Square of an ideal, tensor and quotient

Question

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..

Answer 1

From user347489's helpful comment, you can find the result on page 370 in Dummit and Foote.

And taking $N=J/I$, we get $J/I\otimes_AA/J\cong (A(J/I))/(J(J/I))=(J/I)/(J^2/I)$.