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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..

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    $\begingroup$ 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. $\endgroup$ – user347489 Aug 14 at 1:35
  • $\begingroup$ Thank you. I was just not very sure $J.J/I=(J/I)^2$... $\endgroup$ – CO2 Aug 14 at 8:06
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From user347489's helpful comment, you can find the result on page 370 in Dummit and Foote.

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

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  • $\begingroup$ $J(J/I)=J^2/I=(J/I)^2$? $\endgroup$ – CO2 Aug 15 at 16:40
  • $\begingroup$ 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. $\endgroup$ – user682705 Aug 15 at 20:49

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