Showing a module is finitely generated and projective

Question

Let $R$ be a commutative ring and $V$ be an $R$-module such that $V\otimes_{R}V \cong R$ as $R$-modules. I want to show that $V$ is a finitely generated and projective $R$-module.

What I've considered:

Clearly $V\otimes_{R}V$ is finitely generated and projective, since $R$ is immediately from definitions. For projectivity, given a short exact sequence of $R$-modules:

$$M \xrightarrow{f} V \rightarrow 0$$

we can apply the right exact functor $\_\otimes_{R} V$, yielding:

$$ M\otimes_{R} V\xrightarrow{f\otimes\text{id}} V \otimes_{R} V \rightarrow 0.$$

As $ V \otimes_{R} V$ is projective, this splits, so, morally, I just want to "restrict the splitting map to the pure tensors of the form $v \otimes 1$" and conclude. The only thing I've thought of that could potentially make this precise is to show that the splitting map must be of the form $s=s_{1}\otimes s_{2}$.

For being finitely generated, it seems "intuitively clear" to me that if $V \otimes_{R} V$ is finitely generated, then so must be $V$, but I'm having trouble showing this carefully. In particular, I feel good about the converse statement, as discussed in Tensor product of two finitely generated modules

Answer 1

If you like it categorically, here's another solution: The functor $F_V: -\otimes_R V: R\text{-Mod}\to R\text{-Mod}$ is a (self-inverse) equivalence on $R\text{-Mod}$, and as such, it preserves (a) projective and (b) finitely generated objects, the latter coinciding with finitely generated modules in the case of $R\text{-Mod}$. As $F_V(R)\cong V$ and $R$ is finitely generated and projective, $V$ therefore has the same properties.

Note: This holds true for any $\otimes_R$-invertible module $V$, it's not necessary to assume that $V$ is self-inverse.

Answer 2

First, let us prove that $V$ is projective, from what you started. Let $F\to V$ be a surjection with $F$ free (possibly infinitely generated). Tensoring by $V$, we have $F\otimes V\to V\otimes V=R$ surjective and hence $F\otimes V=R\oplus M$ for some $M$. Tensor again with $V$ to get $F\otimes V\otimes V=F=V\oplus M\otimes V$, proving that $V$ is a direct summand of a free module and hence projective.

Next we show that $V$ is finitely generated. Again, we have $V\oplus N=F$, $F$ a free module (possibly of infinite rank) and $N$ some module, from the first part. Tensor with $V$ to get $V\otimes V\oplus N\otimes V=F\otimes V$. But $V\otimes V=R$ and so $R$ is a direct summand of $F\otimes V$. Since the element $1\in R$ can have only finitely many non-zero components in $F\otimes V=\oplus V$, we see that $R$ is in fact a direct summand of $\oplus V$, a finite direct sum. Tensoring with $V$, we see that $V$ is a direct summand of $V\otimes (\oplus V)=\oplus R$, again a finite direct sum. That is, $V$ is a direct summand of a free module of finite rank and you are done.