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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: 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.

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.

A 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: 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 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.

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: 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.

A 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: 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 over an arbitrary ground field. For instance, if $k$ is any field which admits a nondegenerate biqquadratic 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.

A 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: 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 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.