Let $R$ be a ring. If $M, N$ and $L$ are $R$-modules, with, $\ell(M), \ell(N), \ell(L) < \infty,$ and $M \times N \cong M \times L,$ it is true that $N \cong L?$

The same occurs replacing $\times$ by $\oplus$ or $\otimes?$

When the converse is true? That is, if $M \times N \cong M \times L, $$N \cong L$ and (additional hypothesis), then $\ell(M), \ell(N), \ell(L) < \infty$?

For example, we know that, if $\ell(M), \ell(N) < \infty,$ then $\ell(M \otimes N) \le \ell(M) \ell(N),$ but this is not useful in a general case.


First, a remark: finite direct products and finite direct sums of modules are the same. Thus $M\times N = M\oplus N$.

The Krull-Schmidt Theorem states that any finite length module is isomorphic to a direct sum of finitely many indecomposable modules, and that the indecomposable modules occuring in any such decomposition are uniquely determined up to isomorphism and permutation. Thus $M\oplus N \cong M\oplus L$ implies that $N \cong L$ if $L,M$ and $N$ have finite length.

For the tensor product, things are not so well-behaved. For example, if one takes $R=\mathbb{Z}$, $M = \mathbb{Z}/2\mathbb{Z}$, $N = \mathbb{Z}/3\mathbb{Z}$ and $L = \mathbb{Z}/5\mathbb{Z}$, then $M\otimes_{\mathbb{Z}} N = 0 = M\otimes_{\mathbb{Z}} L$, even though $L$ and $N$ are not isomorphic.

I don't think anything can be said for a converse statement.


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