, continuation). Assume that A is local, with residue field k, and let M be an A-module of finite type, and N an A-module. (1) If N is of finite type over A and M bA N “ t0u, show that M “ t0u or N “ t0u. (2) Let f : N Ñ M be an A-linear map such that Idk bf : k bA N Ñ k bA M is surjective. Show that f is surjective. Exercise 13.10. Let A be a ring and I Ă A be an ideal of finite type such that I2 “ I. Show that I is generated by an element e P I such that e2 “ e.

Exercise 13.11. Let A be a ring, M an A-module of finite type and f P EndApMq a surjective endomorphism. Show that f is injective. Exercise 13.12. Let A be a ring. Show the following: (i) if An » Am then n “ m; (ii) if there exists a surjective A-linear map An Ñ Am then n ě m; (iii) [difficult] if there exists an injective A-linear map An Ñ Am, then m ď n.

Exercise 13.13. Let A be a ring. An A-module P is projective if the functor HomApP,.q is exact, i.e. whenever a sequence 0 Ñ M 1 Ñ M Ñ M 2 Ñ 0 is exact, so is the sequence 1 2 0 Ñ HomApP,M q Ñ HomApP,Mq Ñ HomApP,M q Ñ 0. (1) Show that a free module is projective. (2) Show that an A-module is projective if and only if it is a direct factor of a free module. Definition 13.14. Let A be a ring. An A-module M is of finite presentation if there exists an exact sequence L1 Ñ L Ñ M Ñ 0 i.e. if there exists a surjective A-linear map u: L Ñ M such that L is free of finite rank and Kerpuq of finite type. Being of finite presentation implies being of finite type, but the converse is false in general. It holds true when A is noetherian. Exercise 13.15. (Snake lemma). Let A be a commutative ring. (1) Assume there is a commutative diagram of A-modules

a b M 1 / M / M 2 / 0

u v w

 c  d  0 / N 1 / N / N 2 with exact rows. Show that there is an exact sequence of A-modules a b δ b d Kerpuq ÝÑ Kerpvq ÝÑ Kerpwq ÝÑ Cokerpuq ÝÑ Cokerpvq ÝÑ Cokerpwq. a b (2) Let 0 Ñ M 1 ÝÑ M Ñ M 2 ÝÑ 0 be an exact sequence of A-modules with M of finite type and M 2 of finite presentation. Show that M 1 is of finite type. Exercise 13.16. Let A be a local ring, with maximal ideal m and residue field k “ A{m. (1) Let 0 Ñ M 1 Ñ M Ñ M 2 Ñ 0 be an exact sequence of A-modules with M 2 flat over A. Show that the 1 2 sequence 0 Ñ k bA M Ñ k bA M Ñ k bA M Ñ 0 is exact. (2) Let M be an A-module. Show that the following are equivalent: (i) M is flat of finite presentation; (ii) M is free of finite rank. (in particular, when A is noetherian, then M is free of finite rank if and only if it is flat of finite type). (3) Deduce that an A-module is projective of finite type if and only if it is free of finite rank. Exercise 13.17. Let A be a ring and M an A-module. Show that the following are equivalent: Number theory 45

(i) M is projective of finite type over A; (ii) M is flat and finitely presented over A. Exercise 13.18. Let A be a local ring with maximal ideal m and k “ A{m its residue field. Let u: M Ñ N be an A-linear map such that M is of finite type, N is projective, and k b u: k bA M Ñ k bA N is injective. (1) Show that M is free of finite rank. (2) Show that u is left invertible (i.e. there exists an A-linear map v : N Ñ M such that v ˝ u “ IdM ).

Z 0 Exercise 13.19. Let R be a ring, M “ R ą and A “ EndApMq: this is a noncommutative ring. Use the maps

ϕ1 : M Ñ M; px1, x2,...q ÞÑ px1, x3, x5,...q

ϕ2 : M Ñ M; px1, x2,...q ÞÑ px2, x4, x6,...q

ψ1 : M Ñ M; px1, x2,...q ÞÑ px1, 0, x2, 0,...q

ψ2 : M Ñ M; px1, x2,...q ÞÑ p0, x1, 0, x2,...q to show that A2 » A (as left A-modules), so that the rank of a free module is not well defined in the non commutative setting. k k`1 Exercise 13.20. Let K be a field and A the sub-K-algebra of KrX,Y s generated by tX Y ukPZě0 . Show that ArXY s is included in a sub-A-module of KrX,Y s of finite type, but that XY is not integral over A. Exercise 13.21. Let A Ă B be a ring extension with A noetherian, x P Bˆ, and y P Arxs X Arx´1s. Show n that there exists n P Zě0 such that the sub-A-module M “ A ` Ax ` ¨ ¨ ¨ ` Ax Ă B is stable under multiplication by y, and that y is integral over A. Exercise 13.22. Let A be a domain and α P Azt0u. Assume that A{αA is reduced and that Arα´1s is integrally closed. Show that A is integrally closed.

Exercise 13.23. Let A Ñ B be an integral morphism of rings, p1 Ă p2 prime ideals in B such that p1 X A “ p2 X A. Show that p1 “ p2. Exercise 13.24. Let A be a ring, A Ă B a finitely generated integral extension, and p Ă A a prime ideal. Show that B has only a finite number of prime ideals lying over p.

Exercise 13.25. (1) Let pXn, ρnqnPZě0 be an inverse system of finite and non empty sets. Show that X lim X is non empty. [Hint: reduce to the case where the maps ρ are surjective.] “ ÐÝ n n n (2) Give an example of an inverse system (indexed by Zě0) of non empty sets whose inverse limit is empty. Exercise 13.26. Find examples where Artin-Rees lemma’s conclusion does not hold because one of its i assumptions is not fulfilled [Hint: try A “ QrX,Z,Y1,Y2,...s{xX ´ Z YiyiPZą0 for the non noetherian case.] Exercise 13.27. Let A be a ring, I Ă A and f : N Ñ M a surjective A-linear map. Show that the map induced on the I-adic completions f : N Ñ M is surjective. Deduce that if M is an A-module of finite type, the natural map A bA M Ñ M is surjective. p p x Exercise 13.28. Let A be a ring, I Ă A and M an A-module. Denote by M the I-adic completion of M. p x (1) Show that M is I-adically separated. (2) Show that the following are equivalent: x (i) the A-modulex M is I-adically complete; n n (ii) for all n P Zą0, the natural map M{I M Ñ M{I M is surjective; n n (iii) for all n P Zą0,x we have I M “ Kerpπnq where πn : M Ñ M{I M is the canonical map. x x (3) Let K be a field, A “ KrXisiPZą0 and I “ xXiyiPZą0 Ă A. Show that A is not I-adically complete. x n x n n (4) Assume that I is finitely generated. Show that I M “ KerpM Ñ M{I Mq “ I M for all n P Zě0 and that M is I-adically complete. p x x z Exercise 13.29. Let A be ring, I Ă A an ideal and M an A-module. (1) Showx that if A is I-adically separated and complete, then I Ă radpAq. (2) Show that if M is I-adically separated and complete and a P I, the multiplication by 1 ` a is an automorphism of M. Exercise 13.30. (Completion is not an exact functor.) Let K be a field, A “ KrXs, M “ ApZą0q 8 and C “ A{XnA. Show that the completion for the X-adic topology of the natural exact sequence n“1 0 Ñ M Ñ ÀM Ñ C Ñ 0 is not exact. 46 Number theory

Exercise 13.31. (Formal Nakayama’s lemma). Let A be a ring, I Ă A an ideal such that A is I-adically separated and complete, and M an A-module of finite type. (1) Show that if M “ IM, then M “ t0u. (2) Assume that f : M 1 Ñ M is an A-linear map such that f bpA{Iq is surjective. Show that f is surjective. Exercise 13.32. Let A be a ring, I Ă A an ideal and M an A-module. Assume that A is I-adically separated and complete, and that M is separated for the I-adic topology. Assume there are m1, . . . , mr P M whose images m1,..., mr P M{IM generate M{IM. Show that m1, . . . , mr generate the A-module M. Exercise 13.33. Let A be a noetherian local ring, with maximal ideal m and residue field k “ A{m. Show that the m-adic completion A of A is a local ring with maximal ideal mA, and residue field k. Exercise 13.34. Let A be a DVR, m its maximal ideal, and A the m-adic completion of A. Show that A is p p a DVR. p p Exercise 13.35. (Krull intersection theorem). Let A be a noetherian ring and I Ă A an ideal. 8 (1) Let M be an A-module of finite type and N “ InM. Then there exists a P A such that a ” 1 n“0 mod I and aN “ 0. Ş (2) If I Ă radpAq, then any A-module of finite type is I-adically separated, and its submodules are all closed. 8 (3) If A is a domain and I a proper ideal, then In “ t0u. n“0 Ş Exercise 13.36. Let A be a noetherian ring and I “ xξ1, . . . , ξny be an ideal. Let A be the I-adic completion of A. Then there is a isomorphism „ p ArrX1,...,Xnss{xX1 ´ ξ1,...,Xn ´ ξny Ñ A that maps Xi to ξi for all i P t1, . . . , nu. p Exercise 13.37. Let A be a noetherian ring and I,J Ă A ideals. Assume that A is both I-adically and J-adically separated and complete. Show that A is I ` J-adically separated and complete. Exercise 13.38. Let A be a noetherian ring and J Ă I Ă A ideals such that A is I-adically separated and complete. Show that A is also J-adically separated and complete.

Exercise 13.39. Let A be a ring, I “ xf1, . . . , fry Ă A a finitely generated ideal, M an A-module and M its I-adic completion. (1) Show that if M lim M f nM is surjective for each i 1, . . . , r , then M M is surjective. x Ñ ÐÝ { i P t u Ñ n (2) Let J Ă A be an ideal such that I Ă J. Show that if M is J-adically complete, then M is I-adically x complete.

References [1] M. Kashiwara & P. Schapira – Sheaves on manifolds, Grundlehren der Mathematischen Wissenschaften, vol. 292, Springer-Verlag, 1990. [2] H. Matsumura – Commutative ring theory, Cambridge university Press, 1986. [3] J.-P. Serre – Algèbre locale, multiplicités, Lecture notes in mathematics, vol. 11, Springer Verlag, 1965. [4] T. Stacks project authors – “The stacks project”, https://stacks.math.columbia.edu, 2018.

Institut de Mathématiques de Bordeaux, Université Bordeaux, 351, cours de la Libération, 33405 Talence, France