If the tensor product of two modules is free of finite rank, then the modules are finitely generated and projective 
If over a commutative ring $R$ we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ are finitely generated projective?

We have finite generation, because if $M\otimes N$ is generated by $\sum_i a_{ij} x_i^j\otimes y_i^j$, then we have $x_{i_1}^{j_1}\otimes y_{i_2}^{j_2}\otimes x^{j_3}_{i_3}$ generating $M^n$. Projecting onto $M$, we see that $x_i^j$ generate $M$. My question is can we show that this map splits?
I am able to verify if in the case that $n=1$. I can't personally think of an example where it is not true, but I am having problems constructing a splitting. Any other solution is appreciated too.
 A: Let $M,N$ be two $R$-modules such that $M \otimes N$ is free of rank $>0$. Let $p : F \to M$ be any epi with $F$ free. Then $N \otimes p : N \otimes F \to N \otimes  M$ is epi, hence it splits. Then also $M \otimes N \otimes p$ splits, which implies that $M \otimes N \otimes M$ is a direct summand of $M \otimes N \otimes F$ and is therefore projective. But since $R$ is a direct summand of $M \otimes N$, it follows that $R \otimes M = M$ is a direct summand of $M \otimes N \otimes M$, hence also projective.
Some additions: If $M \otimes N$ is locally free of finite rank $>0$, then $M$ is locally free of finite rank (i.e. finitely generated projective). One may ask if we can drop the "finite rank" here or, perhaps more interesting, what happens when we assume only that $M \otimes N$ is projective and faithfully flat. In the above proof, then since $M \otimes N \otimes M$ is flat we conclude that $M$ is flat, in fact faithfully flat since $M \otimes N$ is. Now the question is if tensoring with a faithfully flat module (not a ring extension) descends the property of being projective. I only know this in the finitely generated case.
A: $\def\id{\operatorname{id}}$Suppose $M\otimes N$ is isomorphic to $R^n$. Pick a basis $\{x_1,\dots,x_n\}$ of $M\otimes N$, with $x_i=\sum_{j=1}^{r_i}m_{i,j}\otimes n_{i,j}$ for each $i\in\{1,\dots,n\}$. Let $r=r_1+\cdots+r_n$, let $\{e_{i,j}:1\leq i\leq n, 1\leq j\leq r_i\}$ be a basis of $R^r$, and consider the map $f:R^r\to M$ which maps $e_{i,j}$ to $m_{i,j}$. Then $f\otimes\id_N:R^r\otimes N\to M\otimes N$ is obviously surjective. Now $M\otimes N$ is free, so $f\otimes\id_N$ is a split surjection, and therefore so is $f\otimes\id_N\otimes\id_M:R^r\otimes N\otimes M\to M\otimes N\otimes M$. As $R^r\otimes N\otimes M\cong R^r\otimes R^n\cong R^{rn}$ is free, this means that $M\otimes N\otimes M\cong R^n\otimes M\cong M^n$ is projective, as is every one of its obvious $n$ direct summands, each isomorphic to $M$.
Later. We can do this more simply as follows; this is what i wanted to do before out of my comment above, but somehow messed up (in a way, it is a transposed variant of the first paragraph)
Suppose $\phi:M\otimes N\to R^n$ is an isomorphism, let $p:R^n\to R$ be the projection onto the first summand, and suppose $\xi=\sum_{i=1}^rm_i\otimes n_i\in M\otimes N$ is such that $\phi(\xi)=(1,0,\dots,0)$. Then the map $\psi:R^r\otimes N\to R$ such that $\psi(e_i\otimes n)=p(\phi(m_i\otimes n))$ is surjective. Since $R$ is free, $\psi$ is split, so that so is $\psi\otimes\id_M:R^r\otimes N\otimes M\to R\otimes M$. The domain of this last map is free, isomorphic to $R^{rn}$, and the codomain is isomorphic to $M$, so we see that $M$ is projective.
