Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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

1 Answer 1

up vote 1 down vote accepted

(I updated my comment to an answer, so that it doesn't get lost)

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 .

share|improve this answer
    
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
    
Nice. Should have thought of that. –  ronno Oct 2 '12 at 8:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.