Bourbaki

Exercise 4, § 17

Tag 00L7, permanent. Cite it as [Bourbaki, Tag 00L7].

Algebra, VIII, p. 351

Bourbaki marks this one as harder than the rest.

Let D be a field, σ\sigma an automorphism of D, A the ring D[X]σD[X]_{\sigma} (VIII, p. 11),n,n an integer ⩾2\geqslant 2, and aa a nonzero element of D. Denote the element Xn−aX^n-a of A by ff. a) The ideal AfAf is two-sided if and only if we have σ(a)=a\sigma (a) =a and σn(x)=axa−1\sigma^n(x) =axa^{-1} for every x∈Dx\in D. b) Suppose that the group μn\mu_n of nn-th roots of unity contained in the center of D is cyclic of order nn and fixed by σ\sigma. Let g∈Ag\in A be an irreducible monic polynomial (VIII, p. 22, Exercise 27) dividing ff on the right. Prove that the degree dd of gg divides nn and that ff is the product of n/dn/d polynomials of degree dd. (Let hh be the monic polynomial such that Ah=⋂ζ∈μnAgζAh=\bigcap_{\zeta\in\mu_n}Ag_{\zeta}, where gζ(X)=g(ζX)g_{\zeta}(X) =g(\zeta X). Observe that we necessarily have h(ζX)=h(X)h(\zeta X) =h(X) for every ζ∈μn\zeta \in \mu_n, and deduce that h=fh=f. Then, apply loc. cit., b) by constructing, for 1⩽i⩽n/d1\leqslant i\leqslant n/d, an element ζi\zeta_i of μn\mu_n and a polynomial hih_i of degree idid such that Agζ1∩⋯∩Agζi=Ahi.)Ag_{\zeta_1}\cap \cdots \cap Ag_{\zeta_i}= Ah_i.)

If, moreover, the ideal AfAf is two-sided, then the ring A/AfA/Af is semisimple, and its simple components all have the same length (observe that every simple (A/Af(A/Af)-module is isomorphic to one of the modules A/Agζ)A/Ag_{\zeta}).

c) Assume from now on that the hypotheses of b) are satisfied, that the ideal AfAf is two-sided, and that nn is a prime number. Deduce from b) that we are in one of the following situations:

(i) There exists a b∈Db\in D such that a=bna=b^n and that σ\sigma is the inner automorphism associated with bb. The ring C=A/AfC = A/Af is isomorphic to the product ring DnD^n.

(ii) The condition of (i) is not satisfied, but there exists a c∈Dc\in D such that a=σn−1(c)σn−2(c)...σ(c)ca=\sigma^{n-1}(c)\sigma^{n-2}(c). . . \sigma (c)c. The ring C is simple of length nn.

(iii) There is no element cc of D such that a=σn−1(c)σn−2(c)⋯σ(c)ca=\sigma^{n-1}(c)\sigma^{n-2}(c)\cdots \sigma (c)c. The ring C is a field.

In the case when D is commutative, recover the results of Exercise 11 of VIII, p. 330. If, moreover, n=2n= 2, then C is a quaternion algebra over the subfield K of D invariant under σ\sigma.

d) Prove that there exists an automorphism τ\tau of C of order nn whose subfield of invariants is equal to D. In cases (ii) and (iii) above, D is a Galois subfield of C (Exercise 16 of VIII, p. 274).

e) Suppose that σ\sigma is the inner automorphism associated with an element dd of D. Prove that we have a=dnza=d^nz, where zz belongs to the center Z of C, and that the ring C is isomorphic to the tensor product D⊗ZZ[X]/(Xn−z)D\otimes_ZZ[X]/(X^n-z).

f ) Suppose that σ\sigma is not an inner automorphism. Prove that the center of C is the subfield of Z consisting of the elements invariant under σ\sigma.

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