TITLE: Length of the tensor product of two modules
QUESTION [9 upvotes]: I'm trying to prove the following statement:

Let $R$ be a commutative ring. If $M$ and $N$ are finite length $R$-modules, then $M \otimes_{R} N$ has finite length, and $\ell(M\otimes_{R}N)\leq\ell(M)\cdot\ell(N)$.

I've managed to prove the first part (i.e., $M\otimes_{R}N$ has finite length), but I don't know how to prove the inequality (I've tried to find a composition series for $M\otimes_{R}N$, but with no luck so far).
A complete answer, or just a few hints would be much appreciated. Thanks!

REPLY [10 votes]: If you have a short exact sequence $$0\to M'\to M\to M''\to 0$$ you get another one looking like $$M'\otimes N\to M\otimes N\to M''\otimes N\to 0 \tag{$\star$}$$ Suppose that $M$ has finite length and that $M'\neq0\neq M''$, so that $M'$ and $M''$ are also of finite length and, in fact, of length smaller than that of $M$. Inductively, you know that $\ell(M'\otimes N)\leq\ell(M')\ell(N)$ and $\ell(M''\otimes N)\leq\ell(M'')\ell(N)$. Use this and $(\star)$ to conclude what you want.