# Is the Kernel locally free?

Suppose, on a smooth projective complex variety $X$ that we are given an effective divisor $D$ and $A\in\mathrm{Pic}(D)$ a globally generated line bundle on $D$ with $r$ independent sections. Then we have a surjection $\mathcal{O}_X^{\oplus r}\to A$ (here we view $A$ as a torsion sheaf on $X$ with support on $D$) and thus an exact sequence $$0\to F\to\mathcal{O}_X^{\oplus r}\to A \to 0$$ where $F$ is the kernel sheaf. Is $F$ locally free?

What about the same situation but with a general vector bundle $V$ in place of $\mathcal{O}_X^{\oplus r}$ ?

The question is local, so let us look at everything over the local ring at a point. We then have a regular local ring $R$, a nonzero element $f\in R$ (the local equation of $D$), and an $R$-module $A$ which is annihilated by $f$ and is invertible as a module over $R/(f)$. We want to know whether $A$ has projective dimension $\leq 1$ over $R$ (because then the kernel of any surjection from a projective module to $A$ is projective). Since $A$ is invertible over $R/(f)$, it is a direct summand of a free module over $R/(f)$, so it suffices to show that $R/(f)$ has projective dimension $\leq 1$ over $R$. But this is easy, since $0\to R\stackrel{f}{\to} R\to R/(f)\to 0$ is a projective resolution of $R/(f)$ (since $f$ is nonzero and $R$ is a domain). Thus $R/(f)$ has projective dimension $\leq 1$, and hence so does $A$.
More generally, this argument shows that the kernel of any surjection from a locally free sheaf on $X$ to a locally free sheaf on $D$ is locally free.
• Dear Eric, thanks for your answer. A small question: why is the kernel of any surjection from a projective module to $A$ projective when $A$ has projective dimension 1? – Heitor Fontana Dec 3 '15 at 13:04
• If $0\to F\to V\to A\to 0$ is exact and $V$ is projective, then by the long exact sequence in Ext, $\operatorname{Ext}^i(F,M)=\operatorname{Ext}^{i+1}(A,M)$ for all $i>0$ and all modules $M$. So if $\operatorname{Ext}^j(A,M)=0$ for $j>1$, then $\operatorname{Ext}^i(F,M)=0$ for $i>0$. – Eric Wofsey Dec 3 '15 at 14:36