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?

  • 2
    $\begingroup$ "the strong triangle inequality and unique extension of norm is somehow more powerful than my intuition suggests" - this sentence alone sums up large amounts of my experience with nonarchimedean things. The absolute value simply seems to contain unusually large amounts of information about the arithmetic of the field. $\endgroup$
    – Billy
    Commented Apr 23, 2013 at 3:37

1 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 $\alpha$.

  • $\begingroup$ Accidentally, it seems to me that the alternative semantic interpretation ($|\beta-\alpha|<|\beta-\sigma\alpha|$, which follows directly from the strong triangle inequality) seems more invitating to a Galois-theoretic proof. $\endgroup$
    – Fan Zheng
    Commented Nov 27, 2015 at 23:58
  • $\begingroup$ Since the valuation is non archimedean, the two interpretations are equivalent $\endgroup$ Commented May 10, 2018 at 17:59

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