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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 projective module such that $M_{\mathfrak m_i}$ has the same rank for every $i$, then $M$ is free.

I was trying to solve this exercise and I reduced the problem to prove that

there exists an element $x\in M$ such that $\frac{x}{1}\not\in\mathfrak m_iM_{{\mathfrak m}_i}$ for every $i$.

But 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! –  rschwieb May 28 '12 at 23:20
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2 Answers

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.

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I don't know how to reduce to the case you said. However, 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, Suppose $e_1,e_2,\ldots,e_l$ is the canonical basis of $R^l$. Under the map $R^l\to (R/m_1\cdots m_n)^l\to M/m_1\cdots m_n M$, $e_i$ map to $x_i$ say. Pick any preimage $y_i\in M$ of $x_i$, our map $\varphi: R^l\to M$ sending $e_i$ to $y_i$. Then we should show this map is an isomorphism. First, by Nakayama's lemma $\varphi$ is surjective. Then, notice that $M$ is projective, we can show that $\varphi$ is injective. –  wxu Jan 26 '13 at 16:40
    
@YACP, Let $K$ be the kernel. We have a split exact sequence $0\to K\to R^l\to M\to 0$, then tensoring $R/m_1\cdots m_n$. We obtain an exact sequence $0\to K/m_1\cdots m_nK\to (R/m_1\cdots m_n)^l\to M/m_1\cdots m_n M\to 0$. Hence $K/m_1\cdots m_n=0$, but $K$ is finitely generated $R$-module(since it is a direct summand of $R^l$) and apply Nakayama's lemma, we know $K=0$. –  wxu Jan 26 '13 at 16:59
    
Dear @YACP, Where we live has been very late in the night. And I should go to bed. I hope there is no mistakes here :).. –  wxu Jan 26 '13 at 17:21
    
@YACP, $K\oplus M=R^l$! So $K$ can be generated by $l$ elements. Why $K\oplus M=R^l$? Because $M$ is projective by hypothesis. –  wxu Jan 27 '13 at 7:51
    
@YACP, But there might be some confusion about what I write $K\oplus M=R^l$, this exactly the meaning of what I say the first exact sequence of $R$-modules is split. The symbol $"="$ here means there is some isomorpism. So you have a projection $R^l\to K\oplus M\to K$, it follows $K$ is finitely generated. –  wxu Jan 27 '13 at 7:58
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