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 $L_0 \otimes_R N_0$?
Best Answer
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.