# Is the kernel of a morphism of locally free sheaves still a locally free sheaf? [closed]

If we assum the underlying scheme is Notherian and reduced.

-
If $A$ is a Noetherian commutative domain, is the kernel of a map $A^n\to A^m$ always projective? –  Mariano Suárez-Alvarez May 3 '11 at 5:06
Think about k[[x,y,z]] or look up projective dimension of a ring depending upon your tastes. This is true for a ring of projective dimension 1 or 2 though so, in scheme talk, smooth curves or surfaces over a field. –  Daniel Pomerleano May 3 '11 at 5:43
Doesn't this work: take $X = \mathbb C$, the sheaf $L$ as the trivial line bundle, and look at $\phi : L \to L$ which sends $(x,v) \mapsto (x, xv)$. Then $Ker \phi$ is coherent, but not locally free. –  Gunnar Magnusson May 3 '11 at 6:02
Dear Gunnar, no, this doesn't work. The morphism of sheaves $m:\mathcal O \to \mathcal O$ given by multiplication by $x$ is injective, hence the kernel of $m$ is zero and thus free. However if you take the corresponding vector bundle, your $L$, the induced morphism on fibers $m[0]:L[0]\to L[0]$ is no longer injective, as you correctly note. The point is that taking fibers amounts to tensoring with $\mathcal O_0/\frak m_0=\mathbb C$, which is not a flat $\mathcal O_0$-module and so does not preserve injectivity. –  Georges Elencwajg May 3 '11 at 7:17
Why was this closed? It is a valid mathematical question, and I was interested in the answer. –  Drew Sep 16 '13 at 17:15
Let $X$ be a smooth quasi-projective variety over a field of dimension $n$ and $Z\subseteq X$ a subvariety such that the depth of $\mathscr O_Z$ at a fixed (closed) point of $X$ is $d$. Assume that $n-d\geq 3$, i.e., the projective dimension of $\mathscr O_Z$ at that closed point of $X$ is at least $3$. Then consider a locally free sheaf that surjects onto the ideal sheaf of $Z$. That gives a morphism to $\mathscr O_X$ and its kernel cannot be locally free, because then the projective dimension of $\mathscr O_Z$ at that closed point of $X$ would be $2$.