Bourbaki

Exercise 10, § 8

Commutative Algebra, VI, 464

Let w be the 2-adic valuation on the 2-adic field Q2\mathbf{Q}_2; on the field of rational functions Q2(X)\mathbf{Q}_2(X) consider the discrete valuation vv extending ww and such that

v(a0+a1X+⋯+anXn)=inf⁡i(w(ai)) v(a_0 + a_1 X + \cdots + a_n X^n) = \inf_i (w(a_i))
(§ 10, no. 1, Lemma 1); let K be the completion of Q2(X)\mathbf{Q}_2(X) with respect to this valuation and let its valuation also be denoted by vv. Let L be the splitting field of the polynomial (Y2−X)2−2(Y^2 - X)^2 - 2 in K[Y]K[Y] and let v′v' be the unique valuation on L extending vv. Show that [L:K]=8,e(v′/v)=4,f(v′/v)=2[L : K] = 8, e(v'/v) = 4, f(v'/v) = 2; if k (resp. k′k') is the residue field of vv (resp. v′v'), k′k' is a radicial extension of k of degree 2; show that there exists no sub-extension E of L such that [E:K]=2[E : K] = 2 for which the residue field of the restriction of v′v' to E is equal to k′k'. (Of necessity E=K(α)E = K(\sqrt{\alpha}), where α∈K\alpha \in K would not be a square in K but such that α≡X(mod2)\alpha \equiv X \pmod{2}; express &with the aid of a suitable basis of L over K(2)K(\sqrt{2}) and observe that there exists in K(2)K(\sqrt{2}) no element whose square is congruent to X mod. 2.)

Solution

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