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Let $R$ be a PID, let $r \in R$ generate a maximal idal $(r)$ in $R$, and let $k = R/(r).$ Let $M$ be a module over $R$ and let $T = \{ m \in M : \text{$r^n \cdot m = 0$ for some $n > 0$}\}.$ Is it true that $$\dim_k\ M \otimes_R k = \dim_k (M/T) \otimes k \ \text{?}$$

I can prove the $\geq$ part and, for finitely generated $M$, the opposite inequality – but not in general.

Jendrik Stelzner
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Adam
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  • If $R$ is a field, then $r = 0$ and thus $T = M$. – Marktmeister Dec 28 '22 at 20:15
  • @Marktmeister Thanks, indeed, we get a sharp inequality then. But that seems very special. Is there an example when $r\ne 0$? – Adam Dec 28 '22 at 20:31
  • Take any $R$ and $r$ and let $M = R/(r) = k$, then $T = M$ as well, so the right hand side is zero, but the left hand side is not. – Marktmeister Dec 28 '22 at 20:36
  • @Marktmeister Many thanks! The interesting thing about your examples of $M$ is that they have something of an order of $|k|$ genberators. Btw, I just posted a slighly different version of this question: https://math.stackexchange.com/questions/4607307/modules-over-cx – Adam Dec 28 '22 at 21:37
  • $M/T$ appears in $N=M\otimes_R R[r^{-1}]$ ($T=\ker(M\to N)$) not in $M\otimes_R R/(r) = M/rM$ – reuns Dec 28 '22 at 22:35

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