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II.1.21

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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  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.

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