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Somewhere in Mumford's GIT, he seems to imply that any linear algebraic group is rational? This seems strange to me. Is it true?

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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.

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  • $\begingroup$ Can you give me a reference? $\endgroup$
    – user19475
    Commented Jan 10, 2010 at 13:35
  • $\begingroup$ Sure, I edited my response to include one. $\endgroup$
    – Pete L. Clark
    Commented Jan 10, 2010 at 13:52
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    $\begingroup$ You should probably assume the group is connected.... $\endgroup$
    – mdeland
    Commented Jan 10, 2010 at 15:22
  • $\begingroup$ Pete, the Levi decomposition is not true over alg. closed fields of positive characteristic. Typical counterexample is $G(W_2(k))$ for a Chevalley group $G$. See section A.6 of "Pseudo-reductive groups". $\endgroup$
    – BCnrd
    Commented Apr 21, 2010 at 16:28
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    $\begingroup$ The link to eom.springer.de is broken, but the article can now be found at encyclopediaofmath.org/wiki/Rationality_theorems. $\endgroup$
    – The Amplitwist
    Commented Jul 5, 2022 at 0:02
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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.

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    $\begingroup$ Thanks, but I don't know why unipotent groups are isomorphic to affine space. I know that they are a closed subset thereof. $\endgroup$
    – user19475
    Commented Jan 10, 2010 at 16:24
  • $\begingroup$ For me, what was always the definition of a unipotent group, but look at the Springer Reference on "unipotent groups" it seems that things are trickier over characteristic p. Of course, in GL_n, there's an easy proof that upper (or lower) triangular matrices are isomorphic to affine space. The map is just looking at off-diagonal entries. $\endgroup$
    – Ben Webster ♦
    Commented Jan 10, 2010 at 16:32
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    $\begingroup$ In characteristic zero, the exponential map between a nilpotent Lie algebra and the corresponding unipotent Lie group is an algebraic isomorphism. Hence, unipotent groups in characteristic zero are isomorphic to affine space. See mathoverflow.net/questions/10730/… $\endgroup$
    – David E Speyer
    Commented Jan 10, 2010 at 16:39
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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.

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