Nonsurjective map to second dual module

Question

For a free finitely generated module $M$ the canonical map $i: M\to M^{**}$ is an isomorphism. I know an example of a finitely generated module for which this map is not injective. The map is also not surjective, for example, for $M=l_\infty$, which is an infinite-dimensional vector space. But is there a finitely generated module over some (preferably commutative) ring for which this map is not surjective?

Answer 1

Let $k$ be a field and $R=k[x,y]$, and let $M$ be the ideal $(x,y)\subset R$. Note that $M^*\cong R$: given any $r\in R$, we have a homomorphism $f:M\to R$ given by $f(x)=rx$ and $f(y)=ry$, and every homomorphism $M\to R$ has this form (since we must have $yf(x)=xf(y)$ and so $f(x)$ is divisible by $x$ and we can take $r=f(x)/x$). Thus $M^{**}\cong R$. Moreover, the canonical map $M\to M^{**}\cong R$ is just the inclusion map $M\to R$, which is not surjective.

Another example: let $R=k[x,y]/(x^2,xy,y^2)$ and let $M=R/(x,y)$ (so $M$ is $k$ with $x$ and $y$ acting as $0$). Then $M^*\cong M^2$, since a homomorphism $M\to R$ can map a generator of $M$ to any $k$-linear combination of $x$ and $y$. We thus have $M^{**}\cong(M^2)^*\cong M^4$, and the canonical map $M\to M^{**}$ cannot be surjective since $M$ is $1$-dimensional over $k$ and $M^{**}$ is $4$-dimensional.