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Let $k$ be a field and $E/k$ a Galois extension with Galois group $G$ and $[E:k]=n$ and $m\in\mathbb{N}$. Is there any procedure to find a separable extension $F/k$ with $[F:k]=m$ such that

  1. $EF/k$ is Galois.
  2. $[EF:k]=mn$.
  3. $E\cap F=k$?

Or alternatively, finding a separable extension $F$ such that $\text{Gal}(EF/k)\cong \text{Gal}(E/k)\rtimes \text{Gal}(F/k)$?

It does not matter if we restrict the conditions of the problem over a certain characteristic, or adding the restriction for the Galois groups to be cyclic.

(So, my question, again, is if, under some, or no restrictions, we can find that extension, and if that's the case, how).

For example, given $E/\mathbb{Q}$ with Galois group $G=\mathbb{Z}/7\mathbb{Z}$, how can we find an extension $F/\mathbb{Q}$ such that $\text{Gal}(F/\mathbb{Q})=\mathbb{Z}/3\mathbb{Z}$ and $\text{Gal}(EF/\mathbb{Q})\cong\mathbb{Z}/7\mathbb{Z}\rtimes \mathbb{Z}/3\mathbb{Z}$?

Oh, and last. Are the two problems equivalent? For the direct product we replace separable with Galois and it holds. But in this case I fear that something might break.

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Not if $k=\Bbb C$ or $\Bbb R$; these have too few extensions.

Over $\Bbb Q$ the answer is yes, even if you insist that $F$ be cyclic. There will be a prime $p$, not diving the discriminant on $E$, with $p\equiv1\pmod m$. Take $F$ to be the degree $m$ subfield of the $p$-th cyclotomic field.

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