By Jürgen Müller

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Proof. 22). Since CG◦ (x) ≤ CG (x) is a closed subgroup of finite index, we have dim(CG (x)) = dim(CG◦ (x)). Hence to show the last assertion, we may assume that G is connected. Letting ϕ be the action morphism, the orbit map ϕx : G → xG is a dominant morphism between irreducible varieties. Hence there is ∅ = U ⊆ xG such that U ⊆ xG, and such that dim(ϕ−1 x (y)) = dim(G) − dim(xG) for all y ∈ U . For any y ∈ U we have ϕ−1 (y) = {h ∈ G; xh = y} = CG (x)g ⊆ G, where g ∈ G is fixed such x that y = xg, implying dim(ϕ−1 x (y)) = dim(CG (x)).

Show that ϕ is an open map, i. e. maps open sets to open sets. Hint. Use Chevalley’s Theorem. Proof. 5]. 23) Exercise: Upper semicontinuity of dimension. Let V, W be irreducible affine varieties and let ϕ : V → W be a dominant morphism. For any x ∈ V let ϕ (x) ∈ N0 be the maximum of the dimensions of the irreducible components of ϕ−1 (ϕ(x)) ⊆ V containing x. Show that for any n ∈ N0 the set {x ∈ V ; ϕ (x) ≥ n} ⊆ V is closed. Proof. 4]. 24) Exercise: Diagonalisable matrices. Let K be an algebraically closed field, and let n ∈ N.

Since the regular points of Vz form a nonempty open subset, and Lz (G) contains a nonempty open subset of Vz , these sets intersect non-trivially, and hence there is y ∈ G such that Lz (y) ∈ Vz is regular. 32). Being a G-orbit, Lz (G) ⊆ Lz (G) = Vz = G is open. Hence any two such G-orbits intersect non-trivially, thus Lz (G) = G for all z ∈ G. 5) Corollary. Let G be a connected algebraic group with Frobenius homomorphism Φ : G → G. Then GΦ is finite. Proof. Given x, y ∈ G we have L(x) = L(y) if and only if yx−1 = Φ(yx−1 ), which holds if and only if GΦ x = GΦ y.