any linear algebraic group rational? Somewhere in Mumford's GIT, he seems to imply that any linear algebraic group is rational? This seems strange to me. Is it true?
 A: Pete gave the general answer, but let me mention a simple linear algebraic reason this should not surprise you (again, let me work over an algebraically closed field):
Recall that GL_n is almost the product of a torus and two subgroups isomorphic to affine space, because of the Gauss decomposition: a generic invertible matrix is the product of a unique triple consisting of an upper triangular matrix with 1's on the diagonal, a diagonal matrix and a lower triangular matrix.  This is the birational map that Pete was referring to.
A: This is really just Pete Clark's answer -- the new bit is to note that
the Levi decomposition isn't needed.
Let G be a (reduced, connected) linear algebraic group over an
alg. closed k, and let R be the unipotent radical of G. Choose a Borel
group B of G with unipotent radical U < B (so R < U).
There is a dense B-orbit ("big cell") V in G/B which is a rational
variety.
Since U and R are (split) unipotent, [Springer, LAG 14.2.6] shows that there
is a section $s:U/R \to U$ to the natural projection $U \to U/R$.
If $f:G \to G/B$ is the quotient mapping, using $s$ you can find a "local
section" of $f$ over the big cell V. 
This show that $f$ is a locally trivial B-bundle (in the Zariski
topology) and in particular $f^{-1}(V)$ is an open subvariety of G
isomorphic to the rational variety V x B.
A: A (reduced, irreducible) linear algebraic group over an algebraically closed field is rational -- i.e., birational to projective space.  This is a nontrivial result.
Briefest sketch of proof [EDIT: in characteristic 0 only; see the comment below]: use the Levi decomposition to reduce to the case of reductive groups, then use the Bruhat decomposition to handle the reductive case.
This does not hold for geometrically integral linear groups over an arbitrary ground field.  For instance, if $k$ is any field which admits a nondegenerate [i.e., degree $4$] biquadratic extension $l = k(\sqrt{a},\sqrt{b})$, then the norm torus associated to $l/k$ is a three-dimensional nonrational algebraic torus.  I think this example is in some sense minimal.  
See the Springer Online Reference Works for more information, including references.
