Automorphisms of an affine line over finite fields Let $k$ be an infinite field, and consider the affine line $\mathbb{A}_k^1$ over $k$. We know that every isomorphism $\varphi:\mathbb{A}_k^1\longrightarrow\mathbb{A}_k^1$ is of the form $\varphi(x)=ax+b$ where $a,b\in k$ with $a\not=0$. We prove this by describing all possible $k$-algebra isomorphisms $\varphi^*:k[x]\longrightarrow k[x]$.
However, things get complicated when $k$ is a finite field of characteristic $p$. Note that the coordinate ring of $\mathbb{A}_k^1$ is now the quotient $k[x]/I(\mathbb{A}_k^1)$ where $I(\mathbb{A}_k^1)$ is no more the zero ideal. The theory of permutational polynomials, that is, of those where we have that $\varphi$ is at least bijective, seems utterly complicated. Even worse, the so-called Frobenius map $x\mapsto x^p$ is known to be regular and bijective, but it is not an isomorphism of affine algebraic sets since the corresponding $k$-algebra morphism is not surjective.
I'm wondering if we can prove somehow that, even in case of finite fields, linear maps of the form above are the only possible isomorphisms of the affine line.
 A: As Mariano comments, $\Bbb A^1_k=\operatorname{Spec}(k[x])$ for any field $k$ (or indeed, any ring $k$). Thus, an automorphism $\varphi$ of $\Bbb A^1$ is dual to an automorphism of $k$-algebras $\varphi^{\sharp}:k[x]\to k[x].$ The map $\varphi^\sharp$ is determined by $\varphi^\sharp(x)=a_0+a_1x+\cdots+a_dx^d.$ And, the inverse $(\varphi^\sharp)^{-1}$ is also determined by $(\varphi^\sharp)^{-1}(x)=b_0+b_1x+\cdots+b_ex^e.$ We must have $$x=(\varphi^\sharp)^{-1}(\varphi^\sharp(x))=(\varphi^\sharp)^{-1}(a_0+a_1x+\cdots+a_dx^d)=a_0+a_1(\varphi^\sharp)^{-1}(x)+\cdots+a_d(\varphi^\sharp)^{-1}(x)^d=a_0+a_1(b_0+b_1x+\cdots+b_ex^e)+\cdots+a_d(b_0+b_1x+\cdots+b_ex^e)^d$$
which is a polynomial of degree $de.$ Thus, $d=e=1,a_1,b_1\neq 0$ and we find that $\varphi^\sharp(x)=a_0+a_1x.$ 
Edit: Here is a remark that may be more germane to the OPs worry. Consider the case $k=\Bbb F_2.$ The variety $\Bbb A^1$ contains closed points $(x),(x+1)$ corresponding to $0,1\in\Bbb F_2.$ One might think that the product $f(x)=x(x+1)$ lies in "the ideal of" $\Bbb A^1$ since we clearly have $f(0)=f(1)=0,$ and these are the only values in $\Bbb F_2.$ However, classical algebraic geometry is done over algebraically closed fields, which are infinite, and so we cannot run into this problem, by contrivance. To work with nonalgebraically closed fields, or other interesting rings, we pass to the scheme theory of Grothendieck (and many others), taking by definition $\Bbb A^1_k$ to mean the prime spectrum of the (coordinate) ring $k[x].$ One of the beautiful things about this choice of definition is that we are bestowed with an equivalence between the categories of affine schemes/morphisms and (commutative) rings (with unit)/homomorphisms, which means that results such as that which you mention continue to hold, even for spaces that might otherwise exhibit some confusing behaviour. Of course, it is interesting also to note that $\Bbb A^1_k$ now contains many more points than just $0$ and $1,$ for example containing the generic point, along with those points defined by all the higher degree irreducible polynomials over $k=\Bbb F_2.$
