Lemma 2

Let $A_1$ and $A_2$ be algebras over the commutative field K. Let $M_1$ be an $A_1$-module that is finite-dimensional over K and $M_2$ a $A_2$-module of finite length. Then the $A_1\otimes A_2$-module $M_1\otimes M_2$ has finite length.

Set $M = M_1\otimes M_2$. Let $(e_1, . . . , e_n)$ be a basis of $M_1$ over the field K. The mapping $(x_1, . . . , x_n)\mapsto \sum^n_{i=1}e_i\otimes x_i$ is an isomorphism from the $A_2$-module $M^n_2$ to the $A_2$-module M. Since $M_2$ is an $A_2$-module of finite length, so is M. Moreover, every A-submodule of M is an $A_2$-submodule; consequently, M is an A-module of finite length.