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Let $A$ be a local ring and $L$ and $F$ be free $A$-modules of finite rank. Consider bases $e_1,\dots,e_m$ and $f_1,\dots,f_n$ of $L$ and $F$. To a homomorphism $u:L\rightarrow F$ associate the matrix $M=(a_{ij})$, with $u(e_j)=\sum_{i=1}^n a_{ij}f_i$.

Show that if $u$ is surjective, then $M$ has an invertible $n$-minor.

I don't know how to get the invertible $n$-minor. Because $A$ is a local ring, let $\mathfrak m$ be the maximal ideal. Then $F/\mathfrak mF$ is a vector space, and $\overline{u}:L\rightarrow F/\mathfrak mF$ is surjective. But I don't know what to do next.

Peter
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  • The surjectivity of $u$ is equivalent to the fact that the $n$-minors of $M$ generate the unit ideal (see http://math.stackexchange.com/a/133077/586 for a proof). Now use the locality of $A$. – darij grinberg Sep 03 '13 at 20:43

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