Intuition for Krasner's Lemma

Question

From Milne's Algebraic Number Theory, we have (he assumes that $K$ is complete with respect to a discrete nonarchimedian absolute value, but I don't know where the discrete part is being used)

Let $\alpha,\beta\in K^{al}$, and assume that $\alpha$ is separable over $K[\beta]$. If $\alpha$ is closer to $\beta$ than any conjugate of $\alpha$ (over $K$), then $K[\alpha]\subset K[\beta]$.

As a corollary, we have

Let $f(X)$ be a monic irreducible polynomial of $K[X]$. Then any monic polynomial $g(X)\in K[X]$ sufficiently close to $f(X)$ is also irreducible, and each root $\beta$ of $g(X)$ belongs to some root of $\alpha$ of $f(X)$. For such a root $K[\alpha]=K[\beta]$.

For me, this says that we can approximate roots of polynomials over $\mathbb{Q}_{p}$ (or an extension) with polynomials over $\mathbb{Q}$, which seems useful.

I tried to read the proof of the lemma and of the corollary, but all I got was that we play around with bounds and having the strong triangle inequality and unique extension of norm is somehow more powerful than my intuition suggests.

I understand that sometimes we have to just roll up our sleeves, compute, and say it's true because the computation says so. However, is there a more intuitive reason for why Krasner's lemma is true? In particular, is there a way to relate this to the picture of extensions of $\mathbb{Q}_{p}$ given in Daniel Litt's answer here: https://mathoverflow.net/questions/51905/how-to-picture-mathbbc-p?

Answer 1

Be careful - your statement of Krasner's Lemma contains a grammatical ambiguity. The statement should be:

Let $K$ be a local field complete with respect to a nontrivial nonarchimedean valuation, and let $K^\mathrm{al}$ be an algebraic closure of $K$. Let $\alpha, \beta \in K^\mathrm{al}$, where $\alpha$ is separable over $K(\beta)$. If $|\beta - \alpha| < |\sigma\alpha - \alpha|$ as we run through all conjugates $\sigma \alpha \neq \alpha$ over $K$, then $K(\alpha) \subseteq K(\beta)$.

(I don't think discreteness of the valuation is used anywhere in the proof.)

Picture the $\sigma \alpha$ as points in $K^\mathrm{al}$, each with its own neighborhood. Krasner says that, for all $\beta$ in a small enough neighborhood of $\alpha$, the only conjugate of $\alpha$ over the base field $K(\beta)$ is $\alpha$ itself, so that every automorphism in $\mathrm{Gal}(K^\mathrm{al}/K)$ that fixes $\beta$ must also fix $\alpha$.