How to apply the Krasner's lemma to prove $\mathbb C_p$ is algebraically closed?

Question

Krasner's Lemma is the following.

Krasner's Lemma. Let $K$ be a non-archimedian complete field of characteristic $0$ and $a,b \in \overline{K}$. Suppose $$|b-a| < |a-\sigma(a)| \quad \text{for every } \sigma \in \operatorname{Gal}(\overline{K}/K) \text{ with } \sigma(a) \neq a.$$ Then, $K(a) \subseteq K(b)$.

I heard we can apply Krasner's lemma to the proof of $\mathbb C_p$, completion of algebraic closure of $\mathbb Q_p$, is algebraically closed.

Could you tell me how to apply the Krasner's lemma to prove $\mathbb C_p$ is algebraically closed ?

My thoughts.

Arbitrary $α∈$$\mathbb C_p$ is a root of nonzero polynomial $a_nx^n＋・・・＋a_1x＋a_0$. Let $α_1,α_2, ・・・,α_n$ be conjugate of $α$.

Suppose there exists $α_k$ which does not belong to $\mathbb C_p$ . Let's find contradiction. $b_n→α$ be cauchy sequence which converges to $α_k$, then there exists some natural number $N$ such that $b_n$ is closer rather than any other $α_j$($j$ is not $k$) to $α_k$. Then, from Krasner's lemma, we can say $b_n$∈$\mathbb Q_p（α_k)$. But I cannot find contradiction from here.

Thank you in advance.

Answer 1

Just to get this off the unanswered list, I'll give a few hints on top of my comments:

If we want to apply Krasner's Lemma, the only reasonable complete field in sight, which we can take as $K$, is $K=\mathbb C_p$ itself.

Then Krasner would tell us that for certain elements $a,b$ in an algebraic closure of $\mathbb C_p$, we have $\mathbb C_p(a) \subseteq \mathbb C_p(b)$.

Since we want to show that $\mathbb C_p$ actually equals its algebraic closure, i.e. that for any $a$ in such algebraic closure, we have $\mathbb C_p(a) \subseteq \mathbb C_p$, it would be a good idea to find a $b$ like in the lemma but also such that $\mathbb C_p(b) = \mathbb C_p$.

Task: For $a$ algebraic over $\mathbb C_p$, find $b \in \mathbb C_p$ such that $b$ is very close to $a$ (formally: so close that the assumption of Krasner's Lemma is satisfied).

Idea: So far we have not used that the algberaically closed field $\mathbb Q_p^{alg}$ is dense in $\mathbb C_p$. Now, $a$ is by definition the root of some polynomial

$$c_n X^n + \dots + c_1 X + c_0$$

($n \ge 1$, all $c_i \in \mathbb C_p$, $c_n \neq 0$). We can change all coefficients a tiny bit to make them lie in $\mathbb Q_p^{alg}$, and all the roots of this new polynomial would be in $\mathbb Q_p^{alg}$, because that field is algebraically closed. Now changing the coefficients will have changed the roots, but maybe just a tiny bit too! So one of those new roots might be a very good $b \in \mathbb Q_p^{alg} \subset \mathbb C_p$ for our task.