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Let $L$ and $N$ be two Noetherian $R$-modules ($R$ is a commutative ring with 1). Is it right that $L \otimes_R N$ is Noetherian?

If not, what additional conditions on $L$ and $N$ are required in order that the statement will be true?

If every submodule of $L \otimes_R N$ is of the form $L_0 \otimes_R N_0$ where $L_0 \subset L$ and $N_0 \subset N$ are $R$-submodules, then each of them is finitely-generated and thus also their product and we are done. The question is if any submodule of the tensor product can be written as $L_0 \otimes_R N_0$?

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The claim about submodules is not true: consider the ideal (x-y) inside k[x,y]. –  Dylan Wilson Nov 27 '12 at 16:36
    
Also I have doubts about the whole thing being true because it is false for rings: the tensor product of noetherian rings need not be noetherian. Trying to come up with a module counterexample... –  Dylan Wilson Nov 27 '12 at 16:46
    
I thought so. I know that if $M_1 \subset M$ and $M$ is finitely-generated then $M_1$ is not necessarily finitely-generated. However, let $M_1$ be a submodule of $L \otimes_R N$. Suppose it is contained in a submodule of the form $L_1 \otimes_R N_1$ and contains a submodule of the form $L_2 \otimes_R N_2$, that is $$ L_2 \otimes_R N_2 \subset M_1 \subset L_1 \otimes_R N_1 \ . $$ Is $M_1$ is finitely-generated in that case? –  LinAlgMan Nov 27 '12 at 16:47
    
Here there is a claim about rings: math.stackexchange.com/questions/19426/… , it states that if, in addition, one of them is finitely-generated $R$-algebra then the tensor product is Noetherian. This qualifies for the application that one of the modules is $R[t]$. –  LinAlgMan Nov 27 '12 at 16:55
    
"Is $M_1$ finitely generated in that case." What if $L_2 \otimes_R N_2$ is zero? Then you're asking if a submodule of a finitely generated module is finitely generated, and the answer is no (as you said). –  Dylan Wilson Nov 29 '12 at 13:18

1 Answer 1

The tensor product of two Noetherian modules is indeed Noetherian. Even better, if $L$ is finitely generated and $N$ is Noetherian, then $L \otimes_R N$ is Noetherian.

Because $L$ is finitely generated, there is an exact sequence $R^m \to L \to 0$ for some integern $m \geq 0$. Applying the right exact functor $-\otimes_R N$ yields an exact sequence $R^m \otimes_R N \to L \otimes_R N \to 0$. But $R^m \otimes_R N \cong N^m$ is Noetherian, so its homomorphic image $L \otimes_R N$ must also be Noetherian.

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