# Given a surjective linear mapping of free modules how do you show the corresponding matrix has an invertible minor?

The following post can is related to part c) of this problem http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=124137 and boils down to some issues I am having with use of wedge product.

Let $V,W$ be two free $R$ modules. Suppose that $f: V \rightarrow W$ is a linear mapping and let $A$ be the matrix corresponding to the linear map $f$.

How do you show if $f$ is onto and if the ideal $\subset R$ generated by the non-invertible elements is distinct from $R$ there exists a minor of $A$ which is invertible and whose order is the same dimension as $W$ and $dim(V) \geq dim(W)$?

The reason I metioned wedge product is because we are supposed to consider the power $f\wedge f \wedge \ldots \wedge f$ where we have taken $\dim(W)$ wedge products of f.

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