# A torsion-free sheaf of rank 1 on a surface

Let $X$ be a surface and $E$ be coherent sheaf on $X$. Now there is always a natural map $\mu:E\longrightarrow E^{\vee\vee}$. The kernel of this map is precisely the torsion subsheaf of $E$. Now if $E$ is torsion-free, this map $\mu$ is injective. Let us call $E^{\vee\vee}=F$.

Now on a surface, if $E$ is a torsion-free sheaf of rank 1, then we have an injective map $E\longrightarrow E^{\vee\vee}=F$. Now $F$ is a line bundle, therefore, tensoring by $F^{\vee}$, we get an injective map $E\otimes F^{\vee}\longrightarrow F\otimes F^{\vee}=\mathcal{O}_X$. Therefore, $E\otimes F^{\vee}$ is an ideal sheaf of a closed subscheme say $Z$. That is $E$ of the form $I_Z\otimes F$. The claim is that $Z$ is a codimension two subscheme. How do we see this? Any help will be appreciated!

A torsion-free sheaf is in fact free in codimension one, as the local rings are d.v.r.s and over those rings torsion free = free, so the natural map $E\rightarrow E^{\vee\vee}$ is an isomorphism in codimension one, implying that $E\otimes F^{\vee}\hookrightarrow \mathcal{O}_X$ is an isomorphism in codimension one. But then $E\otimes F^{\vee}$ must be the ideal sheaf of a codimension two subscheme.
• Ah! I see! $E\otimes F^\vee\longrightarrow\mathcal{O}_X$ is an isomorphism is codimension one. Why does that mean that $E\otimes F^\vee$ is an ideal sheaf of a codimension two subscheme? The subscheme is given by Supp$(\mathcal{O}_X/E\otimes F^\vee)$, why does that have codimension two? – gradstudent Apr 20 '15 at 12:25
• If we denote $E\otimes F^{\vee}$ by $I$, then we have $I_p\cong \mathcal{O}_{X,p}$ for every point $p$ of codimension one. So $\mathcal{O}_X/I$ is supported at finitely many points, which have codimension two in a surface. – Bernie Apr 20 '15 at 13:32