# Finite Length Modules

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.

## 1 Answer

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.