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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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    @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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Let $\phi: D[x]\to D[x]$ be an automorphism. By definition there is $\psi: D[x]\to D[x]$ such that $\phi\circ \psi=\mathrm{id}_{D[x]} .$ Suppose $\phi(x)=a_0+a_1x+\cdots +a_nx^n$ and $\psi(x)=b_0+b_1x+\cdots +b_mx^m,$ with $a_n,b_m\neq 0.$ Then we easily see that the degree of $\phi(\psi(x))$ equals $mn.$ Thus $m=n=1.$ Clearly $a_1,b_1$ must be units, again by examining the coefficient of $x^{mn}=x$ in $\phi(\psi(x)).$

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    Yes, this seems implicit in the question, especially considering Jacob's comment re. extending an automorphism of the base ring.2012-08-06