How can I find an element $x\not\in\mathfrak mM_{\mathfrak m}$ for every maximal ideal $\mathfrak m$

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?

Solution 1:

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.

Solution 2:

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.