MAT 4200 Commutative Algebra Part 3, Chain Conditions, Noetherian Rings and Modules (Chapt

MAT 4200 Commutative Algebra Part 3, Chain Conditions, Noetherian Rings and Modules (Chapt

MAT 4200 Commutative algebra Part 3, Chain conditions, Noetherian rings and modules (Chapt. 9) October, 1, - October, 8, 2020 1 / 58 October, 1, 2020 We define Noetherian and Artinian modules and rings and look at some basic properties. 9.1-9.2 2 / 58 October, 1, 2020 Definition An A-module M is said to be finitely generated if there exists elements m1;:::;mr in M such that any element m 2 M can be written as a linear combination m = a1m1 + :::ar mr of the mi 's with coefficients ai in A. Notice that the coefficients ai are not unique in general; there might be non-trivial relations between the mi 's, and for most modules this occurs for any set generators. The set of generators fmi g gives rise to a surjective map Ar ! M ! 0 which sends (ai ) to ∑i ai mi . 3 / 58 October, 1, 2020 Given a ring A and an A-module M. Definition An ascending chain of submodules is a (countable) chain of submodules Mi shaped like M0 ⊆ M1 ⊆ :::: ⊆ Mi ⊆ Mi+1 ⊆ :::; or phrased differently, a totally ordered countable subset of I (M). In similar fashion, a descending chain is one shaped like ::: ⊆ Mi+1 ⊆ Mi ⊆ ::: ⊆ M1 ⊆ M0: 4 / 58 October, 1, 2020 Definition A chain is said to be eventually constant if there is an index i0 so that the members of the chain are equal from i0 and on; that is, Mi = Mj for i;j ≥ i0. Common usage is also to say the chain stabilizes at i0. M0 ⊆ M1 ⊆ :::: ⊆ Mi0 = Mi0+1 = Mi0+2 = :::; 5 / 58 October, 1, 2020 Example Given an ideal a in the ring A the powers ai of a form a descending chain ::: ⊆ ai+1 ⊆ ai ⊆ ::: ⊆ a2 ⊆ a For an A-module M we have a descending chain of submodules; ::: ⊆ ai+1M ⊆ ai M ⊆ ::: ⊆ a2M ⊆ aM 6 / 58 October, 1, 2020 Example Concider the abeiain group Q (which is a Z-module). For every −i natural number i one has the submodule of Q generated by p ; −i −i that is, the subgroup Z · p whose elements are of the form np . Then there is an ascending chain −1 −i −(i+1) Z ⊂ Z · p ⊂ ::: ⊂ Z · p ⊂ Z · p ⊂ ::: 7 / 58 October, 1, 2020 Definition The A-module M is Noetherian if every ascending chain of submodules is eventually constant. Definition The A-module M is Artinian if every descending chain is eventually constant. 8 / 58 October, 1, 2020 Definition i) A ring A is called Noetherian if it is a Noetherian as module over itself. ii) It is called Artinian if it is an Artinian as module over itself. The submodules of A are precisely the ideals, so A being Noetherian amounts to ideals of A satisfying the ACC, and similarly, A is Artinian precisely when the ideals comply with the DCC. 9 / 58 October, 1, 2020 Proposition (Main Theorem of Noetherian modules) Let A be a ring and let M be a module over A. The following three statements are equivalent. i) M is Noetherian. ii) Every non-empty set of submodules of M has a maximal member; iii) Every submodule of M is finitely generated over A; 10 / 58 October, 1, 2020 Proof. i) ) ii) Assume that M is Noetherian and let Σ be a non-empty set of submodules. Assuming there is no maximal elements in Σ, one proves easily, by induction on the length, that every finite chain in Σ can be strictly extended upwards, which is not the case since M is Noetherian. ii)) iii) Suppose that every non-empty set of submodules has a maximal element, and let N be a submodule. We shall see that N is finitely generated. Let Σ be the set of submodules of N finitely generated over A. It is clearly non-empty (Am ⊆ N for any m 2 N) and hence has a maximal element N0. Let m 2 N be any element. The module Am + N0 is finitely generated and contained in N, and by the maximality of N0, it follows that m 2 N0. Hence N = N0, and N is finitely generated. 11 / 58 October, 1, 2020 Proof. [continues] iii)) i) Assume that all submodules of M are finitely generated, and let M0 ⊆ M1 ⊆ :::: ⊆ Mi ⊆ Mi+1 ⊆ ::: S be an ascending chain. The union N = i Mi is finitely generated, say by m1;:::;mr . Each mj lies in some Mnj and hence they all lies in Mn with v = maxj nj . Therefore N = Mn , and the chain is constant above n. 12 / 58 October, 1, 2020 Proposition Let M0, M and M00 be A-modules that fit in the short exact sequence b 0 ! M0 −! M −! M00 ! 0 Then M is Noetherian (respectively Artinian) if and only if M0 and M00 are. Remember that f(f −1(x)) = x if f is surjective f −1(f(x)) = x if f is injective 13 / 58 October, 1, 2020 Proof. ) Every chain in M0 is a chain in M, and if M is Noetherian, so is M0. A chain in M00 can be lifted to M and surjectivity of b implies b(b −1(N)) = N for any submodule N ⊆ M00. It follows that M00 is Noetherian as well. ( Assume that the two extreme modules M0 and M00 are Noetherian and let fMi gi be a chain in M. The chain 0 0 0 fMi \ M g is eventually constant, hence Mi \ M = Mj \ M for i;j >> 0. 00 00 Mapping the Mi 's into M gives the chain fb(Mi )gi in M and since M00 is Noetherian by assumption, it is eventually constant. Hence b(Mi ) = b(Mj ) for i;j >> 0. This gives 0 0 Mi =Mi \ M = Mj =Mj \ M and hence Mi = Mj . 14 / 58 October, 1, 2020 Proposition Let S be a multiplicative set in the ring A and let M be an A-module. If M is Noetherian, respectively Artinian, the localized module MS is Noetherian, respectively Artinian, as well. Proof. −1 −1 −1 m 1. Let N ⊆ S M. Observ that S (i (N)) = N. In fact, since 1 2 N m is equivalent to s 2 N for any s 2 S, we have −1 −1 −1 m S (i (N)) = S fm 2 M j 1 2 Ng m m = f s j 1 2 Ng = N −1 2. Any chain fNi g in S M, whether ascending or descending, induces −1 a chain fi (Ni )g in M, and if this chain stabilizes, say −1 −1 i (Ni ) = i (Nj ) for i;j ≥ i0, it holds true that −1 −1 −1 −1 Ni = S (i Ni ) = S (i Nj ) = Nj , and the original chain stabilizes at i0 as well. 15 / 58 October, 1, 2020 Example (Vector spaces) A vector space V over a field k is Noetherian if and only if it is of finite dimension. Indeed, if V is of finite dimension it is the direct sum of finitely many copies of k, hence Noetherian. If V is not of finite dimension one may find an infinite set v1;:::;vi ;::: of linearly independent vectors, and the subspaces Vi =< v1;:::;vi > will form a strictly ascending chain of subspaces; hence V is not Noetherian. A similar argument shows that neither is V Artinian: The spaces Wi =< vi ;vi+1 ··· > form a strictly decreasing chain of subspaces. 16 / 58 October, 1, 2020 Proposition (Main Theorem of Noetherian rings) Let A be a ring. The following three statements are equivalent. i) A is Noetherian. ii) Every non-empty family of ideals in A has a maximal element; iii) Every ideal in A is finitely generated. 17 / 58 October, 1, 2020 Proposition Let A be a noetherian ring and M an A-module. Then M is Noetherian if and only if M is finitely generated. Proof. A finitely generated A-module M can be realized as the quotient of a finite direct sum nA of n copies of A. When A is Noetherian, it follows that nA is Noetherian; indeed, one obtains nA by successive extensions of A by itself. All quotients of nA, in particular M, will be Noetherian. 18 / 58 October, 1, 2020 Example An example of a non-noetherian ring is the ring A[x1;x2;:::] of polynomials in infinitely many variables over any ring A. The chain of ideals (x1) ⊂ (x1;x2) ⊂ ::: ⊂ (x1;x2;:::;xi ) ⊂ ::: does obviously not stabilize. 19 / 58 October, 1, 2020 Example One might be mislead by the previous example to belive that non-noetherian rings are monstrously big. There are, however, non-noetherian rings contained in the polynomial ring Q[x]. The −1 simplest example is even a subring of Z[p ][x] where p is a natural number greater than one; it is formed by those polynomials in −1 Z[p ][x] taking an integral value at zero. In this ring A one has the following ascending chain of ideals (p−1x) ⊂ (p−2x) ⊂ ::: ⊂ (p−i x) ⊂ ::: which does not stabilize. Indeed, if p−(i+1) 2 (xp−1), one has xp−(i+1) = P(x)xp−i for some polynomials P(x) 2 A. Cancelling xp−i gives us p−1 = P(x), and this is a contradiction since P(0) 2 Z. 20 / 58 October, 1, 2020 Proposition Assume that M is a module over A. If M is Noetherian, then A=Ann(M) is Noetherian as well. Proof. Let x1;:::;xr be generators for M, and consider the map f : A ! rM that sends x to the tuple (x · x1;:::;x · xr ).

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