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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?

user26857
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Mec
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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

2 Answers2

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Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$, and 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.

By CRT we have $R/\mathfrak{m}_1\cdots\mathfrak{m}_n\simeq R/\mathfrak{m}_1\times\cdots\times R/\mathfrak{m}_n$. Tensoring with $M$ we get $M/(\mathfrak{m}_1\cdots\mathfrak{m}_n)M\simeq M/\mathfrak{m}_1M\times\cdots\times M/\mathfrak{m}_nM$. Since $M/\mathfrak{m}_iM\simeq M_{\mathfrak{m}_i}/\mathfrak{m}_iM_{\mathfrak{m}_i}$, the modules $M/\mathfrak{m}_iM$ are free of the same rank $l$, and therefore we have an isomorphism $(R/\mathfrak{m}_1\cdots\mathfrak{m}_n)^l\simeq M/(\mathfrak{m}_1\ldots \mathfrak{m}_n)M$. Lifting this isomorphism to a map $R^l\to M$ and using Nakayama we can show that this map is an isomorphism, too.

user26857
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wxu
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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
  • @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 07:51
  • @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 07:58
  • @YACP, the projectivity(Hence flat) of $M$ gives the second sequence is exact. Another reason to know the second sequence is exact is that the first exact sequence is split. – wxu Jan 27 '13 at 08:01
  • @YACP, Anyway, since we have a surjective $R$-morphism $R^l\to M$. Then as you said we can localize at every maximal prime ideal. So we have a surjective map $R^l_{m_i}\to M_{m_i}$. Now $M_{m_i}$ is free of rank $l$! Then this map must be an isomorphism as a general fact(or a lemma)(e.g., cf. Atiyah's chap3 exercise 15 in "an introduction to commutative algebra"). We might be confused in this proof where do we use the condition that $M$ is projective? But this is another question. – wxu Jan 27 '13 at 08:17
  • If all the localizations of $R^l\to M$ are isomorphisms, then, of course, $R^l\to M$ is also an isomorphism. Where we use now that $M$ is projective? In my opinion we use only that the localizations of $M$ are free of the same rank. –  Jan 27 '13 at 11:10
  • @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." – wxu Jan 27 '13 at 17:27
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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.