Ranks of free submodules of free modules 
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Atiyah-MacDonald, exercise 2.11 

The following question came up during tea today.
Let $R$ be a commutative ring with an identity and let $M \subset R^n$ be a submodule.  Assume that $M \cong R^k$ for some $k$.  Question : Must $k \leq n$?
If $R$ is a domain, then this is obvious.  The obvious approach to proving the general result then is to mod out by the radical of $R$.  If the resulting map $M / \text{rad}(R) M \rightarrow (R / \text{rad}(R))^n$ were injective, then we'd be done.  However, I can't seem to prove this injectivity (I'm not even totally convinced that it's true).
Thank you for any help!
 A: For a proof using multilinear algebra, see Corollary 5.11 at 
http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf
A: This reduces to the question: is there an $R$-module
injection from $R^{n+1}$ to $R^n$. This is a matrix question:
is there a nonzero nullvector for an $n$-by-$n+1$ matrix $M$.
Clearly $M$ has a nullvector formed by the $n$-by-$n$ minors,
the trouble is that it could be zero. In that case we need to show
that an $n$-by-$n$ matrix $N$ with zero determinant has a nonzero nullvector.
Let $r$ be the determinantal rank of $N$: the size of the largest nonzero
subdeterminant of $N$. Then $r < n$. Let's assume the top left $r$ by $r$ submatrix
of $N$ has nonzero determinant. Let $N'$ be the top left $r+1$-by-$r+1$ submatrix
of $N$. Then the adjugate of $N'$ has a nonzero row.
Fill this out to a row vector of length $n$ by adding zeros. Then this
is a nullvector of $N$.
A: Here's a proof by Karl Dahlke:
Math Reference: A Free Submodule Embeds
The result can be generalized to infinite ranks.
