44 1.SEMINORMEDCOMMUTATIVEALGEBRA January 16, 2020, 10:31am 1.4. Seminormed modules Next we turn to the analogues of modules over a ring in commutative algebra. 1.4.1. Definition and examples. Let A be a seminormed ring. A seminormed A-module is an A-module M equipped with a seminorm satisfying am a m for all a A and m M; if equality always holds, then we say thatkM·kis faithfullyk seminormedkk k·.k Whenk the seminorm2 on2M is a norm (resp. complete norm), we say that M is a normed A-module (resp. Banach A-module). Example 1.4.1. A seminormed ring A is a seminormed A-module; it is faithfully seminormed i↵the seminorm on A is a semivaluation. Example 1.4.2. If Z is equipped with the usual norm , then a seminormed Z-module is |·|1 the same thing as a seminormed group. Similarly, if Z is equipped with the trivial norm 0,thena non-Archimedean seminormed Z-module is the same thing as a non-Archimedean seminormed|·| group. Example 1.4.3. Let A be a ring and M an A-module. Equip A with the trivial norm and M with a constant (semi)norm. Then M is a (semi)normed A-module. Example 1.4.4. If k is a valued field, then a seminormed (resp. normed) k-module V is called a seminormed (resp. normed) k-vector space. A seminormed k-vector space is always faithfully normed, see Exercise 1.4.3. A k-Banach space is a Banach k-module, that is, a complete normed k-vector space. Seminormed vector spaces will be studied in 1.10 and 1.12. § § A morphism of seminormed A-modules is a bounded homomorphism of A-modules. As in the case of seminormed groups and rings, there are two natural categories of seminormed rings, using bounded homorphisms and contractive homomorphisms as morphisms, respectively. Each category has several full subcategories in which the objects are normed or even Banach A-modules, and/or are non-Archimedean. Some properties of these categories are developed in the exercises. 1.4.2. Operations on seminormed modules. Operations on seminormed groups typically carry over to seminormed modules. These include products, sums and direct limits, as well as the separation and completion operations. The details are explored in the exercises. Here we content ourselves with a particular but important case: the separated completion. Let A be a seminormed ring and M a seminormed A-module. The separated completion Aˆ of A is a Banach ring, and the separated completion Mˆ of M is a Banach group. There is a natural way in which Mˆ becomes a Banach A-module, faithfully normed whenever M is faithfully seminormed, see Exercise 1.4.7 Using some of these operations, we can construct various seminormed A-modules from a semi- I normed ring A. For example, for any set I we define A as the product i I Ai,whereAi = A for all 2 (I) i I. This seminormed ring is denoted bI (A)in[BGR]. We similarly define A as the direct sum 2 (I) Q i I Ai. As before, we have two possible seminorms on A ; which one we are using will be specified 2 (I) Lwhenever needed. We can also consider the complete direct sum A := i I Ai, where we should specify which of the two seminorms with respect to which we are taking the2 separated completion. max d Lc When A is non-Archimedean, the complete direct sum i I Ai is denoted cI (A)in[BGR]. 2 1.4.3. Localization. In commutative algebra, localization is defined as follows. Let A be a Lc ring and S A a multiplicative set, i.e. 1 S and st S whenever s, t S. For any A-module M, ⇢ 1 2 2 2 the localization S− M is now defined as the set of equivalence classes (m, s), where m M, s S, and where (m, s) is equivalent to (n, t)i↵thereexistsu S such that u(sn tm) = 0. We2 write2 the 1 2 1 1 − 1 equivalence class of (m, s) as s− m; with this notation, s− m + t− n =(st)− (tm + sn). Note that 1 1 if 0 S,thenS− M = 0. When M = A, S− A is a ring. 2 1.4. SEMINORMED MODULES 45 *** CHECK THIS IS CORRECT *** Now let A be a seminormed ring. We say that a subset S A is multiplicative if 1 S, s > 0 for all s S 0, and st S, st = s t for all s, t S. ⇢ 2 k k 2 \ 2 k1 k k k·k k 2 Now consider a seminormed A-module M. Define a seminorm on S− M as follows. If 0 S,then 1 2 S− M = 0 is equipped with the zero seminorm. If 0 S,weset 62 1 1 1 s− m := inf n / t t− n = s− m, t =0 . k k {k k k k| k k6 } 1 We leave it to the reader to verify the following properties (see Exercise 1.4.14): S− A is a seminormed 1 1 ring and S− M is a seminormed S− A-module. Further, if A and M are non-Archimedean, so are 1 1 1 S− A and S− M.IfM is a faithfully seminormed A-module, so is S− A. An important instance of this construction is when A is a valued integral domain and S = A⇥ := 1 A 0 . In this case, K := S− A is the fraction field of A.IfM is a faithfully seminormed A-module, \{ } 1 then S− M becomes a faithfully seminormed K-module, i.e. a seminormed K-vector space, see 1.12. §n Another example is when f A⇥ satisfies f > 0 and is power multiplicative, that is f = f n > 0 for n 1. We can then2 take S = f nk nk 0 . Finally, if p is a prime ideal of A,thenwek k cank k consider the≥ localization of A with respect{ to| S≥:=}A p. Note that the requirement that S be multiplicative is somewhat restrictive, but at least it holds\ when A is valued. 1.4.4. Morphisms and bilinear morphisms. Let A be a seminormed ring, and M and N seminormed A-modules. Denote by HomA(M,N) the set of morphisms ': M N of seminormed A-modules. Recall that this means that ' is a bounded homomorphism of A-modules.! As before, write ' for the smallest nonnegative constant such that '(m) ' m for all m M.Then k k k kk k·k k 2 HomA(M,N) is a seminormed A-module. Now let L be another seminormed A-module. We say that an A-bilinear map Φ: L M N is bounded if there exists C>0 such that Φ(l, m) C l m for l L, m M. The smallest⇥ ! such 2 k k·k k 2 2 C is denoted by Φ . This turns HomA(L, M; N), the set of bounded A-bilinear homomorphisms k k 2 Φ: L M N, into a seminormed A-module. We say that Φ HomA(L, M; N)iscontractive if Φ ⇥1. ! 2 k k 2 Various properties of HomA(M,N) and HomA(L, M; N) are explored in the exercises. 1.4.5. Nakayama’s Lemma. The following result can be viewed as a version of Nakayama’s Lemma. Lemma 1.4.5. Let A be a non-Archimedean Banach ring, and let M 0 M be non-Archimedean ⇢ n normed A-modules. Assume there exist elements m ,...,m M such that M = M 0 + A◦◦m . 1 n 2 i=1 i Then M 0 = M. P Recall that A◦◦ = ⇢<1 ,where⇢ is the spectral radius seminorm. To explain the relationship to the usual version of{ Nakayama’s} Lemma, suppose A is a ring and M an A-module. Equip A with the trivial norm and M with a constant nonzero seminorm. Then A is a Banach ring, and A◦◦ = rad(A) is the nilradical of A. The lemma asserts that if M = M 0 + rad(A)M,thenM 0 = M. At least if rad(A) is replaced by the Jacobson radical of A, this is a version of Nakayama’s Lemma, see [Sta, Tag 07RC]. *** Explain better, see also [Eis95] *** Proof. It suffices to prove that mi M 0 for 1 i n. By assumption there exist elements 2 n m0 M,1 i n, and c A◦◦,1 i, j n, such that m = m0 + c m for 1 i n. i 2 ij 2 i i j=1 ij j Write this as a matrix equation P m0 =(I C)m. − The entries cij of the matrix C all satisfy ⇢(cij) < 1. Since A is complete and non-Archimedean, n it follows (see Exercise 1.2.30) that the matrix I C is invertible. Since m0 (M 0) , we get 1 n − 2 m =(I C)− m0 (M 0) , that is, m M 0 for 1 i n. − 2 i 2 ⇤ 46 1.SEMINORMEDCOMMUTATIVEALGEBRA Exercises for Section 1.4 (1) Let A be a seminormed ring. For any n 1, equip An with the product seminorm. (a) Show that any A-linear map An A≥m is bounded. (b) Show that any A-linear bijection!An An is isometric. (2) Let A be a ring equipped with the zero norm,! and M a seminormed A-module. Prove that the seminorm on M is the zero seminorm.