Lecture 23 Notes: Noetherian Rings

Lecture 23: Noetherian Rings 23 Noetherian Rings Definition 23.1 A ring R is Noetherian if every ideal in R is finitely generated. In other words, every quotient can be obtained by imposing a finite number of relations. 23.1 Submodules over Noetherian Rings The first important property of Noetherian rings is the following: Proposition 23.2 A ring R is Noetherian if and only if every submodule in a finitely generated R-module is itself finitely generated. Corollary 23.3 If R is Noetherian, then every finitely generated module is finitely presented. Before we prove this, we’ll look at a few observations. Lemma 23.4 If we have a surjective homomorphism φ : M → N of R-modules, then: 1. If M is finitely generated, then N is also finitely generated. 2. If N is finitely generated, and K = ker(φ) is also finitely generated, then M is also finitely generated. Note 23.5 This is one place where the intuition from linear algebra is useful: it’s helpful to think about the case where R is a field, so being finitely generated is equivalent to being finite-dimensional. In this case, we know more precisely that dim(M ) = dim(N ) + dim(K). We won’t get information this precise in the general case, but the proof of finiteness is similar. Proof of Lemma 23.4. The first part is obvious — take any set of generators m1 , . . . , mn for M . Then their images φ(m1 ), . . . , φ(mn ) must generate N . For the second part, let k1 , . . . , ka be a set of generators for K, and n1 , . . . , nb a set of generators for N . Now pick ñ1 , . . . , ñb such that φ(ñi ) = ni (which we can do by surjectivity). Now we claim that the ki and ñi generate M : given any x ∈ M , we can find r1 , . . . , rb in R such that φ(x) = r1 n1 + · · · + rb nb (since the ni generate N ). So then we have φ(x − r1 ñ1 − · · · − rb ñb ) = φ(x) − r1 n1 − · · · − rb nb = 0, P P P which means x − ri ñi is in K. So we can express x − ri ñi = sj kj , which gives an expression for x as a linear combination of the ñi and kj . Proof of Proposition 23.2. One direction is clear: an ideal of R, by definition, is the same as a submodule of the free module (which is generated by one element). So if every submodule of a finitely generated module is finitely generated, then R must be Noetherian. For the other direction, we want to show that if every ideal is finitely generated, then so is every submodule of a finitely generated module. 103 Lecture 23: Noetherian Rings The strategy is to reduce to the case of a free module Rn , and use induction on n — we know this is true for n = 1, so we want to reduce to this case. Let M be a finitely generated module. Then by picking a set of generators, we can find a surjective homomorphism φ : Rn → M (fixing such a homomorphism is equivalent to fixing a set of generators). Then by the correspondence theorem and the above lemma, it’s enough to check that every submodule of Rn is finitely generated (since the submodules of M are exactly the images of the submodules of Rn containing ker φ). Now we can argue by induction on n. First, the base case n = 1 follows directly from the definition of a Noetherian ring, since submodules of R are exactly ideals. For the inductive step, consider a submodule N ⊂ Rn , with n > 1. Now split Rn = R × Rn−1 , and take the projection homomorphism π : Rn → Rn−1 , which sends (r1 , . . . , rn ) 7→ (r2 , . . . , rn ). Then π(N ) is a submodule of Rn−1 , so by the induction assumption, it’s finitely generated. Meanwhile, the kernel K of π is the set of elements of the form (r, 0, 0, . . .) which are in N . But this is a submodule of the free rank-1 module R, so K is also finitely generated. So by the lemma, since K and π(N ) are both finitely generated, N is finitely generated as well. Example 23.6 When considering submodules N ⊂ Z2 , we’d take the points on the x-axis as K, and the projections onto the y-axis as π(N ). Note that N is not necessarily K × π(N ). Student Question. Did we use the fact that finitely generated modules are finitely presented in this proof, when we obtained the homomorphism Rn → M ? Answer. No — the surjective homomorphism φ : Rn → M comes just from the definition of being finitely generated. We take some generators m1 , . . . , mn , and then map (r1 , . . . , rn ) to the linear combination r1 m1 + · · · + rn mn . Being finitely presented would only matter if we were looking at the kernel of this homomorphism (which we didn’t need to do here). So now we have that if R is Noetherian, any finitely generated module is also finitely presented: Proof of Corollary 23.3. If the module is finitely generated, then there is a surjective map φ : Rn → M . Then ker φ is a submodule of Rn , so it must be finitely generated as well. This means the classification of finitely presented abelian groups that we saw earlier is actually a classification of all finitely generated abelian groups. Note 23.7 It’s also possible to define Noetherian rings by Proposition 23.2 (as rings with the property that every submodule of a finitely generated module is finitely generated). But we gave the definition using ideals so that the property that’s easier to check is in the definition, and the one that’s more useful is in the proposition. 23.2 Constructing Noetherian Rings So we’ve seen why the notion of Noetherian rings is useful. But in order to use it, we want to see how to produce more examples of Noetherian rings, beyond just fields and PIDs. First, there is a simple observation we can make: Lemma 23.8 A quotient of a Noetherian ring is again Noetherian — if R is a Noetherian ring and I an ideal of R, then the ring S = R/I is also Noetherian. 104 Lecture 23: Noetherian Rings Proof. This is immediate from the correspondence theorem: an ideal in R/I is of the form J/I, where J ⊂ R is an ideal containing I. Then we can just take the images of the generators — if J = (x1 , . . . , xn ), then J/I = (x1 , . . . , xn ). So all ideals of R/I are finitely generated. Note 23.9 A subring in a Noetherian ring is not necessarily Noetherian — so this concept is more subtle than the dimension of a vector space. For example, C[x, y] is Noetherian. But the subring C + xC[x, y] (consisting of polynomials which are constant mod x) is not Noetherian — the ideal xC[x, y] is not finitely generated. 23.2.1 Hilbert Basis Theorem There’s actually a powerful tool that shows many rings are Noetherian: Theorem 23.10 (Hilbert Basis Theorem) If R is Noetherian, then R[x] is also Noetherian. This theorem has useful implications: Corollary 23.11 If R is Noetherian, then R[x1 , . . . , xn ]/I is also Noetherian, for any ideal I. So if we start with a field, this shows us how to produce many examples of Noetherian rings. Corollary 23.12 Any algebraic subset in Cn — a subset given by a collection of polynomial equations — is always given by a finite set of polynomial equations. Proof of Theorem 23.10. Let I ⊂ R[x] be an ideal, so we want to check that I is finitely generated. It’s enough to find a finite collection of polynomials P1 , . . . , Pn in I and a bound d, such that every element in I can be reduced to a polynomial of degree d using the Pi — meaning that I ⊂ (P1 , . . . , Pn ) + R[x]≤d . In other words, once the polynomials and d are fixed, then for every P ∈ I, we need to be able to find Q1 , . . . , Qn in R[x] such that X deg(P − Qi Pi ) ≤ d. If we can find such Pi and d, then I ⊂ (P1 , . . . , Pn ) + (I ∩ R[x]≤d ). But the second term is finitely generated over R (it’s not a submodule of R[x], but it is a submodule of R), since R[x]≤d is a free module of rank d + 1 over R, and R is Noetherian. So if it’s generated by S1 , . . . , Sm , then I is generated by the Pi and Si . So now we want to figure out how to do this — in some sense, the idea is generalized division with remainder. Consider the ideal I in R consisting of the leading coefficients of polynomials in I (along with 0) — we saw in the homework that this is an ideal. Then I is finitely generated. Let P1 , . . . , Pn be polynomials whose leading coefficients generate I, and let d = max(deg Pi ). Now if we P have a polynomial P of degree greater than d, we can cancel its leading coefficient — we can find Qi such that Qi Pi has the same degree and same leading coefficient as P , and then subtract them to decrease the degree of P by at least 1. We can then repeat this until P has degree at most d. Student Question. Why is I ∩ R[x]≤d finitely generated? 105 Lecture 23: Noetherian Rings Answer. The point is that R[x]≤d is a free module of rank d + 1 over R — we forget that we can multiply the polynomials and just add them and scale by elements of R, so the coordinates correspond to the coefficients of 1, x, . . . , xd . Then I ∩ R[x]≤d is a submodule of R[x]≤d , which is a finitely generated module over R; so since R is Noetherian, I ∩ R[x]≤d is also finitely generated. 23.3 Chain Conditions Proposition 23.13 A ring is Noetherian if and only if every increasing chain of ideals stabilizes. In other words, if there is a chain of ideals I1 ⊆ I2 ⊆ · · · , then from some point on, In = In+1 = · · · . In other words, R is not Noetherian if and only if there exists an infinite chain I1 ⊊ I2 ⊊ · · · of ideals. Proof Outline. If R is Noetherian and we have a chain I1 ⊆ I2 ⊆ · · · , then their union I = I1 ∪ I2 ∪ · · · is an ideal. Since R is Noetherian, then I is finitely generated. So each of the generators must have some ideal Ik that it first shows up in, and since there’s finitely many, there’s some ideal Ik containing all the generators of I (which must equal I). For the other direction, we can essentially take an ideal I that isn’t finitely generated, and starting with the empty ideal, we keep adding an element of I which isn’t in any of the previous ideals. 106