Solutions to Hartshorne's Algebraic Geometry/Riemann-Roch Theorem

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Let X have genus g. Since X is dimension 1, there exists a point Q ∈ X, Q =/= P. Pick an n > max{g, 2g − 2, 1}. Then for the divisor D = n(2P − Q) of degree n, l(K − D) = 0 (Example 1.3.4), so Riemann-Roch gives l(D) = n + 1 − g > 1. Thus there is an effective divisor D' such that D' − D = (f). Since (f) is degree 0 (II 6.10), D' has degree n, so D' cannot have a zero of order large enough to kill the pole of D of order 2n. Therefore, f is regular everywhere except at P.