Describing $\mathrm{Ext}^1_R (R/J, R/J )$

Question

Let $J$ be an ideal of a commutative ring with unity $R$. Is it true that $\mathrm{Ext}^1_R (R/J, R/J ) \cong \mathrm{Hom}_R(J/J^2, R/J)$ ?

Since $\mathrm{Tor}_1^R (R/J, R/J) \cong J/J^2$, equivalently I'm asking whether $\mathrm{Ext}^1_R (R/J, R/J ) \cong \mathrm{Hom}_R(\mathrm{Tor}_1^R (R/J, R/J), R/J)$ ?

I tried using the short exact sequence $0 \to J/J^2 \to R/J^2 \to R/J \to 0$ to get a long exact sequence of $\mathrm{Ext}$'s, but got no where.

Please help.

Answer 1

$\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Ext}{Ext}$ Yes, this is true. From the short exact sequence $$0\to J\to R\to R/J\to 0$$ we get a long exact sequence $$\Hom(R,R/J)\to \Hom(J,R/J)\to \Ext^1(R/J,R/J)\to \Ext^1(R,R/J)=0.$$ Now note that for any $R$-module $M$, any homomorphism $M\to R/J$ vanishes on $JM$. Applying this to $M=R$, we see that the map $$\Hom(R,R/J)\to \Hom(J,R/J)$$ in our sequence above is $0$, so that the map $$\Hom(J,R/J)\to \Ext^1(R/J,R/J)$$ is an isomorphism by exactness. Applying the remark above with $M=J$ now tells us that $\Hom(J,R/J)\cong \Hom(J/J^2,R/J)$, and so we have the desired isomorphism $$\Hom(J/J^2,R/J)\cong\Ext^1(R/J,R/J).$$