Another approach to $A/I\otimes_A A/J\simeq A/(I+J)$? The same question appears here $A/ I \otimes_A A/J \cong A/(I+J)$ however I'm looking for a different approach.
Let $A$ be an algebra, $I$ a right ideal and $J$ a left ideal. I'd like to show $$\frac{A}{I}\otimes_A \frac{A}{J}\simeq \frac{A}{I+J}.$$
Indeed I must use the following fact: Given two short exact sequences $$0\longrightarrow L^\prime\stackrel{f}{\longrightarrow} L\stackrel{g}{\longrightarrow} L^{\prime\prime}\longrightarrow 0\quad\textrm{and}\quad 0\longrightarrow M^\prime\stackrel{u}{\longrightarrow} M\stackrel{v}{\longrightarrow} M^{\prime\prime}\longrightarrow 0$$ of left and right modules over $A$, respectively, then the map $$g\otimes v:L\otimes_A M\longrightarrow L^{\prime\prime}\otimes_A M^{\prime\prime},$$ is en epimorphism with kernel $\textrm{im}(f\otimes 1_M)+\textrm{im}(1_L\otimes u)$. 
Well, using the short exact sequences:
$$0\longrightarrow I\stackrel{\imath}{\longrightarrow} A\stackrel{p_I}{\longrightarrow } A/I\longrightarrow 0\quad\textrm{and}\quad 0\longrightarrow J\stackrel{\jmath}{\longrightarrow} A\stackrel{p_J}{\longrightarrow } A/J\longrightarrow 0$$
where $\imath, \jmath$ are the inclusions and $p_I, p_J$ are the canonical projection we find: $$\frac{A}{I}\otimes_A \frac{A}{J}\simeq \frac{A\otimes_A A}{\textrm{im}(\imath\otimes 1_A)+\textrm{im}(1_A\otimes \jmath)}.            (*)$$
Now $$\textrm{im}(\imath\otimes 1_A)=I\otimes_A A\simeq I\quad\textrm{and}\quad \textrm{im}(1_A\otimes \jmath)=A\otimes_A J\simeq J,$$ and we know $A\otimes_A A\simeq A$, but I don't know how to put these informations in $(*)$, can anyone helpe me?
Thanks
 A: I think you want $I$ to be a right ideal and $J$ a left ideal, not the other
way round.
You are almost done. Consider the canonical homomorphism $\mathbf{m}
:A\otimes_{A}A\rightarrow A$ of $\mathbb{Z}$-modules which sends every
$a\otimes b\in A\otimes_{A}A$ to $ab\in A$. It is well-known that $\mathbf{m}$
is an isomorphism. Thus,
$\left(  A\otimes_{A}A\right)  /\left(  \operatorname{im}\left(
\imath\otimes1_{A}\right)  +\operatorname{im}\left(  1_{A}\otimes
\jmath\right)  \right)  $
$\cong\left(  \mathbf{m}\left(  A\otimes_{A}A\right)  \right)  /\left(
\mathbf{m}\left(  \operatorname{im}\left(  \imath\otimes1_{A}\right)
+\operatorname{im}\left(  1_{A}\otimes\jmath\right)  \right)  \right)  $
$=\underbrace{\left(  \mathbf{m}\left(  A\otimes_{A}A\right)  \right)
}_{=A\cdot A=A}/\left(  \underbrace{\mathbf{m}\left(  \operatorname{im}\left(
\imath\otimes1_{A}\right)  \right)  }_{=I\cdot A=I}+\underbrace{\mathbf{m}
\left(  \operatorname{im}\left(  1_{A}\otimes\jmath\right)  \right)
}_{=A\cdot J=J}\right)  $
$=A/\left(  I+J\right)  $.
Combined with your identity (*), this yields $\left(  A/I\right)  \otimes
_{A}\left(  A/J\right)  \cong A/\left(  I+J\right)  $, qed.
