The Stacks project

10.51 More Noetherian rings

Lemma 10.51.1. Let $R$ be a Noetherian ring. Any finite $R$-module is of finite presentation. Any submodule of a finite $R$-module is finite. The ascending chain condition holds for $R$-submodules of a finite $R$-module.

Proof. We first show that any submodule $N$ of a finite $R$-module $M$ is finite. We do this by induction on the number of generators of $M$. If this number is $1$, then $N = J/I \subset M = R/I$ for some ideals $I \subset J \subset R$. Thus the definition of Noetherian implies the result. If the number of generators of $M$ is greater than $1$, then we can find a short exact sequence $0 \to M' \to M \to M'' \to 0$ where $M'$ and $M''$ have fewer generators. Note that setting $N' = M' \cap N$ and $N'' = \mathop{\mathrm{Im}}(N \to M'')$ gives a similar short exact sequence for $N$. Hence the result follows from the induction hypothesis since the number of generators of $N$ is at most the number of generators of $N'$ plus the number of generators of $N''$.

To show that $M$ is finitely presented just apply the previous result to the kernel of a presentation $R^ n \to M$.

It is well known and easy to prove that the ascending chain condition for $R$-submodules of $M$ is equivalent to the condition that every submodule of $M$ is a finite $R$-module. We omit the proof. $\square$

Lemma 10.51.2 (Artin-Rees). Suppose that $R$ is Noetherian, $I \subset R$ an ideal. Let $N \subset M$ be finite $R$-modules. There exists a constant $c > 0$ such that $I^ n M \cap N = I^{n-c}(I^ cM \cap N)$ for all $n \geq c$.

Proof. Consider the ring $S = R \oplus I \oplus I^2 \oplus \ldots = \bigoplus _{n \geq 0} I^ n$. Convention: $I^0 = R$. Multiplication maps $I^ n \times I^ m$ into $I^{n + m}$ by multiplication in $R$. Note that if $I = (f_1, \ldots , f_ t)$ then $S$ is a quotient of the Noetherian ring $R[X_1, \ldots , X_ t]$. The map just sends the monomial $X_1^{e_1}\ldots X_ t^{e_ t}$ to $f_1^{e_1}\ldots f_ t^{e_ t}$. Thus $S$ is Noetherian. Similarly, consider the module $M \oplus IM \oplus I^2M \oplus \ldots = \bigoplus _{n \geq 0} I^ nM$. This is a finitely generated $S$-module. Namely, if $x_1, \ldots , x_ r$ generate $M$ over $R$, then they also generate $\bigoplus _{n \geq 0} I^ nM$ over $S$. Next, consider the submodule $\bigoplus _{n \geq 0} I^ nM \cap N$. This is an $S$-submodule, as is easily verified. By Lemma 10.51.1 it is finitely generated as an $S$-module, say by $\xi _ j \in \bigoplus _{n \geq 0} I^ nM \cap N$, $j = 1, \ldots , s$. We may assume by decomposing each $\xi _ j$ into its homogeneous pieces that each $\xi _ j \in I^{d_ j}M \cap N$ for some $d_ j$. Set $c = \max \{ d_ j\} $. Then for all $n \geq c$ every element in $I^ nM \cap N$ is of the form $\sum h_ j \xi _ j$ with $h_ j \in I^{n - d_ j}$. The lemma now follows from this and the trivial observation that $I^{n-d_ j}(I^{d_ j}M \cap N) \subset I^{n-c}(I^ cM \cap N)$. $\square$

Lemma 10.51.3. Suppose that $0 \to K \to M \xrightarrow {f} N$ is an exact sequence of finitely generated modules over a Noetherian ring $R$. Let $I \subset R$ be an ideal. Then there exists a $c$ such that

\[ f^{-1}(I^ nN) = K + I^{n-c}f^{-1}(I^ cN) \quad \text{and}\quad f(M) \cap I^ nN \subset f(I^{n - c}M) \]

for all $n \geq c$.

Proof. Apply Lemma 10.51.2 to $\mathop{\mathrm{Im}}(f) \subset N$ and note that $f : I^{n-c}M \to I^{n-c}f(M)$ is surjective. $\square$

Proof. Let $N = \bigcap _{n \geq 0} I^ nM$. Then $N = I^ nM \cap N$ for all $n \geq 0$. By the Artin-Rees Lemma 10.51.2 we see that $N = I^ nM \cap N \subset IN$ for some suitably large $n$. By Nakayama's Lemma 10.20.1 we see that $N = 0$. $\square$

Lemma 10.51.5. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Let $N = \bigcap _ n I^ n M$.

  1. For every prime $\mathfrak p$, $I \subset \mathfrak p$ there exists a $f \in R$, $f \not\in \mathfrak p$ such that $N_ f = 0$.

  2. If $I$ is contained in the Jacobson radical of $R$, then $N = 0$.

Proof. Proof of (1). Let $x_1, \ldots , x_ n$ be generators for the module $N$, see Lemma 10.51.1. For every prime $\mathfrak p$, $I \subset \mathfrak p$ we see that the image of $N$ in the localization $M_{\mathfrak p}$ is zero, by Lemma 10.51.4. Hence we can find $g_ i \in R$, $g_ i \not\in \mathfrak p$ such that $x_ i$ maps to zero in $N_{g_ i}$. Thus $N_{g_1g_2\ldots g_ n} = 0$.

Part (2) follows from (1) and Lemma 10.23.1. $\square$

Remark 10.51.6. Lemma 10.51.4 in particular implies that $\bigcap _ n I^ n = (0)$ when $I \subset R$ is a non-unit ideal in a Noetherian local ring $R$. More generally, let $R$ be a Noetherian ring and $I \subset R$ an ideal. Suppose that $f \in \bigcap _{n \in \mathbf{N}} I^ n$. Then Lemma 10.51.5 says that for every prime ideal $I \subset \mathfrak p$ there exists a $g \in R$, $g \not\in \mathfrak p$ such that $f$ maps to zero in $R_ g$. In algebraic geometry we express this by saying that “$f$ is zero in an open neighbourhood of the closed set $V(I)$ of $\mathop{\mathrm{Spec}}(R)$”.

Lemma 10.51.7 (Artin-Tate). Let $R$ be a Noetherian ring. Let $S$ be a finitely generated $R$-algebra. If $T \subset S$ is an $R$-subalgebra such that $S$ is finitely generated as a $T$-module, then $T$ is of finite type over $R$.

Proof. Choose elements $x_1, \ldots , x_ n \in S$ which generate $S$ as an $R$-algebra. Choose $y_1, \ldots , y_ m$ in $S$ which generate $S$ as a $T$-module. Thus there exist $a_{ij} \in T$ such that $x_ i = \sum a_{ij} y_ j$. There also exist $b_{ijk} \in T$ such that $y_ i y_ j = \sum b_{ijk} y_ k$. Let $T' \subset T$ be the sub $R$-algebra generated by $a_{ij}$ and $b_{ijk}$. This is a finitely generated $R$-algebra, hence Noetherian. Consider the algebra

\[ S' = T'[Y_1, \ldots , Y_ m]/(Y_ i Y_ j - \sum b_{ijk} Y_ k). \]

Note that $S'$ is finite over $T'$, namely as a $T'$-module it is generated by the classes of $1, Y_1, \ldots , Y_ m$. Consider the $T'$-algebra homomorphism $S' \to S$ which maps $Y_ i$ to $y_ i$. Because $a_{ij} \in T'$ we see that $x_ j$ is in the image of this map. Thus $S' \to S$ is surjective. Therefore $S$ is finite over $T'$ as well. Since $T'$ is Noetherian and we conclude that $T \subset S$ is finite over $T'$ and we win. $\square$

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