Showing $I \otimes J \cong IJ$ Let $I, J$ be ideals of a ring $R$ and assume $I$ is flat as a right $R$-module.
I would like to show that Showing $I \otimes J \cong IJ$. 
Here is my strategy: Let $j : J \rightarrow R$ be the inclusion map. Then by flatness $I \otimes j$ is injective. I would then like to conclude that $Im (I \otimes j) = IJ$ and thus complete the proof, but I am uncertain about this step - is it true that $Im (I \otimes j) = IJ$? It seems correct, but I am having trouble formulating why this should be true.
 A: Expanding  on @PedroTamaroff's comments: 
Consider the short exact sequence $\;0\to J\to R\to R/J\to 0$, and tensor it by the flat $R$-module $I$ to obtain the short exact sequence
$$0\to I\otimes_R J\to  I\otimes_RR  \to   I\otimes_R R/J\to 0.$$
Now  we have a canonical isomorphism $\;\begin{aligned}[t]  I\otimes_RR &\xrightarrow{~\sim~}I\\ i\otimes r&\longmapsto ir\end{aligned}$, so we can identify the middle tensor product to $I$. In this identification, the image of $I\otimes_R J$, which is generated by the images of the elementary tensors $i\otimes j$, is generated by the $ij$, i.e. it is the product ideal $IJ$.
Last, note the quotient $ I\otimes_R R/J$ is identified to $\; I/IJ$.
A: $IJ$ is the smallest set containing all of the products $ab$ for $a \in I$ and $b \in J$ and is closed under addition.
A: Consider the exact sequence $0\to J\to R$ and the obvious commutative diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> I\otimes J @>>> I\otimes R  \\
@. @VVV @VVV  \\
0 @>>> IJ @>>> I
\end{CD}
$$
A very simple diagram chasing shows that the map $I\otimes J\to IJ$ (such that $x\otimes y\mapsto xy$) is injective (because $I\otimes R\to I$ is an isomorphism). Clearly it is also surjective, because $IJ$ is generated by the elements $xy$, with $x\in I$ and $y\in J$.
