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Let $w \in \mathbb{C}[x,y]-\mathbb{C}$ and let $u \in \mathbb{C}[x,y]-\mathbb{C}[w]$.

Is it possible to find a $\mathbb{C}$-algebra endomorphism $f$ of $\mathbb{C}[x,y]$ such that $f(w)=w$ and $f(u) \neq u$?

There are special cases having a positive answer, for example: $w=x^2+y^2$, $u=x$; in this case, one can take $f=\alpha: (x,y) \mapsto (y,x)$ the exchange involution (more generally, if $w$ is symmetric with respect to some involution $\iota$ on $\mathbb{C}[x,y]$, and $u$ is non-symmetric with respect to that involution $\iota$, then $\iota(w)=w$ and $\iota(u) \neq u$). However, in my question there is no such information about $w$ and $u$.

Is it hopeless to try to find such $f$ or perhaps it is possible to apply one of the many fixed point theorems to solve my question in the affirmative?

Remarks: (1) This quesiton is more general; actually, I am mostly interested in $\mathbb{C}[x,y]$, so I asked the question above. (2) See also this question (unfortunately, the fixed point theorems mentioned there are not relevant for $\mathbb{C}[x,y]$, but maybe there are generalizations of them that are relevant?). (3) This paper is perhaps relevant.

Thank you very much!

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  • $\begingroup$ Are you interested by the special case when $u$ is algebraic over $\mathbb{C}(w)$ or do you want to discard it? $\endgroup$
    – YCor
    Commented Jun 4, 2017 at 1:37
  • $\begingroup$ Good comment. Truly, in what I had in mind $u$ is algebraic over $\mathbb{C}(w)$ (more precisely, $u$ is integral over $\mathbb{C}[w]$). $\endgroup$
    – user237522
    Commented Jun 4, 2017 at 1:41
  • $\begingroup$ Actually, I wish to obtain that there is a conjugate $v \in \mathbb{C}(x,y)$ of $u$ which is different from $u$ (conjugate= an element that has the same minimal polynomial as $u$), where it is known that $u$ is integral over $\mathbb{C}[w]$. The existence of such $f$ will guarantee that $\mathbb{C}[x,y] \ni f(u)=:v$ is the desired conjugate. $\endgroup$
    – user237522
    Commented Jun 4, 2017 at 2:01
  • $\begingroup$ @YCor, please do you have an answer to one of the two cases ($u$ is algebraic over $\mathbb{C}(w)$ or not)? $\endgroup$
    – user237522
    Commented Jun 4, 2017 at 10:03
  • $\begingroup$ I would have told you if I had one. Possibly the case when $u$ is algebraic over $\mathbf{C}(w)$ deserves a separate question. $\endgroup$
    – YCor
    Commented Jun 4, 2017 at 10:28

1 Answer 1

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If we view $w$ as a map $\mathbb A^2 \to \mathbb A^1$ and the geometric generic fiber has a trivial automorphism group, then there will be no nontrivial automorphisms of $\mathbb C[x,y]$ fixing $w$.

If $w$ is a general quartic polynomial, say, then the geometric generic fiber is a general curve of genus $3$, which has trivial automorphism group.

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  • $\begingroup$ Thank you very much for your answer (I am not familiar enough with the notions in it, so it will take me some time to accept it). Please, there is something in your answer that I do not understand (probably because my lack of knowledge in algebraic geometry): Did you claim that if $w$ is of degree $4$, then there if no endomorphism $f$ of $\mathbb{C}[x,y]$ that fixes $w$? But, for example, if $w=x^4+y^4$ then $\alpha: (x,y) \mapsto (y,x)$ fixes $w$. What am I missing? $\endgroup$
    – user237522
    Commented Jun 4, 2017 at 6:26
  • $\begingroup$ @user237522. Your degree $4$ polynomial is not a "general" quartic polynomial. $\endgroup$
    – Jason Starr
    Commented Jun 4, 2017 at 9:38
  • $\begingroup$ @Jason Starr, please, what is a general quartic polynomial? $\endgroup$
    – user237522
    Commented Jun 4, 2017 at 9:52
  • $\begingroup$ Is it possible to describe all polynomials $w$ for which there do exist $f$ as I wish? Actually, I prefer to further assume that $w$ is reducible, like $x^4+y^4=(x^2−iy^2)(x^2+iy^2)=\ldots$. $\endgroup$
    – user237522
    Commented Jun 4, 2017 at 10:05
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    $\begingroup$ @user237522 For a specific family of $w$s, the first step would be to understand the genus of the geometric generic fiber and try to calculate its automorphism group. The next would be to determine whether the automorphisms extend to the whole space. $\endgroup$
    – Will Sawin
    Commented Jun 4, 2017 at 15:13

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