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Let $D$ be an integral domain. Prove that every automorphism of $D[x]$ is of the form: $\phi_{a,b} : D[x] \rightarrow D[x]$

$f$ $\rightarrow$ $f(ax+b)$

where a is a unit of $D$ and $b \in D$.

Not sure where to exactly jump in on this problem.

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    The second part means that $\phi_{a,b}(f(x)) = f(ax+b)$, right?2012-08-06
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    Do you require automorphisms to fix the base ring?2012-08-06
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    @ Eric..yes..I meant that. Thanks.2012-08-06
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    @Jacob...I don't know.2012-08-06
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    @Melky, I ask because any automorphism of the base ring extends to an automorphism of the polynomial ring. In particular an automorphism fixing $x$.2012-08-06

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