3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.