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.

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    $\begingroup$ The affine line over a field k has coordinate ring k[x] for all fields. $\endgroup$ Dec 7, 2012 at 2:37
  • $\begingroup$ I am not convinced. For any subvariety V of the affine line A_k, the coordinate ring k[V] is defined as the quotient k[x]/I(V). According to this, for V=A_k, we have k[A_k]=k[x]/I(A_k). Of course, when k is an infinite field, I(A_k)=(0), and the coordinate ring is just k[x]. In case of a finite field, we may have different polynomials with the same values on each element of k. $\endgroup$
    – user37391
    Dec 7, 2012 at 3:04
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    $\begingroup$ It is not a matter of your being convinced, really: it is simply the definition. $\endgroup$ Dec 7, 2012 at 3:18
  • $\begingroup$ @Gruff: s'pose $k$ is finite and $A_k = k[x]/I(V)$ as in your comment (where $V$ is the finites set $k = A_k$). Now let $K$ be an algebraic closure of $k$. Then $A_k \otimes_k K$ is just $K[x]/K.I(V)$, which is the coordinate ring of this finite subset of the affine line over $K$, and not of the affine line over $K$ itself. The need/desire for "scalar extension" to work correctly perhaps motivates the definition to which Mariano refers. $\endgroup$ Dec 7, 2012 at 12:54
  • $\begingroup$ You seem to be confusing $\mathbb{A}^1(k)$ -- the set of $k$-points of the affine line -- with $\mathbb{A}^1_k$ -- the affine line over $k$. $\endgroup$
    – user14972
    Mar 5, 2014 at 7:33

1 Answer 1


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 $k=\Bbb F_2.$

  • $\begingroup$ Is there a classical theory for this without using the language of schemes, or it is just not interesting or perhaps difficult to deal with? $\endgroup$
    – user37391
    Dec 7, 2012 at 4:06
  • $\begingroup$ Another method is to start with $\Bbb A^n$ defined as $\{(a_1,\ldots,a_n)|a_i\in\overline k\},$ where $\overline k$ is a fixed algebraic closure of $k,$ and then study "rationality" questions, i.e., what are the $k$-rational points of varieties over $\overline k.$ However, again we are back to the algebraically closed/infinite field case as a basis for the work. $\endgroup$
    – Andrew
    Dec 7, 2012 at 4:13
  • $\begingroup$ And, by the way, I think it's very interesting, and that you've asked a good question, which has made me think! Cheers $\endgroup$
    – Andrew
    Dec 7, 2012 at 4:14
  • $\begingroup$ Very nice answer, Andrew:+1 . $\endgroup$ Dec 7, 2012 at 7:49
  • $\begingroup$ Dear @GeorgesElencwajg, that means a lot coming from you, whose contributions to this site I respect very much. Thank you! $\endgroup$
    – Andrew
    Dec 7, 2012 at 16:06

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