# Verify proof that if $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian

I have to prove that

If $M,N$ are $R$-modules and $M$ is Noetherian, $N$ is finitely generated, then $M\otimes_R N$ is Noetherian

We let $S$ be a non-finitely generated submodule of $M\otimes_R M$. Then there are elements $$g_j=\sum_{i=0}^{n_j} m_{ji}\otimes n_{ji}$$ s.t. $$\left< g_1\right>\subsetneq \left< g_1,g_2\right>\subsetneq \dots$$ However, if we let $n_1,\cdots,n_N$ be a generating set for $N$, we can write: \begin{align} g_j&=\sum_{i=0}^{n_j} m_{ji}\otimes n_{ji}\\ &=\sum_{i=0}^{n_j} m_{ji}\otimes\left( \sum_{k=1}^{N}c_{jik}n_k\right)\\ &=\sum_{k=1}^N\left(\sum_{i=0}^{n_j}c_{jik}m_{ji}\right)\otimes n_k \end{align} However this means that for an arbitrary element $x$ in $\left<g_1,g_2,\dots,g_M\right>$ we can write \begin{align} x&=\sum_{j=1}^{M}d_jg_j\\ &=\sum_{j=1}^{M}d_j\sum_{k=1}^N\left(\sum_{i=0}^{n_j}c_{jik}m_{ji}\right)\otimes n_k\\ &=\sum_{k=1}^N\left(\sum_{j=1}^{M}d_j\sum_{i=0}^{n_j}c_{jik}m_{ji}\right)\otimes n_k \end{align} However, the elements in the first tensor slots of the $k$th term in the above sum are elements in the submodule $$\left< \sum_{i=0}^{n_1}c_{1ik}m_{1i},\dots, \sum_{i=0}^{n_1}c_{Mik}m_{Mi}\right>$$ And since $M$ is Noetherian, this sequence of ideals eventually stabilizes. Hence for every $1\leq k\leq N$ there is an $M_k$ s.t. the above sequence has stabilized; set $M=\max\{M_k\}$. Then for $K>M$ we also have $$\left<g_1,\dots,g_M\right>=\left<g_1,\dots,g_K\right>$$ contradicting $M\otimes_R N$ not being finitely generated.

I would appreciate someone going over this proof to verify it is correct, and if not tell me where I made a mistake. Also welcome are alternative proofs/other remarks. Thanks.

## 1 Answer

There is no provision for $N$ to have a free basis, but your idea is good, assuming $R$ is commutative.

Consider an epimorphism $R^n\to N$, which gives an epimorphism $$M\otimes_R R^n\to M\otimes_R N$$ Since $M\otimes_R R^n\cong M^n$ is noetherian, we are done.

• Thanks, thats really nice. Just to be sure: replacing 'free basis' by 'generating set' my proof goes through right? – user2520938 Jan 2 '16 at 16:22
• @user2520938 I don't think so. You're using heavily that $N$ is freely generated. – egreg Jan 2 '16 at 16:25
• Are you sure? I don't think I ever use that the $c_{jik}$ are unique, which would be the only difference between free basis and generating set. – user2520938 Jan 2 '16 at 16:28
• @user2520938 Maybe you're right; the idea is well hidden in heavy notation, though. ;-) – egreg Jan 2 '16 at 16:31
• I agree that the indexes make it fairly difficult to read. Thanks again – user2520938 Jan 2 '16 at 16:37