Bourbaki

Exercise 23, § 10

Algebra, V, p. 160

Let k be a field of characteristic 2. a) Let A1,…,AnA_1, \ldots, A_n be indeterminates, K the field k(A1,...,An)k(A_1, ..., A_n), E a splitting field of the polynomial P=Xn+A1Xn−1+...+An∈K[X]P = X^n + A_1 X^{n-1} + ... + A_n \in K[X] and X1,...,XnX_1, ..., X_n the roots of P in E, so that E=k(X1,...,Xn)E = k(X_1, ..., X_n) (V, p. 21). We identify the Galois group of E/K with the symmetric group SnS_n acting by permutations of the variables X1,...,XnX_1, ..., X_n (V, p. 59). Put

β(X1,...,Xn)=∑i<jXi/(Xi+Xj)∈EandC(A1,...,An)=∑i<j(XiXj)/(Xi2+Xj2)∈K. \beta(X_1, ..., X_n) = \sum_{i < j} X_i/(X_i + X_j) \in E \quad \text{and} \quad C(A_1, ..., A_n) = \sum_{i < j} (X_i X_j)/(X_i^2 + X_j^2) \in K.
Then β∉K\beta \notin K and
β2+β+C=0. \beta^2 + \beta + C = 0.
We have σ(β)=β\sigma(\beta) = \beta for σ∈Un\sigma \in U_n and σ(β)=β+1\sigma(\beta) = \beta + 1 for σ∈Sn−Un\sigma \in S_n - U_n. Deduce that K(β)K(\beta) is the unique quadratic subextension of E/KE/K. b) Let D∈Z[T1,...,Tn]D \in \mathbf{Z}[T_1, ..., T_n] be the discriminant of the polynomial Xn+T1Xn−1+...+TnX^n + T_1 X^{n-1} + ... + T_n. Show that there exist Q, R, S ∈Z[T1,...,Tn]\in \mathbf{Z}[T_1, ..., T_n] with
D=Q2−4R+8S D = Q^2 - 4R + 8S
and
C=C(A1,...,An)=R(A1,...,An)Q2(A1,...,An)=R(A1,...,An)D(A1,...,An). C = C(A_1, ..., A_n) = \frac{R(A_1, ..., A_n)}{Q^2(A_1, ..., A_n)} = \frac{R(A_1, ..., A_n)}{D(A_1, ..., A_n)}.
For every triple of elements U, V, W of Z[T1,...,Tn]\mathbf{Z}[T_1, ..., T_n] such that D=U2−4V+8WD = U^2 - 4V + 8W there exists M∈Z[A1,...,An]M \in \mathbf{Z}[A_1, ..., A_n] such that
C=V(A1,...,An)D(A1,...,An)+(MD(A1,...,An))2+MD(A1,...,An) C = \frac{V(A_1, ..., A_n)}{D(A_1, ..., A_n)} + \left( \frac{M}{D(A_1, ..., A_n)} \right)^2 + \frac{M}{D(A_1, ..., A_n)}
c) Let f=Xn+a1Xn−1+...+anf = X^n + a_1 X^{n-1} + ... + a_n be a separable polynomial of k[X]k[X] and G the Galois group of a splitting extension. Show that (a1,...,an)(a_1, ..., a_n) is substitutable in C and that the following conditions are equivalent (i) every element of G induces an even permutation of the roots of f ; (ii) there exists a∈ka \in k such that C(a1,...,an)=a2+aC(a_1, ..., a_n) = a^2 + a. \* d) Henceforth assume that k is finite and choose the elements U, V, W of Z[T1,...,Tn]\mathbf{Z}[T_1, ..., T_n] arbitrarily to satisfy the condition stated in b). Show that (i) and (ii) of c) are equivalent to the following conditions : (iii)
Tr⁡k/F2(C)=0; \operatorname{Tr}_{k/\mathbf{F}_2}(C) = 0;
(iv)
Tr⁡k/F2(V(a1,...,an)D(a1,...,an))=0. \operatorname{Tr}_{k/\mathbf{F}_2}\left( \frac{V(a_1, ..., a_n)}{D(a_1, ..., a_n)} \right) = 0.
(v) The number of irreducible factors of f is congruent to n modulo 2 (use Ex. 22 of V, p. 165). \*

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.