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This construction comes from the chapter VII of "the local Langlands conjecture for $GL(2)$":

Let $F$ be a local field. We know that for any $n \in \mathbb{N}$ there exists a unique unramified extension of degree $n$. We can denote by $F_{\infty}$ the composite of all these fields that is clearly the maximal unramified extension of $F$. Now for each integer $n\geq 1$ such that $p \not\mid n$ (where $p$ is the characteristic of the residue field of $F$), the field $F_{\infty}$ has a unique extension $E_n/F_{\infty}$ of degree $n$.

My question is: why has $F_{\infty}$ a unique extension of degree $n$?

This is a totally tamely ramified extension and so $E_n$ is generated, over $F_\infty$, by an $n$-th root of a uniformizer. Moreover I know that there exists a finite number of extensions of fixed degree for a local field. But now I don't know how to proceed.

KCd
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Mario
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1 Answers1

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By your last paragraph, it appears that you know how to show each tamely totally ramified extension of a local field $K$ with degree $n$ is $K(\sqrt[n]{\pi})$ where $\pi$ is some choice of uniformizer in $K$. That is an important background step.

Because $F_\infty$ is the maximal unramified extension of $F$, its finite extensions are all totally ramified, so a finite extension of $F_\infty$ with degree $n$ that is not divisible by $p$ is a tamely totally ramified extension of $F_\infty$ with degree $n$. By the same type of reasoning used to get the result in the previous paragraph, the extension is $F_\infty(\sqrt[n]{\pi})$ where $\pi$ is some uniformizer in $F_\infty$. We want to show this extension is independent of the choice or $n$th root of $\pi$ and on the choice of $\pi$.

  1. The field $F_\infty(\sqrt[n]{\pi})$ is independent of the choice of $n$th root of $\pi$ because any two $n$th roots of $\pi$ have a ratio that is an $n$th root of unity and all $n$th roots of unity are in $F_\infty$ due to $F_\infty$ being the maximal unramified extension of $F$ and $p \nmid n$.

  2. The field $F_\infty(\sqrt[n]{\pi})$ is independent of the choice of $\pi$ because if $\pi'$ is another uniformizer then $\pi'/\pi$ is in $\mathcal O_{F_\infty}^\times$ and each unit $u$ in $\mathcal O_{F_\infty}^\times$ is an $n$th power in $\mathcal O_{F_\infty}^\times$ by Hensel's lemma. (Although $F_\infty$ is not complete, $F(u)$ is a finite extension of $F$ and thus is complete, so you apply Hensel's lemma to $x^n - u$ in $F(u)[x]$ to find an $n$th root of $u$ in ${\mathcal O}_{F(u)}^\times$.) So $F_\infty(\sqrt[n]{\pi}) = F_\infty(\sqrt[n]{\pi'})$.

That answers your question.

Similar reasoning goes through with $F_\infty$ replaced by the Laurent series field $\mathbf C((t))$, but it leads to a stronger result since $\mathbf C((t))$ contains all roots of unity: $\mathbf C((t))$ has a unique extension of each degree $n \geq 1$, namely $\mathbf C((t^{1/n}))$ and that field is independent of the choice of $n$th root of $t$. Thus the algebraic closure of $\mathbf C((t))$ is $\bigcup_{n \geq 1} \mathbf C((t^{1/n}))$, and this is true with $\mathbf C$ replaced by any algebraically closed field of characteristic $0$.

KCd
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