Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
@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

1 Answer 1

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 $\ell$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.