Prove that all automorphisms of the line $\Bbb A^1$ are of the form $f(x) = ax + b$ with $a\neq 0$. An isomorphism $f : X → X$ of a closed set $X$ to itself is called an automorphism. Prove that all automorphisms of the line $\Bbb A^1$ are of the form $f(x) = ax + b$ with $a\neq 0$.
I think I can take the co-ordinate ring and again closed subsets are finite here.So do I use these efficiently?
 A: The problem requires additional assumption that automorphism are over the base field of $\mathbb{A}^1$. Consider $\mathbb{A}^1_{\mathbb{C}}$. It has an automorphism corresponding to a map $u: \mathbb{C}[x] \to \mathbb{C}[x], u(x) = x, u(a) = \overline{a}$ for $a \in \mathbb{C}$. This is automorphism of $\mathbb{C}[x]$ not having a form $x \mapsto ax + b$.
Anyway, such automorphisms $f: \mathbb{A}^1 \to \mathbb{A}^1$ are in 1-1 correspondence with isomorphisms $g: k[x] \to k[x]$ of the coordinate ring that keeps $k$ still. Any map $k[x] \to  k[x]$ over $k$ is determined by the image of $x$. I claim any automorphism $g: k[x] \to k[x]$ must satisfy $\deg g(x) = 1$. Otherwise, either $\deg g(x) = 0$, so $g(x) \in k$, from which it follows that the image of $g$ is $k$, and so $g$ is not isomorphism. On the other hand, if $\deg g(x) > 1$, it can be proved easily that that for any $w \in k[x]$, $\deg g(w) = \deg w \deg g(x)$. This is never equal to 1 -- either $\deg w \geq 1$, then $\deg w \deg g(x)  \geq \deg g(x) > 1$, or $\deg w = 0$, then $\deg w \deg g(x) = 0$. In any case, there is no $w \in k[x]$ such that $\deg g(w) = 1$, which means that $g$ is not surjective.
We obtained that $\deg g(x) = 1$, so $g$ must have form $g(x) = ax + b$ for $a \ne 0$.
A: Since $\mathbb{A}^1$ is affine each automorphism $f: \mathbb{A}^1 \to \mathbb{A}^1$ is induced by a $k$-automorphism $\varphi: k[x] \to k[x]$, which is determined by the image of $x$. It is clear then that $\varphi$ is an automorphism if and only if $\varphi(x)$ is a polynomial of degree 1. Going through the standard conversion process to produce a morphism $f: \mathbb{A}^1 \to \mathbb{A}^1$ we get precisely the 'affine transformations' as they are called, which map $x \in \mathbb{A}^1$ to $ax+b, a\neq 0$.
