# 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

• You mixed up "left ideal" and "right ideal", right? (Or left?) – darij grinberg Jul 3 '15 at 1:02

## 1 Answer

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.