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Let $f:X\to Y$ be a non-constant morphism of smooth projective connected curves over $\mathbf{C}$ (or compact connected Riemann surfaces).

Suppose that $X=Y$ and that $f$ is not the identity.

Why are the fixed points of $f$ isolated?

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If they weren't isolated, they would have an accumulation point, so by the identity theorem $f$ would be the identity. –  Qiaochu Yuan Feb 4 '12 at 17:55
    
Alternately, a smooth connected projective curve over $\mathbb{C}$ is irreducible, and if $f$ is not the identity then its set of fixed points is a proper subvariety, so a finite set of points. –  Qiaochu Yuan Feb 4 '12 at 18:01
    
So what if I replace $X$ by a surface (a two-dimensional variety)? Then, a proper subvariety could be an entire curve so they don't have to be isolated. I don't see how the argument in the complex case breaks down. The identity theorem still holds to give that $f$ is the identity..right? –  Ali Feb 4 '12 at 18:22
    
Actually, come to think about it, to see that the fixed points form a proper subvariety one uses the fibre product interpretation, right? The scheme of fixed points is $X_f\times_X X$, where the first $X$ goes with $f$ and the second with the identity. –  Ali Feb 4 '12 at 18:28
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By surface what do you mean exactly? (the term "Riemann surface" refers to a complex curve, a complex variety of complex dimension one) There are tons of smooth connected projective complex surfaces (varieties of complex dimension two) with automorphisms with non-isolated sets of fixed points! –  Mariano Suárez-Alvarez Feb 4 '12 at 18:29
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