Grothendieck group of a smooth curve

Question

Let $X$ be a smooth projective and geometrically irreducible curve over a field $k$. It an exercise in Hartshorne's book on "Algebraic Geometry" (p.149) that in the case that $k$ is algebraically closed the Grothendieck group $K(X)$ is isomorphic to $\textrm{Pic}(X)\oplus\mathbb{Z}$ (and the isomorphism is given by determinant and rank). Is this also the case when $X$ is not algebraically closed? It seems to me on the first glimpse that the same proof would work but then I am a bit insecure because why does Hartshorne then not state it for arbitrary fields?

Answer 1

Yes, the proof from exercise II.6.11 will show that for a regular projective curve $X$ over any field $k$ the Grothendieck group is $\operatorname{Pic}(X)\oplus\Bbb Z$ (regular is weaker than smooth and there is no need to add a geometrically irreducible hypothesis). One slight adjustment to the strategy as outlined by Hartshorne is over a non-algebraically closed field one may have to be a touch more careful with the skyscraper sheaf section of the argument - instead of saying

For any divisor $D=\sum n_iP_i$ on $X$, let $\psi(D)=\sum n_i \gamma(k(P_i))\in K(X)$, where $k(P_i)$ is the skyscraper sheaf $k$ at $P_i$ and 0 elsewhere...

one should amend this statement to $k(P_i)$ being the skyscraper sheaf with value the residue field of $P_i$ at $P_i$ and 0 elsewhere.

As to why Hartshorne would state things this way, Hartshorne spends a lot of time over algebraically closed fields, especially when doing more concrete geometry, and doesn't always seem to care so much about the non-algebraically-closed case. You'd have to ask him why, but it's possible that's just where his head was 50 years ago when he wrote the text.

(Let me also point out that the result holds in much greater generality: the result is true for any separated noetherian normal connected scheme of dimension one via essentially the same proof - some improved argument is needed to justify that such a scheme has the resolution property, but it's not such an obscure proof.)