# Krasner’s lemma

###### Lemma 1.

(Krasner’s Lemma) Let $K$ be a field of characteristic $0$ complete       with respect to a nontrivial nonarchimedean absolute value   . Assume $\alpha,\beta\in\bar{K}$ (where $\bar{K}$ is some algebraic closure  of $K$) are such that for all nonidentity embeddings    $\sigma\in\operatorname{Hom}_{K}(K(\alpha),\bar{K})$ we have $\left\lvert\alpha-\beta\right\rvert<\left\lvert\sigma(\alpha)-\alpha\right\rvert$. Then $K(\alpha)\subset K(\beta)$.

This says that for any $\alpha\in\bar{K}$, 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 $\sigma\in\operatorname{Hom}_{K(\beta)}(K(\alpha,\beta),\bar{K})$, 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

 $\left\lvert\sigma(\alpha)-\beta\right\rvert=\left\lvert\sigma(\alpha)-\sigma(% \beta)\right\rvert=\left\lvert\sigma(\alpha-\beta)\right\rvert=\left\lvert% \alpha-\beta\right\rvert$

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

 $\left\lvert\sigma(\alpha)-\alpha\right\rvert=\left\lvert(\sigma(\alpha)-\beta)% +(\beta-\alpha)\right\rvert\leq\max(\left\lvert\sigma(\alpha)-\beta\right% \rvert,\left\lvert\alpha-\beta\right\rvert)=\left\lvert\alpha-\beta\right\rvert$

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

###### Proposition 2.

With $K$ as above, let $P(X)\in K[X]$ be a monic irreducible polynomial  of degree $n$ with (distinct) roots $\alpha_{1},\ldots,\alpha_{n}$. Then any monic polynomial  $Q(X)\in K[X]$ of degree $n$ that is “sufficiently close” to $P(X)$ will be irreducible  over $K$ with roots $\beta_{1},\ldots\beta_{n}$, and (after renumbering) $K(\alpha_{i})=K(\beta_{i})$.

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 $\left\lvert\cdot\right\rvert$.

###### Proof.

Since $P(X)$ has distinct roots, we may choose $0<\gamma<\min(\left\lvert\alpha_{i}-\alpha_{j}\right\rvert)$ for $i\neq j\leq n$. 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 $\beta_{1},\ldots,\beta_{n}$ with $\left\lvert\alpha_{i}-\beta_{i}\right\rvert<\gamma$. But $\{\alpha_{j}\}_{j\neq i}$ are all the Galois conjugates of $\alpha_{i}$, and $\left\lvert\alpha_{i}-\beta_{i}\right\rvert<\gamma<\left\lvert\alpha_{i}-% \alpha_{j}\right\rvert$ by construction, so by Krasner’s lemma, $K(\alpha_{i})\subset K(\beta_{i})$. But

 $[K(\beta_{i}):K]\leq\deg Q=\deg P=[K(\alpha_{i}):K]$

so that $K(\beta_{i})=K(\alpha_{i})$. In addition  , we see that $\deg Q=[K(\beta_{i}):K]$ and thus that $Q(X)$ is irreducible. ∎

###### Corollary 3.

Let $K$ be a finite extension of $\mathbb{Q}_{p}$ of degree $n$. Then there is a number field $E$ and an absolute value $\left\lvert\cdot\right\rvert$ on $E$ such that $\hat{E}\cong K$.

###### 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 $\left\lvert\cdot\right\rvert_{E}$, the restriction    of the absolute value on $\mathbb{Q}_{p}(\alpha)=\mathbb{Q}_{p}(\beta)$. Then $\hat{E}$ 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 $\hat{E}=\mathbb{Q}_{p}(\beta)=\mathbb{Q}_{p}(\alpha)=K$. ∎

###### Lemma 4.

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

###### Proof.

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

Title Krasner’s lemma KrasnersLemma 2013-03-22 19:03:02 2013-03-22 19:03:02 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 12J99 msc 11S99 msc 13H99