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Let $R$ be a commutative ring with identity. I'd like to know how to determine the set $\text{Aut}_R(R[X])$ of all $R$-automorphisms of $R[X]$.

I've proved that all $\sigma\in\text{Aut}_R(R[X])$ must satisfy $\sigma(X)=f(X)(X-a)$ (where the constant term of $f$ is invertible and the coefficients of other terms are nilpotent). Thus if $R$ is reduced, $\text{Aut}_R(R[X])=\{X\mapsto aX+b|\ a\in R^\times,b\in R\}$ ($R^\times$ is the group of units of $R$). How about the general cases?

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    $\begingroup$ If $R$ is not reduced, pick nonzero $b$ with $b^2=0$ in $R$, and let $P$ be any polynomial such that $bP\neq 0$, for instance $P(X)=X^2$. Define an $R$-algebra endomorphism by $f_b:X\mapsto X+bP(X)$. Then it is an automorphism with inverse $f_{-b}$. (Because $bP(X+bP(X))=bP(X)$.) $\endgroup$
    – YCor
    Commented Mar 31, 2015 at 12:43
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    $\begingroup$ All $R$-automorphisms of $R[X]$ are substitutions $X\mapsto a_0+a_1X+a_2X^2+\dots+a_nX^n$ with $a_i\in R$ and $a_0$ arbitrary, $a_1$ invertible, $a_i$ nilpotent for $i\ge 2$. This is either an exercise, or (I believe) stated somewhere in Demazure-Gabriel. $\endgroup$
    – Matthieu Romagny
    Commented Mar 31, 2015 at 13:13

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This is solved by R. Gilmer in his paper $R$-automorphisms of $R[X]$ appared in the Proceedings of London Mathematical Society, (3) 18, pp. 328-33, (1968).

If you want to replace the polynomal ring $R[X]$ by the formal power series $R[[X]]$, look at M. O'Malley and C. Wood's $R$-endomophisms of $R[[X]]$ in Journal of Algebra (15) 314-327 (1970).

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    $\begingroup$ Since Gilmer's paper is not easily available to everybody, let me just mention that it says that the $R$-automorphisms of $R[X]$ are indeed the $R$-endomorphisms mapping $X$ to a polynomial $\sum_{n\ge 0}a_nX^n$ with $a_1$ invertible and $a_n$ nilpotent for every $n\ge 2$ (as in Matthieu Romagny's comment). $\endgroup$
    – YCor
    Commented Mar 31, 2015 at 18:40

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