How to compute $\operatorname{Ext}^1_R(R/(x),M)$ where $R$ is a commutative ring with unit, $x$ is a nonzerodivisor and $M$ an $R$-module?



There is an alternative way to doing this problem than taking a projective resolution. Consider the ses of $R$ - modules

$$0 \to R \stackrel{x}{\to} R \to R/(x) \to 0$$

where the multiplication by $x$ map is injective because it is not a zero divisor in $R$. Now we recall a general fact from homological algebra that says any SES of $R$ - modules gives rise to an LES in Ext. We need only to care about the part

$$0 \to \textrm{Hom}_R(R/(x),M) \to \textrm{Hom}_R(R,M) \stackrel{f}{\to} \textrm{Hom}_R(R,M) \to \textrm{Ext}^1_R(R/(x),M) \to 0 \to 0\ldots $$

where the zeros appear because $R$ as a module over itself is free (and hence projective) so that $\textrm{Ext}^1_R(R,M) = 0$. Now we recall that $\textrm{Hom}(R,M) \cong M$ because any homomorphism from $R$ to $M$ is completely determined by the image of $1$. It is easily seen now that under this identification, $\textrm{im} f \cong xM$ so that

$$\textrm{Ext}^1_R(R/(x),M) \cong M/xM.$$

  • 3
    $\begingroup$ Who downvoted my answer? Why is it wrong? $\endgroup$ – user38268 Nov 21 '12 at 9:57
  • $\begingroup$ I made a typo in writing $\textrm{Ext}(R,R)$ instead of $\textrm{Ext}(R,M)$. Still the fact remains that the latter $R$ - module is zero. $\endgroup$ – user38268 Nov 21 '12 at 10:01

$0 \to R \xrightarrow{~x} R \to R/(x) \to 0$ is a projective resolution. Applying $\hom(-,M)$ gives the complex $0 \to \hom(R/(x),M) \to \hom(R,M) \xrightarrow{x} \hom(R,M) \to 0$, which identifies with $0 \to \{m \in M : xm=0\} \to M \xrightarrow{x} M \to 0$. It follows

$$ \mathrm{Ext}^n(R/(x),M) = \left\{\begin{array}{l} \{m \in M : xm=0\} & n =0 \\ M/xM & n=1 \\ 0 & n >1 \end{array}\right.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.