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Let $R$ be a commutative ring with only finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_r$. Let $M$ be a finitely generated $R$-module. Then $$\mu_R(M)=\max\{\dim_{R/\mathfrak m_i}M/\mathfrak m_iM\mid 1\leq i\leq r\},$$ where $\mu_R(M)$ is the minimum number of generators of $M$ as $R$-module.

How can I prove this?

Of course the inequality $\geq$ is trivial; what I want to prove is that $\dim_{R/\mathfrak m_i}M/\mathfrak m_iM\leq\mu_R(M)$.

I was trying to prove it first if $R$ is a finite product of fields but I wasn't succesful; any help?

  • 1
    I have a strong feeling [wxu's answer](http://math.stackexchange.com/questions/150944/how-can-i-find-an-element-x-not-in-mm-m-for-every-maximal-ideal-m) in your previous problem might hold the key, although I have not checked.2012-05-29
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    I change \mathrm{dim} to \dim and \mathrm{max} to \max. Both are standard usage. One of the differences in the outcome is that in a "displayed" setting, \max_{x\in S} will look like this: $\displaystyle\max_{x\in S}$, with the subscript directly below $\max$ (that last feature doesn't work in an "inline" setting). Another difference is that in things like $a\max b$, you automatically get proper spacing before and after $\max$, so you don't need to add spacing by hand.2012-05-29

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