By Jürgen Müller

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We show that Ad is a morphism: A 0 ∈ ∗ ∗ GL(gln ); A ∈ GL(g)} ≤ GL(gln ) closed, we have the morphism of algebraic groups π : H → GL(g) : A → A. For x ∈ G we have an extension κx−1 : GLn → Ad(x) 0 GLn , inducing Ad : G → H ≤ GL(gln ) : x → , hence Ad = π Ad. ∗ ∗ Thus extending Ad to GLn it suffices to consider Ad : GLn → GL(gln ) ∼ = GLn2 : Let G ≤ GLn closed, hence we have g ≤ gln . Letting H := {A = Letting X := {X11 , . . , Xnn } we for i, j ∈ {1, . . , n} have κ∗x−1 (Xij )(y) = n n n n Xij (xyx−1 ) = [xyx−1 ]ij = k=1 l=1 xik ykl xlj = k=1 l=1 xik Xkl (y)xlj = (x · [Xkl ] · x−1 )ij (y), for all y = [yij ] ∈ GLn , where x = [xij ] ∈ GLn and x−1 = [xij ] ∈ GLn .

C) Let λ = [λ1 , . . , λn ] n and µ = [µ1 , . . , µn ] n. Then µ is called to k k dominate λ, if for all k ∈ {1, . . , n} we have i=1 λi ≤ i=1 µi ; we write λ µ. 25). d) We have λ max µ if and only if µ = [λ1 , . . , λr−1 , λr + 1, λr+1 , . . , λs−1 , λs − 1, λs+1 , . . 26): II Algebraic groups 32 If λ max µ, let r := min{i ∈ {1, . . , n}; λi = µi } and r < s := min{k ∈ k k {r + 1, . . , n}; i=1 λi = i=1 µi } ≤ n. Hence we have λr < µr , and µr ≤ µr−1 = λr−1 if r > 1, as well as λs > µs ≥ µs+1 ≥ λs+1 .

Conclude that R ⊆ S is a finite ring extension, i. e. S is a finitely generated R-algebra and integral over R, if and only if S is a finitely generated R-module. b) Show that the integral closure R := {s ∈ S; s integral over R} ⊆ S of R in S is a subring of S, and that R = R holds. 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.

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