Bourbaki

Exercise 20, § 2

Algebra, VI, p. 42

Let K be a maximal ordered field and

f(T)=a0+(n1)a1T+(n2)a2T2+⋯+anTn f(T) = a_0 + \binom{n}{1} a_1 T + \binom{n}{2} a_2 T^2 + \cdots + a_n T^n
a polynomial in K[T]K[T]. For all integers p,qp, q such that 0≤p<p+q≤n0 \leq p < p + q \leq n, the number of roots of the polynomial
ap+(q1)ap+1T+(q2)ap−2T2+⋯+ap+qTq a_p + \binom{q}{1} a_{p+1} T + \binom{q}{2} a_{p-2} T^2 + \cdots + a_{p+q} T^q
counted with multiplicity, which do not belong to K is at most equal to the number of roots of ff, counted with multiplicity, which do not belong to K (use Ex. 12). Deduce that if b1,…,bnb_1, \ldots, b_n are nn distinct elements >0> 0 in K, and if
(T+b1)⋯(T+bn)=Tn+(n1)m1Tn−1+⋯+mn (T + b_1) \cdots (T + b_n) = T^n + \binom{n}{1} m_1 T^{n-1} + \cdots + m_n
then
mk1/k>mk+11/(k+1)for1≤k≤n−1. m_k^{1/k} > m_{k+1}^{1/(k+1)} \quad \text{for} \quad 1 \leq k \leq n-1 .

Solution

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Transcribed from the printed volume. Not translated.