Krasner’s lemma

Krasner’s lemma (along with Hensel’s lemma) connects valuations on fields to the algebraic structure of the fields, and in particular to polynomial roots.

Lemma 1.

(Krasner’s Lemma) Let $K$ be a field of characteristic $\mathrm{0}$ complete with respect to a nontrivial nonarchimedean absolute value. Assume $\alpha \mathrm{,}\beta \mathrm{\in }\overline{K}$ (where $\overline{K}$ is some algebraic closure of $K$) are such that for all nonidentity embeddings $\sigma \mathrm{\in }{\mathrm{Hom}}_K\mathit{}\mathrm{(}K\mathit{}\mathrm{(}\alpha \mathrm{)}\mathrm{,}\overline{K}\mathrm{)}$ we have . Then $K\mathit{}\mathrm{(}\alpha \mathrm{)}\mathrm{\subset }K\mathit{}\mathrm{(}\beta \mathrm{)}$.

This says that for any $\alpha \in \overline{K}$, there is a neighborhood of $\alpha$ each of whose elements generates at least the same field as $\alpha$ does.

Proof.

It suffices to show that for every $\sigma \in {\mathrm{Hom}}_{K(\beta )}(K(\alpha ,\beta ),\overline{K})$, we have $\sigma (\alpha )=\alpha$, for then $\alpha$ is in the fixed field of every embedding of $K(\beta )$, so $\alpha \in K(\beta )$. Note that

$$\left|\sigma (\alpha )-\beta \right|=\left|\sigma (\alpha )-\sigma (\beta )\right|=\left|\sigma (\alpha -\beta )\right|=\left|\alpha -\beta \right|$$

where the final equality follows since $\left|\sigma (\cdot )\right|$ is another absolute value extending ${|\cdot |}_K$ to $K(\alpha ,\beta )$ and thus must be equal to $|\cdot |$. But then

$$\left|\sigma (\alpha )-\alpha \right|=\left|(\sigma (\alpha )-\beta )+(\beta -\alpha )\right|\le \max (\left|\sigma (\alpha )-\beta \right|,\left|\alpha -\beta \right|)=\left|\alpha -\beta \right|$$

But this is impossible by the bounds on $\alpha ,\beta$ unless $\sigma (\alpha )=\alpha$. ∎

The first application of Krasner’s lemma is to show that splitting fields are “locally constant” in the sense that sufficiently close polynomials in $K[X]$ have the same splitting fields.

Proposition 2.

With $K$ as above, let $P\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\in }K\mathit{}\mathrm{[}X\mathrm{]}$ be a monic irreducible polynomial of degree $n$ with (distinct) roots ${\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_n$. Then any monic polynomial $Q\mathit{}\mathrm{(}X\mathrm{)}\mathrm{\in }K\mathit{}\mathrm{[}X\mathrm{]}$ of degree $n$ that is “sufficiently close” to $P\mathit{}\mathrm{(}X\mathrm{)}$ will be irreducible over $K$ with roots ${\beta }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathit{}{\beta }_n$, and (after renumbering) $K\mathit{}\mathrm{(}{\alpha }_i\mathrm{)}\mathrm{=}K\mathit{}\mathrm{(}{\beta }_i\mathrm{)}$.

Here “sufficiently close” means the following: consider the space of degree $n$ polynomials over $K$ as homeomorphic to $K^n$ as a topological space; close then means close in the obvious metric induced by $|\cdot |$.

Proof.

Since $P(X)$ has distinct roots, we may choose for $i\ne j\le n$. Since the roots of a polynomial vary continuously with its coefficients, we say that a degree $n$ polynomial $Q(X)\in K[X]$ is sufficiently close to $P(X)$ if $Q(X)$ has roots ${\beta }_1,\mathrm{\dots },{\beta }_n$ with . But ${\{{\alpha }_j\}}_{j\ne i}$ are all the Galois conjugates of ${\alpha }_i$, and by construction, so by Krasner’s lemma, $K({\alpha }_i)\subset K({\beta }_i)$. But

$$[K({\beta }_i):K]\le \mathrm{deg}Q=\mathrm{deg}P=[K({\alpha }_i):K]$$

so that $K({\beta }_i)=K({\alpha }_i)$. In addition, we see that $\mathrm{deg}Q=[K({\beta }_i):K]$ and thus that $Q(X)$ is irreducible. ∎

We use this fact to show that every finite extension of ${\mathbb{Q}}_p$ arises as a completion of some number field.

Corollary 3.

Let $K$ be a finite extension of ${\mathrm{Q}}_p$ of degree $n$. Then there is a number field $E$ and an absolute value $\mathrm{|}\mathrm{\cdot }\mathrm{|}$ on $E$ such that $\widehat{E}\mathrm{\cong }K$.

Proof.

Let $K={\mathbb{Q}}_p(\alpha )$ and let $P$ be the minimal polynomial for $\alpha$ over ${\mathbb{Q}}_p$. Since $\mathbb{Q}$ is dense in ${\mathbb{Q}}_p$, we can choose $Q(X)\in \mathbb{Q}[X]$ (note: in $\mathbb{Q}[X]$, not ${\mathbb{Q}}_p[X]$), and $\beta$ a root of $Q(X)$, as in the proposition, so that ${\mathbb{Q}}_p(\alpha )={\mathbb{Q}}_p(\beta )$. Let $E=\mathbb{Q}(\beta )$. Clearly $E$ is a number field which, when regarded as embedded in ${\mathbb{Q}}_p(\beta )$, has absolute value ${|\cdot |}_E$, the restriction of the absolute value on ${\mathbb{Q}}_p(\alpha )={\mathbb{Q}}_p(\beta )$. Then $\widehat{E}$ is a complete field with respect to that absolute value; ${\mathbb{Q}}_p(\beta )$ is as well, and $E$ is dense in both, so we must have $\widehat{E}={\mathbb{Q}}_p(\beta )={\mathbb{Q}}_p(\alpha )=K$. ∎

Finally, we can prove the following generalization of Krasner’s Lemma, which is also given that name in the literature:

Lemma 4.

Let $K$ be a field of characteristic $\mathrm{0}$ complete with respect to a nontrivial nonarchimedean absolute value, and $\overline{K}$ an algebraic closure of $K$. Extend the absolute value on $K$ to $\overline{K}$; this extension is unique. Let $\widehat{\overline{K}}$ be the completion of $\overline{K}$ with respect to this absolute value. Then $\widehat{\overline{K}}$ is algebraically closed.

Proof.

Let $\alpha$ be algebraic over $\widehat{\overline{K}}$ and $P(X)$ its monic irreducible polynomial in $\widehat{\overline{K}}[X]$. Since $\overline{K}$ is dense in $\widehat{\overline{K}}$, by proposition 2 we may choose $Q(x)\in \overline{K}[X]$ with a root $\beta \in \widehat{\overline{K}}$ such that $\widehat{\overline{K}}(\alpha )=\widehat{\overline{K}}(\beta )$. But $\widehat{\overline{K}}(\beta )=\widehat{\overline{K}}$ so that $\alpha \in \widehat{\overline{K}}$. ∎

Title	Krasner’s lemma
Canonical name	KrasnersLemma
Date of creation	2013-03-22 19:03:02
Last modified on	2013-03-22 19:03:02
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	6
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 12J99
Classification	msc 11S99
Classification	msc 13H99