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Let $R$ be a noetherian local ring. Is it true that if $M$ is a finite length $R$-module, then for all $k \geq 0$, $\mathrm{Ext}_R^k(M, R)$ is finite length as well?

I can show the $k = 0$ case, since $\mathrm{Hom}_R(M, R)$ is a (finitely generated) ideal of $R$ which is annihilated by a power of the maximal ideal. However, I'm not sure how to boot-strap this up to all $k$.

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  • $\begingroup$ The support of $\mathrm{Ext}_R^k(M, R)$ contains at most the maximal ideal, and this is easily seen by localizing at a prime and taking into account that in your setting the localization commutes with $\mathrm{Ext}$. $\endgroup$
    – user26857
    Commented Jun 10, 2017 at 22:37

2 Answers 2

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Let $(R,\mathfrak{m})$ be a Noetherian local ring. If $M$ is a finite length $R$-module, the only associated prime of $M$ is $\mathfrak{m}$ so there is a filtration $$ 0 = M_{0} \subset \dotsb \subset M_{n} = M $$ of $M$ where each successive quotient $M_{i}/M_{i-1}$ is isomorphic to the residue field $R/\mathfrak{m}$. Using the long exact sequence in Ext, we may reduce to the case where $M = R/\mathfrak{m}$. In this case $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is a finitely generated $R$-module which contains $\mathfrak{m}$ in its annihilator. To see that $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is annihilated by $\mathfrak{m}$, note that one way to compute $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is to take a resolution $R \to I^{\bullet}$ of $R$ by injective $R$-modules, apply $\mathrm{Hom}_{R}(R/\mathfrak{m},-)$ to $I^{\bullet}$, then take cohomology. We have that $\mathrm{Hom}_{R}(R/\mathfrak{m},N)$ is annihilated by $\mathfrak{m}$ for any $R$-module $N$.

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  • $\begingroup$ Why is $\mathrm{Ext}_R^k(R/\mathfrak{m}, R)$ annihilated by $\mathfrak{m}$? $\endgroup$
    – Dorebell
    Commented Jun 12, 2017 at 9:23
  • $\begingroup$ One way to compute $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is to take a resolution $R \to I^{\bullet}$ of $R$ by injective $R$-modules, apply $\mathrm{Hom}_{R}(R/\mathfrak{m},-)$ to $I^{\bullet}$, then take cohomology. We have that $\mathrm{Hom}_{R}(R/\mathfrak{m},N)$ is annihilated by $\mathfrak{m}$ for any $R$-module $N$. $\endgroup$ Commented Jun 12, 2017 at 20:06
  • $\begingroup$ Perfect - Would you mind adding this comment to the answer before I accept? $\endgroup$
    – Dorebell
    Commented Jun 12, 2017 at 20:13
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Over a Noetherian ring $R$, finitely generated modules $M,N$ and $f\in R$, one has $(\mathrm{Ext}^k_R(M,N))_f=\mathrm{Ext}^k_{R_f}(M_f,N_f)$. Can you finish the rest?

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