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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!

user237522
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  • Are you interested by the special case when $u$ is algebraic over $\mathbb{C}(w)$ or do you want to discard it? – YCor Jun 04 '17 at 01:37
  • Good comment. Truly, in what I had in mind $u$ is algebraic over $\mathbb{C}(w)$ (more precisely, $u$ is integral over $\mathbb{C}[w]$). – user237522 Jun 04 '17 at 01:41
  • 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. – user237522 Jun 04 '17 at 02:01
  • @YCor, please do you have an answer to one of the two cases ($u$ is algebraic over $\mathbb{C}(w)$ or not)? – user237522 Jun 04 '17 at 10:03
  • I would have told you if I had one. Possibly the case when $u$ is algebraic over $\mathbf{C}(w)$ deserves a separate question. – YCor Jun 04 '17 at 10:28
  • Thanks for your comment. (Perhaps I will think a little more about that case myself, before asking again). – user237522 Jun 04 '17 at 13:51

1 Answers1

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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.

Will Sawin
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  • 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? – user237522 Jun 04 '17 at 06:26
  • @user237522. Your degree $4$ polynomial is not a "general" quartic polynomial. – Jason Starr Jun 04 '17 at 09:38
  • @Jason Starr, please, what is a general quartic polynomial? – user237522 Jun 04 '17 at 09:52
  • 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 $x^4+y^4=(x^2−iy^2)(x^2+iy^2)=\ldots$. – user237522 Jun 04 '17 at 10:05
  • @user237522 A quartic polynomial where all monomials of degree at most $4$ appear, and all are independent transcendentals, does the trick. However a polynomial with random integer coefficients in some large box also does the trick with high enough probability. Most likely something reducible works - I doubt there are any nontrivial automorphisms of $(y^2- x^3 - x - 7) (3y-5x + 12)$, for instance. – Will Sawin Jun 04 '17 at 12:26
  • Thanks for your explanation. Please, do you see any hope to find such $f$ for a specific family of $w$'s? – user237522 Jun 04 '17 at 13:55
  • BTW, perhaps the reducible $w$ is not so arbitrary, because (in what I had in mind) there exists (an irreducible) $u$ which is integral over $\mathbb{C}[w]$. – user237522 Jun 04 '17 at 13:59
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    @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. – Will Sawin Jun 04 '17 at 15:13
  • @Will Sawin, thank you! (if one of MO users can "translate" the algebraic geometry answer to a pure algebraic answer, that would be great! although I know that it is important to try to understand the algebraic geometry answer). – user237522 Jun 04 '17 at 15:47