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By Jürgen Müller

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Let G be a connected algebraic group, and let Φ : G → G be a Frobenius homomorphism. Then the Lang map L : G → G : x → x−1 Φ(x) is surjective. Proof. We consider the Φ-conjugation action G×G → G : [z, x] → x−1 zΦ(x). For z ∈ G we have the orbit map Lz : G → G : x → x−1 zΦ(x) = x−1 zΦ(x)z −1 z, where in particular we have L1 = L: We have Lz = ρz µ(ι × (κz−1 Φ)), where κz : G → G : x → z −1 xz, implying d1 (Lz ) = d1 (ρz )(d1 (κz−1 Φ) − idT1 (G) ). Letting d ∈ N such that d1 (Φ)d = 0, we have (κz−1 Φ)d = κz−1 Φd , where z := z · Φ(z) · Φ2 (z) · · · · · Φd−1 (z) ∈ G.

Thus dimK (TEn (GLn )) = n2 , and TEn (GLn ) ∼ = DerK (K[X ]detn , KEn ) ∼ = DerK (K[X ], KEn ) ∼ = TEn (Kn×n ) ∼ = Kn×n , where δ = n n n×n . i=1 j=1 δ(Xij )∂ij ∈ DerK (K[X ], KEn ) is mapped to [δ(Xij )]ij ∈ K n For all i, j ∈ {1, . . , n} we have µ∗ (Xij ) = k=1 Xik ⊗ Xkj ∈ K[GLn ] ⊗K n K[GLn ], hence for δ, δ ∈ DerK (K[X ], KEn ) we get (δ · δ )(Xij ) = k=1 δ(Xik ) ⊗ δ (Xkj ), which hence transported to Kn×n yields the usual matrix product. Hence we have TEn (GLn ) ∼ = gln := Kn×n as Lie algebras, where the latter is endowed with the usual Lie product.

Show that a factorial domain R is integrally closed, i. e. we have R = R ⊆ S := Q(R). c) Let R ⊆ S be an integral ring extension, and let S be a domain. Show that R is a field if and only S is a field. d) Let R ⊆ S be an integral ring extension, and let J S and I := J ∩ R R. . Show that dim(I) = dim(J) ∈ N0 ∪ {∞} and ht(I) = ht(J) ∈ N0 ∪ {∞}. Proof. 9]. 16]. 10]. 14) Exercise: Infinite dimension (Nagata, 1962). Let K be a field and R := K[Xi ; i ∈ N], let d0 := 0, and for i ∈ N let di ∈ N such that di < di+1 , and Pi := Xdi−1 +1 , .

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