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Let $K$ be a field. Let $p$ be any prime number.

Can one always construct an algebraic extension $K_p$ of $K$ with the following properties?

(1) If $L$ is a finite extension of $K$ contained in $K_p$, then $[L:K]$ is coprime to $p$.

(2) If $L$ is a finite extension of $K_p$, then $[L:K_p]$ is a power of $p$.

If yes, how?

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I think this is true whenever $\text{Gal}(K_s/K)$ is abelian and $K$ is perfect, since you can take $K_p$ to be the fixed field of the pro-$p$ part. In general I am not sure what happens. – Qiaochu Yuan Jun 26 '11 at 23:46
One would be tempted to take $K_p$ to be the compositum of all finite extensions of $K$ of degree prime to $p$, but it's not obvious that works. – Arturo Magidin Jun 26 '11 at 23:48
@Arturo I am pretty sure that it doesn't. As we all know, two cubic extensions can together generate a degree 6 extension. – Alex B. Jun 27 '11 at 0:11
up vote 8 down vote accepted

Yes, this is true for all fields $K$.

Step 1: Suppose $K$ is perfect, so that $\overline{K}/K$ is Galois. Then the result follows from the (straightforward) extension of the usual Sylow theory from finite groups to profinite groups. Namely, for $G = \operatorname{Aut}(\overline{K}/K)$, one has for each prime $p$ a Sylow p-subgroup $G_p$, a closed pro-$p$-subgroup of $G$ such that the index $[G:G_p]$ is a supernatural number which is prime to $p$. What this means is precisely what is desired: put $K_p = \overline{K}^{G_p}$. Then every finite subextension of $K_p/K$ has degree prime to $p$ and every finite subextension of $\overline{K}/K_p$ has degree a power of $p$.

For the construction and basic facts about profinite Sylow subgroups, consult any text where profinite groups are systematically discussed, for instance Serre's Galois Cohomology, $\S 1.4$.

Step 2: Suppose $K$ is not perfect of characteristic $p$, and the prime number in question is also equal to $p$. Then one applies the same construction with $K^{\operatorname{sep}}$ instead of $\overline{K}$: this works since every finite subextension of $\overline{K}/K^{\operatorname{sep}}$ has $p$-power degree.

Step 3: Suppose $K$ is imperfect of characteristic $p$ and the prime in question $\ell$ is different from $p$. Then we pass first from $K$ to its perfect closure $L = K^{p^{-\infty}}$, which is a perfect field and a purely inseparable extension of $K$. Thus $\overline{L} = L^{\operatorname{sep}}$, so Step 1 applied to $L$ gives us a field extension $L_{\ell}$ such that every finite subextension of $L_{\ell}/L$ has degree prime to $\ell$ and every finite extension of $\overline{L}/L_{\ell}$ has degree a power of $\ell$. We may therefore take $K_{\ell} = L_{\ell}$ since every finite subextension of $L/K$ also has degree prime to $\ell$.

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