Let $M$ be a finitely generated reflexive module over a regular local ring $(R,\mathfrak m,k)$ such that $\operatorname {Ext}^1_R( \operatorname {Hom}_R(M,M),R)=0$. Then how to show that $M$ is a free $R$-module ?

Here reflexive means $\operatorname {Hom}_R(\operatorname {Hom}_R(M,R),R)\cong M$ .

My try: since we're in a regular local ring, so $pd(M)+depth M=depth R=\dim R$. So to show $M$ is free, it is enough to show $depth(M)\ge depth(R)$. Also, since our module is finitely generated over a Noetherian local ring, so $M$ is free if and only if $\operatorname{Ext}^1_R(M,k)=0$. Unfortunatelyy I don't know how to show any of these.

Please help.

  • $\begingroup$ Do you know if $\text{Ext}^{i}(\text{Hom}(M,M),R)=0$ for all $i\geq1$? $\endgroup$
    – Zeek
    Apr 26, 2020 at 20:11
  • $\begingroup$ @Zeek: I do not ... $\endgroup$
    – user521337
    Jun 23, 2020 at 13:34

1 Answer 1


Seems like this is just Theorem 2.1 in https://doi.org/10.1007/BF01229753


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .