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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.

user26857
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user521337
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1 Answers1

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$\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Ext}{Ext}$ 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 $$\Hom(J/J^2,R/J)\cong\Ext^1(R/J,R/J).$$

Eric Wofsey
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  • Remark. The above proof of the isomorphism $\operatorname{Hom}(J, R/J) \cong \operatorname{Ext}^1(R/J, R/J)$ does not require $R$ to be commutative. – Orat Mar 04 '19 at 13:31
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    @Orat: That's not true. The map $\operatorname{Hom}(R,R/J)\to \operatorname{Hom}(J,R/J)$ may be nonzero if $R$ is noncommutative, since $R/J$ may not be annihilated by $J$ as a left module. For instance, if $R=M_n(k)$ for a field $k$ and some $n>1$, then $R$ as a left module over itself is isomorphic to $V^n$ where $V=k^n$ with the canonical action of $R$. Taking $J$ to be one of the copies of $V$ in $V^n$, then $J$ is a direct summand of $R$ so every homomorphism $J\to R/J$ extends to $R$, but there are plenty of nonzero homomorphisms $J\to R/J$ since $J\cong V$ and $R/J\cong V^{n-1}$. – Eric Wofsey Mar 04 '19 at 18:21
  • In that example, the Ext is trivial since $R$ is semisimple, but $\operatorname{Hom}(J,R/J)$ is nontrivial. – Eric Wofsey Mar 04 '19 at 18:22
  • My comment was totally wrong. Obviously not true for semisimple cases like that. Thank you for your correction! – Orat Mar 04 '19 at 18:56