# Automorphisms of $\mathbb{F}_2[x,y]$

What are the automorphisms of the 2-variable polynomial ring over $\mathbb{F}_2$, the field with 2 elements? Are they generated by $(x \mapsto y, y \mapsto x)$, $(x\mapsto x+ p(y), y\mapsto y)$, and $(x \mapsto x, y \mapsto y + p(x))$ where $p$ runs over all polynomials over $\mathbb{F}_2$? These are automorphisms, right?

I can see that any automorphism must fix the constants, but not much more.

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Well, these are isomorphisms to something, and they seem to be surjective, since the image has $F_2$, $y$, $x+p(y)$ and hence $x$. –  ronno Sep 18 '12 at 6:23
Getting late!${}{}{}$ –  André Nicolas Sep 18 '12 at 6:28
I asked google: Jung - van der Kulk theorem says that that you are right. See ams.org/journals/tran/1992-331-01/S0002-9947-1992-1038019-2/… (sect. 2.3 and 2.4)for a statement applicable over any field. –  user8268 Sep 18 '12 at 6:59
The keyword here is "affine Cremona group." The answer might be known for two variables but I think not much is known in general. –  Qiaochu Yuan Sep 18 '12 at 6:59

You are right, the group of automorphisms is generated by the two types of automorphisms you suggested. It follows from Jung - van der Kulk theorem (for a general field you need affine transformations as generators too, but for $\mathbb{F}_2$ they are already generated by your automorphisms). For a precise statement valid over any field see sect 2.3 and 2.4 of http://www.ams.org/journals/tran/1992-331-01/S0002-9947-1992-1038019-2/S0002-9947-1992-1038019-2.pdf .
Its not true that we don't need affine transforms at all, $x \mapsto y, y\mapsto x$ is needed. –  ronno Oct 1 '12 at 2:56
@ronno: actually $(x,y)\mapsto(x+y,y)\mapsto(x+y,y+(x+y))=(x+y,x)\mapsto(x+y+x,x)=(y,x)$. –  user8268 Oct 1 '12 at 19:16