Lemma 10.50.1. 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.

## 10.50 More Noetherian rings

**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 10.50.2 (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 10.50.1 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 10.50.3. 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

for all $n \geq c$.

**Proof.**
Apply Lemma 10.50.2 to $\mathop{\mathrm{Im}}(f) \subset N$ and note that $f : I^{n-c}M \to I^{n-c}f(M)$ is surjective.
$\square$

Lemma 10.50.4 (Krull's intersection theorem). Let $R$ be a Noetherian local ring. Let $I \subset R$ be a proper ideal. Let $M$ be a finite $R$-module. Then $\bigcap _{n \geq 0} I^ nM = 0$.

**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 10.50.2 we see that $N = I^ nM \cap N \subset IN$ for some suitably large $n$. By Nakayama's Lemma 10.19.1 we see that $N = 0$.
$\square$

Lemma 10.50.5. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Let $N = \bigcap _ n I^ n M$.

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

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 10.50.1. 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 10.50.4. 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 10.22.1. $\square$

Remark 10.50.6. Lemma 10.50.4 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 10.50.5 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 10.50.7 (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 a 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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)