Bourbaki

Exercise 22, § 2

Algebra, VI, p. 42

Let K be a maximal ordered field and let

f(T)=a0+a1T+…+anTn f(T) = a_0 + a_1 T + \ldots + a_n T^n
be a polynomial in K[T]K[T] such that a0≠0≠ana_0 \neq 0 \neq a_n and
ap=ap+1=⋯=ap+2m−1=0(1≤p<p+2m−1<n) a_p = a_{p+1} = \cdots = a_{p+2m-1} = 0 \quad (1 \leq p < p + 2m - 1 < n )
Show that ff has at most n−2mn - 2m roots in K, counted with multiplicity (apply Descartes' rule). Deduce that if g(T)=1+c1T+…+cnTng(T) = 1 + c_1 T + \ldots + c_n T^n has all its roots in K, and if h(T)=1+b1T+…+b2mT2mh(T) = 1 + b_1 T + \ldots + b_{2m} T^{2m} is the polynomial consisting of the first 2m+12m + 1 terms of the formal power series 1/g∈K[[T]]1/g \in K[[T]], then h has no roots in K.

Solution

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Transcribed from the printed volume. Not translated.