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The completion ${\mathbb{C}}_p$ of an algebraic closure of ${\mathbb{Q}}_p$ is an algebraically closed field. Approximate a polynomial over ${\mathbb{C}}_p$ by one over ${\overline{\mathbb{Q}}}_p$, use continuity of roots over a non-Archimedean field to obtain a nearby algebraic root, and then use Krasner's lemma to show that the original root already belongs to ${\mathbb{C}}_p$.

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.

Krasner's lemma says that if $K$ is complete, $\alpha$ is separable over $K$, andthen $K(\alpha )\subseteq K(\beta )$.

To prove it, let $\tau$ be an automorphism of $\overline{K}$ fixing $K(\beta )$. The absolute value has a unique extension to every finite extension of the complete field $K$, so it is invariant under $\tau$. If $\tau (\alpha )\ne \alpha$, the ultrametric inequality and the displayed strict inequality giveBut $\tau (\beta )=\beta$ and invariance gives $|\tau (\alpha )-\beta |=|\alpha -\beta |$, a contradiction. Every automorphism fixing $K(\beta )$ therefore fixes $\alpha$, which proves the field inclusion by Galois correspondence.

Now let $L/{\mathbb{Q}}_p$ be finite. By the primitive element theorem, write $L={\mathbb{Q}}_p(\alpha )$ with separable minimal polynomial $f\in {\mathbb{Q}}_p[X]$. Approximate the coefficients of $f$ closely by those of a polynomial $g\in \mathbb{Q}[X]$ of the same degree. Continuity of roots over a non-Archimedean field gives a root $\beta$ of $g$ arbitrarily close to $\alpha$. Choose it close enough for Krasner's lemma; thenThe degree bound from $\mathrm{deg}g=\mathrm{deg}f$ forces equality. If $F=\mathbb{Q}(\beta )$ and $\mathfrak{p}$ is the prime selected by the embedding $F\hookrightarrow {\overline{\mathbb{Q}}}_p$, its completion isThus every finite extension of ${\mathbb{Q}}_p$ is the completion of a number field at a prime ideal.