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Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$. Let $M$ be a finitely generated module. Then there exists an element $x\in M$ such that $\frac{x}{1}\not\in\mathfrak m_iM_{{\mathfrak m}_i}$ for every $i=1,\dots,n$.

I cannot prove that such an element there exists. I was trying to prove it by induction on $n$. If $n=1$ it is true by Nakayama, so suppose it is true for $n-1$. Then for every $i$ I can find an $x_i\in M$ such that $\frac{x_i}{1}\not\in \mathfrak m_jM_{\mathfrak m_j}$ for every $j\neq i$. If $\frac{x_i}{1}\not\in \mathfrak m_iM_{\mathfrak m_i}$ for some $i$ we are done, so suppose $\frac{x_i}{1}\in\ m_iM_{m_i}$ for every $i$. I don't know how to go on, could you help me?

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    Hmm, well no unit is in any proper ideal in any ring, so I begin to wonder about this reduction. Can you put up the details of your reduction, as well? Thanks!2012-05-28

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We can use CRT.

We have $M/\mathfrak{m}_iM\cong M_{\mathfrak{m}_i}/\mathfrak{m}_iM_{\mathfrak{m}_i}$ and $R/\mathfrak{m}_1\cdots\mathfrak{m}_n\cong R/\mathfrak{m}_1\times\cdots\times R/\mathfrak{m}_n$. Tensoring this with $M$ we get $M/\prod_i\mathfrak{m}_iM\cong M/\mathfrak{m}_1M\times\cdots\times M/\mathfrak{m}_nM$. Since $M/\mathfrak{m}_iM$ has the same rank $l$, we may find a map $(R/\mathfrak{m}_1\cdots\mathfrak{m}_n)^l\cong M/\mathfrak{m}_1\ldots \mathfrak{m}_nM$. Lifting this map to $R^l\to M$ and applying Nakayama we may show this map is an isomorphism. Done.

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    @YACP, Yes, you are right. In fact, we have the following true statement. "Let $M$ be an $R$-module. Then $M$ is finite projective if and only if $M$ is finite and for every prime $\mathfrak{p}$ the module $M_{\mathfrak{p}}$ is free and the function $\rho_M: Spec(R)\to \mathbb{Z}$, $\mathfrak{p}\mapsto dim_{\kappa(\mathfrak{p})}M\otimes_R\kappa(\mathfrak{p}) $ is locally constant in the Zariski topology."2013-01-27
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The question about the finitely generated projective module is exercise 2.40(1) from Lam, Exercises in Modules and Rings, and a proof can be found there.