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Let $R$ be a commutative Noetherian $F$-algebra, where $F$ is a field (perfect, say). Assume that $R \otimes_F \overline F$ is a polynomial ring over the algebraic closure $\overline F$. Does it follow that $R$ was already a polynomial ring over $F$?

I doubt it, but haven't had any luck constructing a counterexample.

The question arises because I'm trying to understand Bruhat cells in flag manifolds for non-split algebraic groups. In split ones, Bruhat cells are affine spaces.

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Sorry, I have to run, so this very rough, but isn't the first Galois cohomology of the affine group trivial? By write it as an extension of $GL$ by an additive group, and using Hilbert 90. So maybe it's true? – Donu Arapura Jan 27 '11 at 19:33
The automorphism group of the affine space is a whole lot bigger than the affine group. The answer is certainly positive in dimension 1, and might be known in dimension 2, but I do believe that this is a hard problem in higher dimension. By the way, over non-perfect fields this is false, even in dimension 1. – Angelo Jan 27 '11 at 19:40
I would be surprised if the Galois cohomology of the automorphism group of a polynomial ring in three variables were known! – Mariano Suárez-Alvarez Jan 27 '11 at 19:44
Angelo: in dimension 2 it is the amalgamated product of the linear automorphisms and the triangular automorphisms (say, those mapping $k[x]$ into itself, and mapping $y$ to $y+f(x)$ for some $f$) – Mariano Suárez-Alvarez Jan 27 '11 at 19:44
@Angelo: it is now known that it is not an amalgamated product in dimension 3: Shestakov and Umirbaev have shown there are wild automorphisms. (IIRC the answer is not stable w.r.t. adding more variables, so I don't think this decides the question of all dimensions) (By the way, your "certainly" is quite difficult to prove :) ) – Mariano Suárez-Alvarez Jan 27 '11 at 19:56
up vote 15 down vote accepted

In his Bourbaki talk of 1994 Kraft tells us that the complex affine plane does not have non-trivial forms and that the corresponding question in higher dimensions is open. 1994 is an age ago, though!

(He also shows that the automorphism groups in high dimension are not amalgamated products of the two obvious subgroups by exhibiting two expressions of an automorphism as a product, contradicting the uniqueness implicit in amalgamation)

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Nice reference! It has a lot of other surprisingly hard problems about ${\mathbb A}^n$. Unfortunately, the 33 papers on MathSciNet that reference it seem to be about those other problems. – Allen Knutson Jan 27 '11 at 22:47
My guess: your problem is quite harder than the other ones ;) – Mariano Suárez-Alvarez Jan 27 '11 at 23:14
Kraft tells me there has been no progress, so this seems likely to be the best answer. – Allen Knutson Jan 28 '11 at 17:22

First, since the OP is interested in Bruhat cells in flag varieties over nonsplit groups (over perfect fields), I think the question is probably unnecessary. I'd recommend Borel and Tits IHES publication "Groupes Reductifs," where they describe the theory of relative root systems for nonsplit groups, and much more. In and around Theorem 5.15, you can find the basic results on parabolic subgroups and Bruhat decomposition, in this relative setting. From that section, one can deduce that the appropriate Bruhat cells are isomorphic to unipotent groups (as varieties) and in characteristic zero, they are just affine spaces, not twists thereof.

So, I doubt that twists of affine spaces have any place in studying Bruhat cells for nonsplit groups reductive groups over perfect fields.

Now I have a few remarks regarding the issue of existence of nontrivial twists of affine space (though this is outside my expertise). As the comments indicate, this dances around the famous Jacobian conjecture, and demonstrates how far we must go to understand the group of polynomial automorphisms of affine $n$-space (over $\bar Q$, for example).

To understand this group of automorphisms $G(\bar Q)$, it helps to filter out the easy part. One finds a translation subgroup $T(\bar Q)$ of $G(\bar Q)$, given by automorphisms which are translations of the affine space; the quotient of $G$ modulo this translation subgroup can be identified with the subgroup $G^0(\bar Q)$ of automorphisms preserving the origin. Among these, one can look at the induced automorphism on the tangent space at $0$, an element of $GL_n(\bar Q)$. The origin-preserving automorphisms which act as the identity on the tangent space at $0$, is called the group of principal automorphisms, or $G^1(\bar Q)$. So there are exact sequences of $Gal(\bar Q / Q)$-modules: $$1 \rightarrow G^1(\bar Q) \rightarrow G^0(\bar Q) \rightarrow GL_n(\bar Q) \rightarrow 1.$$ $$1 \rightarrow T(\bar Q) \rightarrow G(\bar Q) \rightarrow G^0(\bar Q) \rightarrow 1.$$

(I'm working over $\bar Q$ for convenience). By standard Galois cohomology, one finds that the possibility of twisted forms of affine $n$ space boils down to the Galois cohomology with coefficients in $G^1(\bar Q)$, the principal automorphisms.

Now the group $G^1(\bar Q)$ is very complicated. I think Shafarevich was the first to study such automorphism groups, as points of Ind-varieties (or Ind-affine varieties), and there are a few papers of Kambayashi (see one in J. of Algebra, for example) describing such an algebraic structure on $G^1$ which I'm more familiar with.

Here, in my opinion, is why dimension $2$ is so much easier to deal with. There's a famous result of Gabber, which states that if $P: A^n \rightarrow A^n$ is a polynomial automorphism (a polynomial map with polynomial inverse) of degree $d$, then the inverse of $P$ has degree at most $d^{n-1}$. For this reason, in dimension $2$, invertible polynomial maps have inverses of the same (or less) degree. This makes it easy to "filter" the automorphism group $G^1$ by degree -- Galois cohomology will have image in a subgroup $G_d^1(\bar Q)$ of $G^1(\bar Q)$ consisting of automorphisms of degree $\leq d$ for some $d$. This filtration makes computation of Galois cohomology easier, I'd guess.

In dimension $\geq 2$, I don't think anyone knows enough about the structure of $G^1(\bar Q)$ to compute its Galois cohomology.

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Thanks for the, indeed, more relevant reference! Does $G^1$ act on something like 2-jets at the origin, such that we could continue this sequence? – Allen Knutson Jan 28 '11 at 15:30

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