# Show that $M$ is a Noetherian $A$-module.

This is a question from Atiyah and Macdonald, Introduction to Commutative Algebra.

Problem: Let $M$ be a Noetherian $A$-module. Show that $M[x]$ is a Noetherian $A[x]$-module.

Solution:

So, I can solve the problem with an extra assumption. That is, if we assume that $M$ is faithful (i.e., $Ann(M)=0$). In this case, it follows that $A$ is necessarily Noetherian as well. Hence, by Hilbert’s Basis Theorem, it follows that $A[x]$ is Noetherian as well.

It can easily be shown that $M[x]\cong A[x]\bigotimes_A M.$ I can also show that the tensor product of two Noetherian modules is Noetherian, hence the result.

I am wondering though, does this result hold without this extra assumption? I guess, I’m not also sure whether the ring $A$ is always necessarily Noetherian if we do not require that our module $M$ be faithful? All I can show is that if $M$ is Noetherian as an $A$-module then $A/Ann(M)$ is necessarily Noetherian as a ring.

Thanks!!

#### Solutions Collecting From Web of "Show that $M$ is a Noetherian $A$-module."

Set $I=\hbox{Ann}_A(M)$.

Since $M$ is a noetherian $A$-module, we get that $A/I$ is a noetherian ring and by Hilbert’s Basis Theorem $(A/I)[x]$ is a noetherian ring, too.

On the other side, since $M$ is a noetherian $A$-module, $M$ is a finitely generated $A$-module, and thus a finitely generated $A/I$-module. It follows that $M[x]$ is a finitely generated $(A/I)[x]$-module, hence noetherian. But $(A/I)[x]\cong A[x]/I[x]$, and since $I[x]=\hbox{Ann}_{A[x]}(M[x])$ we get that $M[x]$ is a noetherian $A[x]/\hbox{Ann}_{A[x]}(M[x])$-module. It is well known that the submodules of $M[x]$ as an $A[x]$-module coincide with the submodules of $M[x]$ as an $A[x]/\hbox{Ann}_{A[x]}(M[x])$-module, and we are done.

Notice that Hilbert’s Basis Theorem is the special case $M=A$. But the same proof works here:

Let $U$ be an $A[x]$-submodule of $M[x]$. Consider the subset $N \subseteq M$ of all elements which appear as the leading coefficient of a polynomial in $U$. It is easily checked that this is actually an $A$-submodule of $M$. By assumption, $N$ is finitely generated, say by $n_1,\dotsc,n_s$. Choose polynomials $p_i \in U$ with leading coefficient $n_i$, i.e. $p_i = n_i x^{d_i} + \text{lower terms}$. Let $d = \max_i d_i$. For $k < d$ we similarly have the submodule $N_k \subseteq M$ of all elements which appear as the leading coefficient of a polynomial in $U$ which has degree $\leq k$. It is finitely generated, say by $n_{k1},\dotsc,n_{ks}$ with corresponding polynomials $p_{i1},\dotsc,p_{is}$. We may choose $s$ uniformly (this simplifies notation).

Claim. The $A[x]$-module $U$ is generated by the $p_i$ and the $p_{ki}$ for $k<d$ and $0 \leq i \leq s$.

Proof. Let $q \in U$, say $q = m x^k + \text{lower terms}$. We make an induction on the degree $k$. If $k \geq d$, we may write $m = a_1 n_1 + \dotsc + a_s n_s$ with $a_i \in A$. Then $q-\sum_i a_i x^{k-d_i} p_i \in U$ is of lower degree, so that we are done by induction. If $k<d$, we may write $m=a_1 n_{k1} + \dotsc + a_s n_{ks}$ with $a_i \in A$. Then $q-\sum_i a_{ki} x^{k-\deg(p_{ki})} p_{ki} \in U$ is of lower degree, so that we are done by induction. $\square$

The module structure on M[x] is coincided with the A/a [x]- module structure where a is the ideal ann(M). Since you have proved that M[x] is a Noetherian A/a[x ] module, M[x] is automatically a Noetherian A[x] module.