I'm reading a book on algebraic curves, and at one point it says that if C is a smooth curve and f belongs to K(bar)(C)* for perfect field K, and if div(f)=0, then f has no poles. It's my understanding, that div(f) is the sum of order of f over various points P in C. So isn't it possible to have a function f with a pole and a zero of the same order at two different points P, and thus wouldn't the order still be 0?
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$\begingroup$ I have voted to close as no longer relevant. Reasonable, but very easy, question gets two spot-on answers. Let's move on... $\endgroup$– Pete L. ClarkCommented Mar 10, 2010 at 21:10
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2 Answers
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No, $div(f)$ is an element of the free abelian group generated by symbols $[P]$ as $P$ runs through the points of the curve. So a function $f$ having a simple pole at a point $P$ and a simple zero at a point $Q$ and is nonzero elsewhere satisfies $div(f)=-[P]+[Q]$. You are mistaking $div(f)$ for $\deg(div(f))$, the degree of the divisor of $f$ which is the integer obtained by adding up the order of $f$ at every point of the curve.
answered Mar 10, 2010 at 20:27
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div(f) is not the sum of the orders (in fact the sum of the orders is always zero). div(f) is a formal sum $\sum_P ord_P(f)P$ which is an element of the free abelian group generated by the points of C. The fact that div(f)=0 implies f has no poles is a trivial consequence of the definition.
answered Mar 10, 2010 at 20:26