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The Stacks project

10.31 Noetherian rings

A ring R is Noetherian if any ideal of R is finitely generated. This is clearly equivalent to the ascending chain condition for ideals of R. By Lemma 10.28.10 it suffices to check that every prime ideal of R is finitely generated.

Lemma 10.31.1.slogan Any finitely generated ring over a Noetherian ring is Noetherian. Any localization of a Noetherian ring is Noetherian.

Proof. The statement on localizations follows from the fact that any ideal J \subset S^{-1}R is of the form I \cdot S^{-1}R. Any quotient R/I of a Noetherian ring R is Noetherian because any ideal \overline{J} \subset R/I is of the form J/I for some ideal I \subset J \subset R. Thus it suffices to show that if R is Noetherian so is R[X]. Suppose J_1 \subset J_2 \subset \ldots is an ascending chain of ideals in R[X]. Consider the ideals I_{i, d} defined as the ideal of elements of R which occur as leading coefficients of degree d polynomials in J_ i. Clearly I_{i, d} \subset I_{i', d'} whenever i \leq i' and d \leq d'. By the ascending chain condition in R there are at most finitely many distinct ideals among all of the I_{i, d}. (Hint: Any infinite set of elements of \mathbf{N} \times \mathbf{N} contains an increasing infinite sequence.) Take i_0 so large that I_{i, d} = I_{i_0, d} for all i \geq i_0 and all d. Suppose f \in J_ i for some i \geq i_0. By induction on the degree d = \deg (f) we show that f \in J_{i_0}. Namely, there exists a g\in J_{i_0} whose degree is d and which has the same leading coefficient as f. By induction f - g \in J_{i_0} and we win. \square

Lemma 10.31.2. If R is a Noetherian ring, then so is the formal power series ring R[[x_1, \ldots , x_ n]].

Proof. Since R[[x_1, \ldots , x_{n + 1}]] \cong R[[x_1, \ldots , x_ n]][[x_{n + 1}]] it suffices to prove the statement that R[[x]] is Noetherian if R is Noetherian. Let I \subset R[[x]] be a ideal. We have to show that I is a finitely generated ideal. For each integer d denote I_ d = \{ a \in R \mid ax^ d + \text{h.o.t.} \in I\} . Then we see that I_0 \subset I_1 \subset \ldots stabilizes as R is Noetherian. Choose d_0 such that I_{d_0} = I_{d_0 + 1} = \ldots . For each d \leq d_0 choose elements f_{d, j} \in I \cap (x^ d), j = 1, \ldots , n_ d such that if we write f_{d, j} = a_{d, j}x^ d + \text{h.o.t} then I_ d = (a_{d, j}). Denote I' = (\{ f_{d, j}\} _{d = 0, \ldots , d_0, j = 1, \ldots , n_ d}). Then it is clear that I' \subset I. Pick f \in I. First we may choose c_{d, i} \in R such that

f - \sum c_{d, i} f_{d, i} \in (x^{d_0 + 1}) \cap I.

Next, we can choose c_{i, 1} \in R, i = 1, \ldots , n_{d_0} such that

f - \sum c_{d, i} f_{d, i} - \sum c_{i, 1}xf_{d_0, i} \in (x^{d_0 + 2}) \cap I.

Next, we can choose c_{i, 2} \in R, i = 1, \ldots , n_{d_0} such that

f - \sum c_{d, i} f_{d, i} - \sum c_{i, 1}xf_{d_0, i} - \sum c_{i, 2}x^2f_{d_0, i} \in (x^{d_0 + 3}) \cap I.

And so on. In the end we see that

f = \sum c_{d, i} f_{d, i} + \sum \nolimits _ i (\sum \nolimits _ e c_{i, e} x^ e)f_{d_0, i}

is contained in I' as desired. \square

The following lemma, although easy, is useful because finite type \mathbf{Z}-algebras come up quite often in a technique called “absolute Noetherian reduction”.

Lemma 10.31.3. Any finite type algebra over a field is Noetherian. Any finite type algebra over \mathbf{Z} is Noetherian.

Proof. This is immediate from Lemma 10.31.1 and the fact that fields are Noetherian rings and that \mathbf{Z} is Noetherian ring (because it is a principal ideal domain). \square

Lemma 10.31.4. Let R be a Noetherian ring.

  1. Any finite R-module is of finite presentation.

  2. Any submodule of a finite R-module is finite.

  3. Any finite type R-algebra is of finite presentation over R.

Proof. Let M be a finite R-module. By Lemma 10.5.4 we can find a finite filtration of M whose successive quotients are of the form R/I. Since any ideal is finitely generated, each of the quotients R/I is finitely presented. Hence M is finitely presented by Lemma 10.5.3. This proves (1).

Let N \subset M be a submodule. As M is finite, the quotient M/N is finite. Thus M/N is of finite presentation by part (1). Thus we see that N is finite by Lemma 10.5.3 part (5). This proves part (2).

To see (3) note that any ideal of R[x_1, \ldots , x_ n] is finitely generated by Lemma 10.31.1. \square

Lemma 10.31.5. If R is a Noetherian ring then \mathop{\mathrm{Spec}}(R) is a Noetherian topological space, see Topology, Definition 5.9.1.

Proof. This is because any closed subset of \mathop{\mathrm{Spec}}(R) is uniquely of the form V(I) with I a radical ideal, see Lemma 10.17.2. And this correspondence is inclusion reversing. Thus the result follows from the definitions. \square

Lemma 10.31.6.slogan If R is a Noetherian ring then \mathop{\mathrm{Spec}}(R) has finitely many irreducible components. In other words R has finitely many minimal primes.

Proof. By Lemma 10.31.5 and Topology, Lemma 5.9.2 we see there are finitely many irreducible components. By Lemma 10.26.1 these correspond to minimal primes of R. \square

Lemma 10.31.7. Let R \to S be a ring map. Let R \to R' be of finite type. If S is Noetherian, then the base change S' = R' \otimes _ R S is Noetherian.

Proof. By Lemma 10.14.2 finite type is stable under base change. Thus S \to S' is of finite type. Since S is Noetherian we can apply Lemma 10.31.1. \square

Lemma 10.31.8. Let k be a field and let R be a Noetherian k-algebra. If K/k is a finitely generated field extension then K \otimes _ k R is Noetherian.

Proof. Since K/k is a finitely generated field extension, there exists a finitely generated k-algebra B \subset K such that K is the fraction field of B. In other words, K = S^{-1}B with S = B \setminus \{ 0\} . Then K \otimes _ k R = S^{-1}(B \otimes _ k R). Then B \otimes _ k R is Noetherian by Lemma 10.31.7. Finally, K \otimes _ k R = S^{-1}(B \otimes _ k R) is Noetherian by Lemma 10.31.1. \square

Here are some fun lemmas that are sometimes useful.

Lemma 10.31.9. Let R be a ring and \mathfrak p \subset R be a prime. There exists an f \in R, f \not\in \mathfrak p such that R_ f \to R_\mathfrak p is injective in each of the following cases

  1. R is a domain,

  2. R is Noetherian, or

  3. R is reduced and has finitely many minimal primes.

Proof. If R is a domain, then R \subset R_\mathfrak p, hence f = 1 works. If R is Noetherian, then the kernel I of R \to R_\mathfrak p is a finitely generated ideal and we can find f \in R, f \not\in \mathfrak p such that IR_ f = 0. For this f the map R_ f \to R_\mathfrak p is injective and f works. If R is reduced with finitely many minimal primes \mathfrak p_1, \ldots , \mathfrak p_ n, then we can choose f \in \bigcap _{\mathfrak p_ i \not\subset \mathfrak p} \mathfrak p_ i, f \not\in \mathfrak p. Indeed, if \mathfrak {p}_ i\not\subset \mathfrak {p} then there exist f_ i \in \mathfrak {p}_ i, f_ i \not\in \mathfrak {p} and f = \prod f_ i works. For this f we have R_ f \subset R_\mathfrak p because the minimal primes of R_ f correspond to minimal primes of R_\mathfrak p and we can apply Lemma 10.25.2 (some details omitted). \square

Lemma 10.31.10. Any surjective endomorphism of a Noetherian ring is an isomorphism.

Proof. If f : R \to R were such an endomorphism but not injective, then

\mathop{\mathrm{Ker}}(f) \subset \mathop{\mathrm{Ker}}(f \circ f) \subset \mathop{\mathrm{Ker}}(f \circ f \circ f) \subset \ldots

would be a strictly increasing chain of ideals. \square


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