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Let K be an algebraically closed field. Is it true that any endomorphism $\mathbb{P}_K^n\to \mathbb{P}_K^n$ has a fixed closed point?

For $K=\mathbb{C}$ the (oriented) topological version is a classical result and the proof uses Lefschetz fixed point theorem.

If it's true for any algebraically closed field, can it also be generalized to arbitrary field or perhaps to any ring?

Kamil
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  • I added a proof but deleted it because it assumed $n=1$. It's true for all $n$ though (the point is that the diagonal and the graph of your map have to intersect in $\mathbb{P}^n \times \mathbb{P}^n$) and false for non-algebraically closed fields (e.g. when $n = 1$ and over $\mathbb{F}_2$ just shuffle the only three rational points). – hunter Aug 11 '22 at 14:49
  • @hunter that still has a fixed closed point, it's just not a $\Bbb F_2$-rational fixed point. – KReiser Aug 11 '22 at 18:20
  • @KReiser if he's not asking about rational points the result is the same since you can compute after base change to the algebraic closure – hunter Aug 11 '22 at 18:21
  • @hunter I meant any closed point. So how do you prove that those two graphs in $\mathbb{P}^n \times \mathbb{P}^n$ intersect? – Kamil Aug 11 '22 at 23:14
  • @Kamil there is a sketch of the computation of the Chow ring of that product in Example 1 here. From there you can compute that any two graphs meet. I'll bet there is an easier proof that I don't know, though! http://www.math.columbia.edu/~chaoli/docs/IntersectionTheory.html#sec2 – hunter Aug 12 '22 at 18:33

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The Chow ring of $\mathbb{P}^n \times\mathbb{P}^n$ can be written as $$ \mathrm{CH}^\bullet(\mathbb{P}^n) \cong \mathbb{Z}[h_1]/h_1^{n+1} \otimes \mathbb{Z}[h_2]/h_2^{n+1}. $$ The class of the diagonal in it is $$ [\Delta] = 1 \otimes h_2^n + h_1 \otimes h_2^{n-1} + \dots + h_1^n \otimes 1. $$ Similarly, the graph $\Gamma$ of an endomorphism of degree $d$ has the class $$ [\Gamma] = 1 \otimes h_2^n + d h_1 \otimes h_2^{n-1} + \dots + d^n h_1^n \otimes 1. $$ Their intersection product is equal to $$ [\Delta] \cdot [\Gamma] = d^n + \dots + d + 1. $$ As it is non-zero, the intersection is nonempty, hence the endomorphism has a fixed point.

Sasha
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