II.1.21

This has H = ?, V = ?

HAPPY
Just parts d,e

Work on the level of presheaves. You clearly have an injection $$O \to K$$. And there are natural restriction maps $$ K \to K/O_p$$. So there is at least a map $latex K \to \prod_p K/O_p

Now let $$f$$ be a rational function on an open set $$U$$. Since we're dealing with $$\mathbb{P}^1$$, after some automorphism we can assume there are no poles at infinity, so write $$f = \frac{\prod (x - a_i)}{\prod (x - b_i)}$$ the point is that is can poles at only finitely many points $$ p_1, ..., p_n$$ so if $$p$$ is not one of these points, then $$f$$ restricts to an element of $$O_p$$ hence maps to zero in $$K/O_p$$.

I.e. the map $$K \to \prod K/O_p$$ is nonzero at only finitely many places, so factors through the direct sum, so there is a factorization on the level of presheaves, so by universal property of sheafification you get $$ O \to K \to \oplus K/O_p$$ and its exact on stalks which gives the result.

For part e, just take a finite number of prescribed principle parts, and sum them, then this will constitute a lift, showing the required map is surjective.