Let $V$ be a real vector space and let $S(V^*)$ be the symmetric algebra of $V^*$. Suppose $L$ and $N$ are two $S(V^*)$-modules.
How to prove the statement in corollary 16 in the article by Kumar and Vergne on equivariant cohomology, which says that
If $L$ is a free $S(V^*)$-module, then the tensor product space $L \otimes_{\mathbb R} N$ is also free.
According to Lemma 15, whenever you consider a tensor product of the form $S(V^*)\otimes N$ where $S(V^*)$ is the regular module, this is isomorphic to $S(V^*)\otimes N'$ where $N'$ has the trivial module structure. If $M$ is free, it is of the form $S(V^*)\otimes W$ and their argument in Lemma 15 shows that $M\otimes N$ is isomorphic to $M\otimes N'$, which is ---unless I am missing something obvious--- free over $W\otimes N'$.