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Let $k$ be a field of characteristic zero (I do not mind to assume that $k \in \{\mathbb{R},\mathbb{C}\}$), and let $R=k[x,y]$ be the $k$-algebra of polynomials in two variables $x,y$. Let $f$ be a $k$-algebra endomorphism of $R$.

(I) Is there a fixed-point theorem that can be applied to $f$? (II) If so, then is it possible to obtain a fixed point $\in R-k$? (the elements of $k$ are always fixed points for a $k$-algebra endomorphism $f$).

Remarks:

(1) I have tried to consider $k[x,y]$ as a complete metric space, see this question, but I guess that $f$ is not a contraction in that metric. So, is there another metric that makes $f$ a contraction? (probably yes? But it may be difficult to guarantee that $f$ has a fixed point $\in R-k$).

(2) The answer probably depends on the given $f$, since, for example, it seems that $f:(x,y) \mapsto (x^2,y^2)$ does not have fixed points other than the elements of $k$.

(3) Here is a list of fixed point theorems.

(4) The following questions seem relevant: a and b (what if we consider our given endomorphism as a multiplicative group morphism?).

(5) A similar question of mine is this (the fourth remark there says that the existence of such a fixed point implies the two-dimensional Jacobian Conjecture).

Thank you very much!

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  • $\begingroup$ I don't think it is possible in any nice way, because you cannot keep the $k$-vectorspace structure compatible with a complete metric (Since $k[x,y]$ is a countable union of nested finite-dimensional subspaces contradicting Baire). $\endgroup$ – user10354138 Sep 20 '18 at 14:21
  • $\begingroup$ Thank you very much. Can you please elaborate (in an answer)? $\endgroup$ – user237522 Sep 20 '18 at 14:34
  • $\begingroup$ (If I am not wrong, if we work in the formal power series $k[[x,y]]$, then the 'natural' metric is complete, but $f$ is not a contraction). $\endgroup$ – user237522 Sep 20 '18 at 14:40
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I very much doubt you can approach it in this fashion.

If there is a fixed point, then the fixed set would have to be a subalgebra, so in particular, a vector subspace, and this endomorphism of course respects this structure so you probably want a metric that also respects this.

In particular, you want proper vector subspaces to be closed and nowhere dense. But this yields a contradiction when you demand the metric to be complete, because we can write $k[x,y]$ as a countable union of proper (hence closed nowhere dense) subspaces $$ k[x,y]=\bigcup_{d=0}^\infty \operatorname{span}_k\{x^iy^j\mid i+j\leq d\} $$ but Baire category theorem says a complete metric space is not meagre.

A few words about the formal power series ring

If you decide to work with the formal power series ring $k[[x,y]]$ instead, then obviously your endomorphism $f$ must send the nilpotent $x,y$ to some nilpotent elements. This means in particular that we can reduce to looking at fixed elements of each $(x,y)/(x,y)^m$ and patching each graded piece together, like in Hansel's lemma.

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