Exercise 15, § 8

(a) Let $K$ be a field, $v_1, v_2$ two independent valuations (§ 7, no. 2) on $K$ and $K_1, K_2$ the completions of $K$ with respect to $v_1$ and $v_2$ respectively. Let $L_1, L_2$ be two fields such that $K \subset L_1 \subset K_1, K \subset L_2 \subset K_2$, which are Henselian (with the extensions by continuity of $v_1$ and $v_i$ respectively). Let $g_1 \in L_1[X], g_2 \in L_2[X]$ be two separable polynomials with the same degree $n$. Show that there exists a polynomial $h \in K[X]$ which, in $L_i[X]$, has the same type of decomposition (Exercise 12 (c)) as $g_i$ ($i = 1, 2$). (Reduce it to the case where $g_1$ and $g_2$ are in $K[X]$ and are monic; consider the polynomial

where $v_1(a) \leq 0, v_2(a) > 0$ is arbitrarily large and $v_1(b) > 0$ is arbitrarily large.)

(b) Deduce from (a) that, if $K$ is Henselian with $v_1$, the algebraic closure of $L_i$ is radicial over $L_i$ and $K_2$ is algebraically closed (take $g_2$ to be irreducible and $g_1 \in K[X]$ to be the product of $n$ distinct factors of prime degree).

(c) Deduce from (b) that, if $K$ is Henselian with $v_1$ and $v_2$, the algebraic closure of $K$ is radicial over $K$.

(d) Show that, if K is Henselian with a discrete valuation v, the algebraic closure of K cannot be radicial over K and therefore K cannot be Henselian with any valuation independent of v (use Algebra, Chapter V, § 11, Exercise 12).