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In the book "Algebraic Geometry" of Hartshorne, in corollary 6.10 chapter 2, he wants to prove that principal divisors have degree zero. To do this he takes $f\in K(X)^*$ and this gives a morphism $\phi: X\rightarrow \mathbb{P}^1$. Now he says that $(f)=\phi^*(0-\infty)$, why is this true?

SC30
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    That's the definition of the divisor $(f)$. :P If you prefer, it's $\phi^(0) - \phi^(\infty)$. – Ted Shifrin Apr 21 '17 at 22:01
  • $(f)=\sum v_P(f)\cdot P$, why it is equal to $\sum_{P\mapsto 0}v_p(t)\cdot P -\sum_{P\mapsto \infty}v_p(t')\cdot P $ where t,t' are local parameter at 0 and $\infty$? – SC30 Apr 21 '17 at 22:05
  • Write things down in local coordinates at the various points $P$ to check. But the preimage of $0$ is precisely the points $P$ that map to $0$, counted with multiplicity $\nu_P$. – Ted Shifrin Apr 21 '17 at 22:24
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    You're right, it is the definition. What you do is saying $(f)=\sum (order vanishing of f at P)\cdot P$, so when the order is positive we have a zero of f, so those points are $\phi^*(0)$. When the order is negative we have a pole. So it was obvious. – SC30 Apr 22 '17 at 00:14
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    Well, lots of math is obvious once you sort it out. Well done. :) – Ted Shifrin Apr 22 '17 at 00:16

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