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Let $(A,m,k)$ be a local ring and let $M$ be a finite torsion $A$-module. Is ${\rm Tor}^A_1(M,k)$ finite over $k$?

I am aware that the conclusion holds for any finite $A$-module $M$ when $A$ in Noetherian, a reference is e.g. Bourbaki Algebre X.103, Exemple 3.

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    $\begingroup$ If $M=k$. then your Tor is finite dimensional iff $m$ is finitely generated (you can compute the Tor explicitely) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 5, 2011 at 22:25
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    $\begingroup$ No, you can have $m$ not finitely generated but $m=m^2$ (thus $m\otimes _Ak=0$ and $Tor_1(k,k)=0$). $\endgroup$
    – Tom Goodwillie
    Commented Dec 6, 2011 at 1:55
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    $\begingroup$ In some more detail: $Tor_1^A(k,k)=m/m^2$, so the answer to the question is no because $m/m^2$ need not be finite-dimensional over $k$. (But on the other hand there are examples in which $m/m^2=0$ even though $m$ is not a finitely generated module, so Mariano's "iff" should be just "if".) $\endgroup$
    – Tom Goodwillie
    Commented Dec 6, 2011 at 2:31
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    $\begingroup$ @Tom: This is a perfect answer, not just a comment :) $\endgroup$
    – Martin Brandenburg
    Commented Dec 6, 2011 at 7:33
  • $\begingroup$ Thanks to Mariano and Tom, I wanted the finiteness to hold so I forgot to check the special cases :) $\endgroup$
    – Ivan Tomasic
    Commented Dec 6, 2011 at 13:41

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