Exercise 2, § 4

Let K be an algebraically closed field of characteristic 0. $F = K((X^{1/n!}))$ the field of formal power series in $X^{1/n!}$, and $E = K((X))$, where $X = (X^{1/n!})^{n!}$, the subfield of F consisting of formal power series in X. For every formal power series $f \in K((X))$ we denote by $\omega(f)$ its order. a) Let $P(Y) = a_n Y^n + a_{n-1} Y^{n-1} + \ldots + a_1$ be a polynomial of $E[Y]$ of degree n, such that $a_n \neq 0$. Show that there exists a strictly increasing sequence $(i_k)_0, \ldots, r$, of integers in the interval $[0, n]$ such that: 1" $i_0 = 0, i_r = n$; 2" $\omega(a_{i_k})$ is finite for $0 \leq k \leq r$; 3" for each index j such that $0 \leq j \leq n$, distinct from the $i_k$ and such that $\omega(a_j)$ is finite, the point $(j, \omega(a_j)) \in \mathbf{R}^2$ is above the straight line passing through the points $(i_k, \omega(a_{i_k}))$, $(i_{k-1}, \omega(a_{i_{k-1}}))$ and strictly above it if $j < i_{k-1}$ or $j > i_k$ ($1 \leq k \leq r$). The union of the segments joining the points $(i_{k-1}, \omega(a_{i_{k-1}}))$ and $(i_k, \omega(a_{i_k}))$ for $1 \leq k \leq r$ is called the Newton polygon of P, the preceding segments are called its sides and the points $(i_k, \omega(a_{i_k}))$ its vertices. b) Put $\rho_k = i_k - i_{k-1}$, $\sigma_k = (\omega(a_{i_k}) - \omega(a_{i_{k-1}}))/\rho_k$; show that P has exactly $\rho_k$ zeros (counted with the right multiplicity) belonging to the field $K((X^{1/(\rho_k!)})$ and whose orders are equal to $\sigma_k$ (determine by recursion the coefficients of a formal power series of this type satisfying the equation $P(z) = 0$). c) Conclude that the union of the fields $K((X^{1/n}))$ for all integers $n > 0$ is an algebraic closure of the field $E = K((X))$ ("Puiseux's theorem"). d) If $K'$ is a field of characteristic 2 and $E' = K'((X))$, then the polynomial $Y^2 + XY + X$ of $E'[Y]$ does not have a root in the union of the fields $K'((X^{1/n!}))$.