Why do we need the infinite field hypothesis in this cohomology calculation?

Question

I've just finished my very first calculation with sheaf cohomology. It's exercise III.2.1(a) in Hartshorne, and it says

Let $X = \mathbb{A}_K^1$ be the affine line over an infinite field $K$. Let $P, Q$ be distinct closed points of $X$ and let $U = X\backslash\left\{P, Q\right\}$. Show that $H^1(X,\mathbb{Z}_U)\neq 0$ where $\mathbb{Z}_U$ is the sheaf $\mathbb{Z}\vert_U$ extended by zero.

A quick description of my proof: $X$ is a Noetherian topological space of dimension $1$, so Grothendieck's vanishing theorem says $H^i(X,\mathbb{Z}_U) = 0$ for all $i>1$. We have a short exact sequence of sheaves on $X$:

$$0\to \mathbb{Z}_U \to \mathbb{Z} \to \mathbb{Z}_Y \to 0$$

where $\mathbb{Z}_Y$ is the extension by zero of the constant sheaf $\mathbb{Z}$ on the closed set $Y = \left\{P, Q\right\}$. Since $X$ is irreducible the constant sheaf $\mathbb{Z}$ is flabby and therefore $H^1(X, \mathbb{Z}) = 0$. Then the long exact sequence in cohomology is

$$0\to \mathbb{Z}_U (X)\to \mathbb{Z}(X)\to\mathbb{Z}_Y (X) \to H^1 (X,\mathbb{Z}_U)\to 0$$

Now $\mathbb{Z}_Y (X) = \mathbb{Z}\vert_Y (Y) = \mathbb{Z}\oplus\mathbb{Z}$. Furthermore $\mathbb{Z}(X) = \mathbb{Z}$ because $X$ is irreducible. If $H^1(X,\mathbb{Z}_U)=0$ then we'd have a surjection $\mathbb{Z}\to \mathbb{Z}\oplus\mathbb{Z}$. Therefore it must not be zero.

Although this seems reasonable to me, I cannot see why the assumption that $K$ is infinite is required. Can anyone expand on this to show what's gone wrong?

Answer 1

Here are two possible reasons Hartshorne wanted $K$ to be infinite:

1. If you interpret $\mathbf{A}^1_K$ to mean the subspace of the scheme $\operatorname{Spec} K[x]$ consisting of $K$-valued points, then you can think of $\mathbf{A}^1_K$ as a one-dimensional $K$-vector space. If $K$ is infinite, then $\mathbf{A}^1_K$ is indeed irreducible, but if $K$ is finite, it would not be.

2. If you interpret the problem to be about (as you seem to do) the scheme $\operatorname{Spec} K[x]$, then I think in part (a) of the problem you would be okay. The issue comes in part (b) when you need "hyperplanes in suitably general position", since you need the hyperplanes to intersect in a certain way; see the proof of a different answer of mine. If $K$ is finite, then I think there's an issue as to how to find the necessary general configuration of hyperplanes, which would be more obvious if you follow user4571's suggestion to use induction.