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For every finite subset $A $ of $\mathbb C[[X]]$, does there exist a ring automorphism (bijective ring endomorphism ) $f$ of $\mathbb C[[X]]$ such that $f(a)=a, \forall a \in A$ but $f(\mathbb C[X]) \ne \mathbb C[X]$ ?

Related Are $\mathbb C$ , $\mathbb C[X]$ definable in $\mathbb C[[X]]$? because if $D \subseteq M$ is definable by a set $A \subseteq M$ then $f(D)=D, \forall f \in Aut_A M$ , and in the related case we mean definable by some finite subset of $M$.

user
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  • Just by curiosity, can you find one when $A = {X}$ ? – Nicolas Hemelsoet Feb 27 '18 at 18:22
  • @NicolasHemelsoet: Uh no ... I don't have any ..... – user Feb 27 '18 at 18:53
  • Ok. Of course if $f$ is a morphism of $\Bbb C$-algebra, then $f(X) = X$ implies that $f(\Bbb C[X]) = \Bbb C[X]$. On the other hand, one could write $\Bbb C$ as the algebraic closure of (say) $\Bbb C(x,y)$ and construct many weird ring automorphism, in particular $\Bbb C$ might not be in the image. But I'm not exactly sure how to extend it to an automorphism after... – Nicolas Hemelsoet Feb 27 '18 at 21:34

1 Answers1

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I can't answer this question, but I can reduce it to your other question.

Claim: The following are equivalent.

  1. There exists a finite set $A\subseteq \mathbb{C}[[X]]$ such that every automorphism of $\mathbb{C}[[X]]$ which fixes $A$ pointwise fixes $\mathbb{C}$ setwise.

  2. There exists a finite set $B\subseteq \mathbb{C}[[X]]$ such that every automorphism of $\mathbb{C}[[X]]$ which fixes $B$ pointwise fixes $\mathbb{C}[X]$ setwise.

Proof: (1)$\implies$(2). Let $B = A\cup \{X\}$. Then for any automorphism $f$ which fixes $B$ pointwise, $f$ fixes $\mathbb{C}$ setwise and $f(X) = X$, so for any $a_nX^n+\dots+a_1X + a_0\in \mathbb{C}[X]$, $$f(a_nX^n + \dots + a_1X + a_0) = f(a_n)X^n + \dots + f(a_1)X + f(a_0)\in \mathbb{C}[X].$$

(2)$\implies$(1). Let $A = B$. Then any automorphism $f$ which fixes $A$ pointwise fixes the subring $\mathbb{C}[X]$ setwise, so it restricts to an automorphism of $\mathbb{C}[X]$. In particular, it fixes $\mathbb{C} = 0\cup \mathbb{C}^\times$ setwise, since $\mathbb{C}^\times$ is the group of units of $\mathbb{C}[X]$.

Alex Kruckman
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