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Suppose $R$ is a local Noetherian domain, and $M$ is a finitely generated $R$-module. Furthermore, let's suppose there exists $k>0$ such that $$ \dim_{R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}}M_\mathfrak{p}/\mathfrak{p}M_\mathfrak{p}=k $$ for any prime $\mathfrak{p}\in\operatorname{spec}(R)$.

I've been curious though, how does this imply that $M$ is in fact a free module? I figure you want to extract some basis for $M$ from a generating set $\{x_1,\dots,x_n\}$, and this is where the dimension condition comes in. However, after localizing at $\mathfrak p$ and taking quotients, I'm losing sight of how to connect to the two ideas.

Can someone explain why $M$ is free here? Thank you.

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    $\begingroup$ Do you know Nakayama's lemma? It doesn't immediately solve your problem, but it seems related. $\endgroup$ Mar 3, 2012 at 1:38
  • $\begingroup$ @DylanMoreland: You were indeed correct, if my proof is correct. $\endgroup$
    – Rankeya
    Mar 3, 2012 at 4:26
  • $\begingroup$ @Rankeya Ah, my eyes skipped over the domain condition, so I didn't get anywhere. I'll take a look at your proof! $\endgroup$ Mar 3, 2012 at 4:35
  • $\begingroup$ This is given as an exercise in Lang's Algebra, page 444, with a hint. Apparently, it is enough for the ring to have no nilpotent elements. $\endgroup$
    – Mike B
    Mar 3, 2012 at 5:12

1 Answer 1

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Let $m \subset R$ be the unique maximal ideal of the Noetherian ring. By hypothesis, $M/mM$ has dimension $k$ as an $R/m$ vector space. So, let $x_1 + mM, ..., x_k +mM$ be a basis for $M/m$. Then one can show using Nakayama's lemma (see proposition 2.8 in Atiyah-Macdonald) that $M = <x_1,...,x_k>$.

Let $p \in Spec(R)$. Then $M_p/pM_p$ is generated by $\{\frac{x_1}{1} + pM_p, ..., \frac{x_k}{1} + pM_p\}$ as a $R_p/pR_p$ module, hence is a basis for $M/pM_p$.

Let $r_1x_1 + ... + r_kx_k = 0$ in $M$ $(r_i \in R)$. Then, for all $i$, $\frac{r_i}{1} + pR_p = 0$ in $R_p/pR_p$. Thus, $\frac{r_i}{1} \in pR_p$. Hence, $r_i \in p$. But, $p$ was an arbitrary prime ideal, and since $R$ is a domain $(0)$ is prime. So, $r_i \in (0)$, and we are done.

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  • $\begingroup$ If this proof is correct note that this method of proof will work if $R$ is a Noetherian reduced ring. $\endgroup$
    – Rankeya
    Mar 3, 2012 at 4:14
  • $\begingroup$ Also, I do not think one needs to assume that $R$ is Noetherian. I have a feeling I did not use Noetherianity anywhere in this proof. Proposition 2.8 from Atiyah-Macdonald works for local rings in general. So, your statement is true for any finitely generated module over a local reduced ring. $\endgroup$
    – Rankeya
    Mar 3, 2012 at 5:17
  • $\begingroup$ Thanks Rankeya. By the way, in the second paragraph, did you mean to say $R_p/pR_p$ module, not $R_p/pM_p$ module? $\endgroup$
    – Buble
    Mar 4, 2012 at 12:51
  • $\begingroup$ Yes, indeed. Thanks for pointing it out. $\endgroup$
    – Rankeya
    Mar 4, 2012 at 17:10

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