Exercise 13, § 8

(a) Let K be a field and $v$ a non-improper valuation on K. Let E be a set filtered by an ultrafilter $\mathscr{U}$ and $\xi \mapsto x_\xi, \xi \mapsto y_\xi$ two mappings of E to K with the topology $\mathscr{T}_v$. Show that, if the mapping $\xi \mapsto x_\xi y_\xi$ converges to 0 in K with respect to $\mathscr{U}$, so does one of the mappings $\xi \mapsto x_\xi, \xi \mapsto y_\xi$. (Observe that, if $\xi \mapsto x_\xi$ has no cluster point with respect to $\mathfrak{U}$, the mapping $z \mapsto x_\xi^{-1}$ is bounded on a subset of U.)

(b) Let $f$ be a non-constant polynomial in $K[X]$. Show that, if $\xi \in f(\mathbf{K})$ converges to 0 in $K$ with respect to $\mathfrak{U}$, $\xi \mapsto x_\xi$ converges in $\hat{K}$ (decompose $f$ into factors in an algebraic extension of $K$ and use (a)).

(c) Suppose that $K$ is algebraically closed in $\hat{K}$. Then the mapping $x \mapsto f(x)$ of $K$ to itself is closed (use (b)). Deduce that, if $g$ is a rational function in $K(X)$ and $B$ is a bounded closed subset of $K$, then $g(B - P)$ (where $P$ is the set of poles of $g$ in $B$) is closed in $K$.

(d) Suppose that the algebraic closure of $K$ is a radicial extension of $K$. Then show that $\hat{K}$ is algebraically closed. (Apply (b) to an algebraic closure of $\hat{K}$ and a polynomial $f \in \hat{K}[X]$; use also Exercise 12 (a).)