Bourbaki

Exercise 5, § 17

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Algebra, VIII, p. 352

Bourbaki marks this one as harder than the rest.

Let K0K_0 be the field of rational fractions Q(X)\mathbf{Q}(X). Let K be the quadratic extension

√ \surd

K0(ρ)K_0(\rho ), where ρ\rho is a primitive cubic root of unity, D be the cyclic extension K(3X)K(^{^3}X) of K, and σ\sigma be a generator of the Galois group of D over K. Consider the algebra C defined as in Exercise 4, with n=3n= 3 and a=2a= 2.

a) Prove that C is a field with center K, of degree 9 over K. (To see that D has no element of norm 2, reduce to proving that there cannot be any relations of the form 2F3=P32F^3= P^3+XQ3+X2R3^3+X^2R^3 +3XPQR, where F,P,Q,RF,P,Q,R are mutually prime polynomials in Q(ρ)[X]\mathbf{Q}(\rho )[X]. To do this, consider the lowest-degree terms of these polynomials.) b) Prove that there does not exist any automorphism of C extending the unique K0K_0-automorphism of K distinct from the identity (reduce to the case when such an automorphism preserves D). Deduce that K0K_0 is not a Galois subfield of C (VIII, p. 274, Exercise 16) and that there does not exist a K0K_0-automorphism of C of order 2.

c) Let ξ\xi be an element of D such that ξ3=X\xi^3= X, and let η\eta be an element of C such that η3=2\eta^3= 2 and ηξ=ρξη\eta \xi =\rho \xi \eta; set ζ=ξ+η+η2\zeta =\xi +\eta +\eta^2. Prove that K(ζ)K(\zeta ) is a maximal commutative subfield of D and is not Galois over K, whereas the maximal commutative subfields K(ξ)K(\xi ) and K(η)K(\eta ) are Galois over K (calculate ζ3\zeta^3).

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