LEC ALG4 4

Noetherian modules

Proposition 389.1(Proposition).

Let $M$ be an $R$-module. TFAE:

1. The ascending chain condition holds for submodules of $M$.
2. Every nonempty collection of submodules of $M$ contains a maximal element.
3. Every submodule of $M$ is finitely generated.

An $R$-module satisfying any of these conditions is called a noetherian $R$-module.

Proof.

$(1⟹2)$ Let $\Lambda$ be a nonempty collection of submodules of $M$. FTSOC, assume $\Lambda$ does not have a maximal element. Let $N_1\in \Lambda$. Since $N_1$ is not maximal, there exists $N_2\in \Lambda$ such that $N_1⊊N_2$. Repeat indefinitely to get an infinite ascending chain.

$(2⟹3)$ Let $N$ be a submodule of $M$. Let $\Lambda$ be the collection of finitely generated submodules of $N$. Since $0\in \Lambda$, it is nonempty. Let $N^{\prime }$ be a maximal element of $\Lambda$. $N^{\prime }$ is finitely generated. Suppose $N\ne N^{\prime }$. Then, there exists $x\in N$ such that $x\notin N^{\prime }$. Consider the submodule $N^{\prime }+Rx$ of $N$. It strictly contains $N^{\prime }$ and is finitely generated, contradicting the maximality of $N^{\prime }$.

$(3⟹1)$ Let $N_1\subseteq N_2\subseteq \dots$ be a chain of submodules of $M$. Then, $N={\bigcup }_{i⩾1}{N}_i$ is a submodule of $M$, and hence is finitely generated. Let $\{x_1,\dots ,x_n\}$ be a finite generating set of $N$. There exist $i_1,\dots i_n$ such that $x_j\in N_{i_j}$ with $1⩽j⩽n$. Let $k⩾\max \{i_1,\dots ,i_n\}$, so that $\{x_1,\dots ,x_n\}\subseteq N_k$. Thus, $N={\sum }_{i=1}^{n}R{x}_i\subseteq N_k\subseteq N$. Thus, $N=N_k$, and $N_m=N_k$ for all $m⩾k$.□

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Note that $R$ is a Noetherian ring $⟺$ $R$ is a Noetherian $R$-module.

Exercise 389.2(Exercise).

Let $M$ be a module with submodule $N$. Then, $M$ is Noetherian $⟺$ $N$ and $M/N$ are Noetherian.

Proof.

$(⟹)$ Suppose $M$ is Noetherian. We will use Prp 1.3. Since submodules of $N$ are submodules of $M$, $N$ is clearly Noetherian. Similarly, submodules of $M/N$ are projections of submodules of $M$ containing $N$, and hence are finitely generated.

$(⟸)$ Let $N^{\prime }\subseteq M$ be a submodule. If $N^{\prime }\subseteq N$, we are done. Suppose $N^{\prime }\supseteq N$. Since $M/N$ is Noetherian, $N^{\prime }/N$ is finitely generated, say by ${\overline{x}}_1,\dots ,{\overline{x}}_m$. Let $z_1,\dots ,z_n$ generate $N$. Clearly, $x_1,\dots ,x_m,z_1,\dots ,z_n$ generates $N^{\prime }$.

It remains to check the case when $N^{\prime }⊊N$ and $N^{\prime }⊋N$. Let $\pi (N^{\prime })$ be the image of $N^{\prime }$ under the projection $\pi :M\twoheadrightarrow M/N$. Being a submodule of $M/N$, it is finitely generated, say by ${\overline{x}}_1,\dots ,{\overline{x}}_m$. WLOG, let the representatives $x_1,\dots ,x_m$ be contained in $N^{\prime }$. Let $N\cap N^{\prime }$ be generated by $z_1,\dots ,z_n$. Then, $x_1,\dots ,x_m,z_1,\dots ,z_n$ generate $N^{\prime }$: for $m\in N^{\prime }$,

$$\begin{array}{rlr}m+N& =\sum _{i=1}^m{r}_i{\overline{x}}_i& \\ ⟹m& =y+\sum _{i=1}^m{r}_i{x}_i& y\in N\cap N^{\prime }\\ ⟹m& =\sum _{j=1}^n{s}_j{z}_j+\sum _{i=1}^m{r}_i{x}_i.& \end{array}$$ □

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Proposition 389.3(Proposition).

Let $R$ be a Noetherian ring. Then, $R^n$ is a Noetherian $R$-module.

Proof.

We induct on $n$. The case $n=1$ is true by hypothesis. Assume $R^{n-1}$ is Noetherian. Now,

$$\begin{array}{rl}& R^{n-1}=\underset{i=1}{\overset{n-1}{⨁}}R{e}_i\subseteq R^n;\\ & \frac{R^n}{R^{n-1}}=R{e}_n,\end{array}$$

which is free of rank 1¹. We are done by Exr 2⇐.□

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Corollary 389.4(Corollary).

Let $R$ be a Noetherian ring and $M$ be an $R$-module. Then $M$ is Noetherian iff it is finitely generated.

Proof.

$(⟹)$ By definition.

$(⟸)$ Suppose $M$ has a generating set $\{x_1,\dots ,x_n\}$. By Prp 373.14, we have a surjection $R^n\twoheadrightarrow M$ which maps $e_i\mapsto x_i$. By Exr 2⇒, it suffices to show that $R^n$ is a noetherian $R$-module; we are done by Prp 3.□

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Theorem 389.5(Hilbert's basis theorem).

If $R$ is a Noetherian ring, then $R[X]$ is a Noetherian ring ².

Proof.

FTSOC, assume that $R[X]$ has a nonzero ideal $I$ that is not finitely generated. Let $f_1\in I\setminus 0$ be of the smallest possible degree. For all $k⩾2$, choose $f_k\in I\setminus ⟨f_1,\dots ,f_{k-1}⟩$ of smallest possible degree. Write $d_i$ for the degree of $f_i$. We have $d_1⩽d_2⩽\dots$. For $i⩾1$, let $b_i$ be the leading coefficient of $f_i$. Let $J=⟨b_1,b_2,\dots ⟩$ be an ideal of $R$. Since $J$ is finitely generated, there exists $k$ such that $J=⟨b_1,\dots ,b_k⟩$.

Now, there exist $r_1,\dots ,r_k\in R$ such that $b_{k+1}={\sum }_{i=1}^k{r}_i{b}_i$. Consider

$$\begin{array}{r}g:=\sum _{i=1}^k{r}_i{X}^{d_{k+1}-d_i}{f}_i.\end{array}$$

Then $\text{deg}g=d_{k+1}$ and its leading coefficient is $b_{k+1}$. We have $g\in ⟨f_1,\dots ,f_k⟩$. $\text{deg}(f_{k+1}-g)⩽d_{k+1}$, and $f_{k+1}-g\notin ⟨f_1,\dots ,f_k⟩$. This contradicts the choice of $f_{k+1}$. Thus, $R[x]$ is Noetherian.□

Footnotes

1. Note that not all rank n-1 free submodules of $R^n$ give this nice quotient! ↩

2. Again, this means $R[x]$ is a Noetherian $R[x]$-module. $R[x]$ is clearly not Noetherian as an $R$-module. ↩