# Square of an ideal, tensor and quotient

Let $$R$$ be a ring and two ideals $$I\subseteq J$$. Is this true that $$J/I\otimes_AA/J\cong\bar{J}/\bar{J}^2$$ where $$\bar{J}:=J/I$$.

When things get together, I am confused..

• Let $A$ be a commutative ring, $I\subseteq A$ and ideal and $M$ an $A$-module. Then, $$M\otimes_A A/I \cong M/IM.$$ You can find a proof of this result in Dummit and Foote, for example. – user347489 Aug 14 at 1:35
• Thank you. I was just not very sure $J.J/I=(J/I)^2$... – CO2 Aug 14 at 8:06

And taking $$N=J/I$$, we get $$J/I\otimes_AA/J\cong (A(J/I))/(J(J/I))=(J/I)/(J^2/I)$$.
• $J(J/I)=J^2/I=(J/I)^2$? – CO2 Aug 15 at 16:40
• Yes. The first equality is from the A-module structure on $J/I$ and the second is from the definition of products of ideals of a quotient ring. – user682705 Aug 15 at 20:49