Exercise 17, § 14

Keep the notation of Exercise 15, and suppose that V is finite-dimensional over D. A group G of automorphisms of A is called Galois if it satisfies the following conditions:

(i) $(G : G_0)$ and $[U_0(G) : Z]$ are finite.

(ii) The ring $U_0(G)$ is simple.

(iii) For every invertible element $s$ of $U_0$(G), the automorphism $a\mapsto sas^{-1}$ belongs to $G_0$.

Let G be such a group.

a) Prove that the mapping $H\mapsto A^H$ is a decreasing bijection from the set of Galois subgroups of G to the set of simple subrings of A that contain $A^G$ and whose commutant $B'_0$ in A is simple. (Deduce from Exercise 15, f) and g) that this mapping is well defined and injective. Let B be a simple subring of A satisfying the conditions above, and let H be the group of automorphisms of A fixing B. Use Exercises 14, c) and 16 to prove that B is Galois and then Proposition 12 of VIII, p. 205 to prove that we have $B = A^H$ and that H is Galois.)

b) Let $B_1$ and $B_2$ be simple subrings of A containing $A^G$ whose commutants in A are simple, and let $\varphi$ be an isomorphism from $B_1$ to $B_2$ fixing $A^G$. Prove that there exists an element of G that coincides with $\varphi$ on $B_1$.

(First observe that we have long$_{B_1}(V) =$ long$_{B_2}$(V), using the equality $h(B_1,A^G) =h(B_2,A^G)$. Let W be the set of elements $f$ of Ω satisfying $f(bx) =\varphi (b)f(x)$ for $b\in B_1,x\in V$. Prove that W is not reduced to 0, and then apply Exercise 14, c) to it to show that it contains a nonzero semilinear mapping $w$. Show that if S is a simple DB$_1$-submodule of V that is not contained in Ker $w$, then $w$ induces a bijection from S to its image and we have long$_{B_1}(S) =$ long$_{B_2}(w(S))$ (use Exercise 16, b)). Reasoning as in loc. cit., conclude that W contains a semilinear bijection that coincides with $w$ on S.)

c) Specialize the results of a) and b) to the following cases:

(i) $G = G_0$ (cf. VIII, p. 259, Theorem 5).

(ii) $G_0=\{1\}$. (Prove that every finite group of automorphisms of A is Galois and that every simple subring of A containing $A^G$ admits Z as commutant in A.)

(iii) V has dimension 1, so that A is a field isomorphic to $D^o$. (The ring $A^G$ is a field, and every subfield of A containing $A^G$ has a field as commutant. Study the specific case when D is commutative (cf. V, §10, No. 7).)