Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 2

Krasner's lemma states that if $K$ is complete, $\alpha$ is separable over $K$, and an algebraic element $\beta$ satisfiesfor every other $K$-conjugate ${\alpha }^{\prime }$ of $\alpha$, then $K(\alpha )\subseteq K(\beta )$.

Let $f\in {\mathbb{C}}_p[X]$ be nonconstant and let $\alpha$ be a root in an algebraic closure of ${\mathbb{C}}_p$. Because the characteristic is zero, replace $f$ by the separable minimal polynomial of $\alpha$. Approximate its coefficients arbitrarily closely by elements of ${\overline{\mathbb{Q}}}_p$. By continuity of roots over a non-Archimedean field, the approximating polynomial has a root $\beta$ arbitrarily close to $\alpha$. Since ${\overline{\mathbb{Q}}}_p$ is algebraically closed, $\beta \in {\overline{\mathbb{Q}}}_p\subset {\mathbb{C}}_p$.

Choose the approximation so that $\beta$ is closer to $\alpha$ than every other ${\mathbb{C}}_p$-conjugate of $\alpha$. Krasner's lemma givesso $\alpha \in {\mathbb{C}}_p$. Hence completion of an algebraic closure of a p-adic field is algebraically closed proves that ${\mathbb{C}}_p$ is algebraically closed.

For the coefficient $a_n=1/n!$, the p-adic absolute value gives $|a_n{|}_p=p^{v_p(n!)}$. By Legendre formula,where $s_p(n)$ is the sum of the base-$p$ digits of $n$. Thus $v_p(n!)/n\to 1/(p-1)$. The Cauchy-Hadamard theorem now yieldsand therefore the radius of convergence of the p-adic exponential is