Tensor product of two ideals/ Projective module? Suppose $R$ is a commutative ring with unity and $I, J$ be two ideals of R with the property that $R\simeq  I\otimes_R J$ as an $R$-modules. How can I show that $I$ is a projective $R$-module? 
For me, this seems like a blend of all concepts that I have learned so far in module theory and couldn't think of a way to solve this question. Can any one give me a solution with an explanation of whats happening here. Thanks for anything you can provide on this. 
 A: Note that  $R \cong I \otimes_R J$ implies that the functor $R\textrm{-}\mathbf{Mod} \to R\textrm{-}\mathbf{Mod}$ given by $I \otimes_R -$ is a category equivalence, since an inverse up to natural equivalence is given by $J \otimes_R-$.
In particular, $I \otimes_R -$ commutes with all limits, so it is exact, thus $I$ is flat as an $R$-module.
There's also a theorem that says that a module $M$ is finitely presented iff $M \otimes_R -$ commutes with infinite products, thus in our case $I$ is finitely presented.
Finally, finitely presented flat modules are projective.
A: Here's a more elementary solution. I recommend you draw the diagrams.
Let $p: M\to N$ be an epimorphism and $f: I\to N$ a morphism. We want to lift $f$ to $g: I\to M$. Let $l: R\to I\otimes J$ be an isomorphism.
We get $f\otimes id_J : I\otimes J\to N\otimes J$, and so composing with our isomorphism $l$ we get $m=k\circ(f\otimes id_J)\circ l : R\to J\otimes N$, where $k : N\otimes J\to J\otimes N$ is the canonical isomorphism.
Now $p$ "restricts" to $q= id_J\otimes p: J\otimes M \to J\otimes N$ which is still an epimorphism and since $R$ is projective (it's free !)   we may lift $m$ to $n: R\to J\otimes M$ with $q\circ n = m$. 
Final step : tensor with $I$ to get $id_I\otimes n : I\otimes R \to I\otimes (J\otimes M)$. Let $s: I\otimes J\otimes M \to M$ be the natural isomorphism induced by $l$ and the natural isomorphism $R\otimes M\to M$; let $t: I\to I\otimes R$ be the natural isomorphism.
Put $g= s\circ (id_I\otimes n)\circ t$. You may now check that $g$ works using the different naturality conditions : write down the diagrams and it will be clear.
