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What are the automorphisms $\sigma$ of $F[x]$ with the property that $ \sigma(f) =f $ for every $f \in F$?

$F$ is a field here.

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    What is $F$? Is it a ring? I presume so because you have added the "ring theory" tag...but then, does $F$ stand for a field? Or are we looking at a specific ring, like a ring of functions? I just do not know...2011-11-14
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    I guess it is not relevant. In the more general case of $F$ being a ring such an automorphism should be necessarily of the form $\sigma (x) =\alpha x$ for $\alpha$ an invertible element of $F$. Or I´m wrong?2011-11-14
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    (What have you done on this problem? Note that your automorphism is entirely defined by where $x$ is sent to, as $\sigma(f)=f$ for all $f\in F$. So I would start from here and see what you get...)2011-11-14
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    @Student73: Perhaps, but the fact we were dealing with a ring wasn't even stated!2011-11-14
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    @Student73 But $x \mapsto x + 1$ should always work, no?2011-11-14
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    I'm sorry. I forget to to state that $F$ is a field.2011-11-14
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    @Dylan. I´m sorry for answering that late. As pointed out in your answer the class is some more large! Thank you for teaching me something!2011-12-07

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I'm assuming that $F$ is a field. An $F$-homomorphism $\sigma\colon F[x] \to F[x]$ is determined by the polynomial $\sigma(x)$, and the image will be $F[\sigma(x)] \subset F[x]$. Thus $\sigma$ is surjective if and only if we can write $x$ as a polynomial in $\sigma(x)$. Can this be done if $\sigma(x)$ has degree $\geq 2$? Is it possible to write down an inverse automorphism if $\sigma(x) = ax + b$ for some $a \in F^*$ and $b \in F$?