$\mathcal O_X^n$ is a coherent sheaf for each $n$

Question

I am reading a proof on these notes about $\mathcal O_X$ modules but there are some details I cannot understand.

Proposition: The sheaf $\mathcal O_X^n$ is coherent

Obviously, it is of finite type. But then he says that we can suppose $U=X$ and work globally. I believe this is because in order to show that the sheaf is coherent, one has to compare with some $\mathcal O_X^m$, but I am not so sure.

Anyway, let's continue with the proof. He does the case $n=1$, so let $L:\mathcal O_X^a\to \mathcal O_X$ be a morphism of $\mathcal O_X-$modules, given by $L(f_1,dots,f_a)=\sum \lambda_i f_i$, with some $\lambda_i\in \mathcal O_X(X)$. We shall show that its kernel is, locally, of finite type.

Let $x\in X$ and $(V,z)$ a chart centered in $x$. If $L=0$, then $ker L = \mathcal O_X^a$ and we are done.

If not, and here is where almost all my concerns appear, we can suppose that $v_x(\lambda_1)\leq v_x(\lambda_i)$ for all $i$, and then $\lambda_i= z^e\mu_i$ with $\mu_1(x)\neq 0$ and, up to restricting $V$, $\mu_1$ is invertible, and $(f_2,\dots,f_a)\mapsto (\phi, f_2,\dots,f_a)$ is an isomorphism between $\mathcal O_X^{a-1}$ and the kernel, and we would be done.

This has been done rather quickly. I guess the assumption of $\lambda_1$ with minimal valuation has to be something we can assume without loss of generality, and $z$ could act as a generator of the maximal ideal $m_{X,x}$ of the stalk. But I don't get how it follows that $\mu_1(x)\neq 0$... which is quite relevant.

Answer 1

What is your definition of a coherent sheaf, and do you work in algebraic, or in analytic category?

Assume that your category is algebraic.

One of the two (equivalent) definitions of a coherent sheaf I know is that this is such a sheaf of $O_X$-modules that there is an open affine cover $U_i$ such that

(a) $F$ restricted to $U_i$ is a sheaf associated with the module $M_i = F(U_i)$ on $U_i = \mathop{\rm Spec} k[U_i]$;

(b) $M_i = F(U_i)$ is a $k[U_i]$-module of finite type.

In this definition, your question seems tautological? $M_i = k[U_i]^n$

Reference: Mumford, Lectures on curves on algebraic surface.

Answer 2

Locally, one can work in a chart that looks like a complex disk, in which your point $x$ is identified with 0. Then $\lambda_i$ are just holomorphic functions in $z$. The assumption that $\lambda_1$ vanishes to the lowest order at $0$ can be achieved by simply reordering the $\lambda_i$. Then if we take $e$ equal to the order of the vanishing of $\lambda_1$, then $\lambda_1$ can be written in the form $z^e\mu_1(x)$, where $\mu_1(0)\neq 0$: you just write $\lambda_1$ as a series in z, its lowest term is some $az^e$ and you factor the $z^e$ out, then $\mu_1(0)=a$.