Let $K$ be a Galois extensions over $\mathbb{Q}$.

Is $K$ always a splitting field of some $P\in \mathbb{Q}[X]$? in which case K would be a finite extension.

I don't know where to start. I tried to use the primitive element theorem but I can't prove there are finitely many intermediate fields.

Thanks for your help, hints.

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    $\begingroup$ There are infinite Galois extensions, for instance the field of all algebraic numbers. $\endgroup$ – Lord Shark the Unknown Dec 27 '18 at 20:38
  • 1
    $\begingroup$ No, the whole algebraic closure is an example of an infinite Galois extension $\endgroup$ – Wojowu Dec 27 '18 at 20:38
  • $\begingroup$ Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials? $\endgroup$ – PerelMan Dec 27 '18 at 20:40
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    $\begingroup$ @PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial $\endgroup$ – Hagen von Eitzen Dec 27 '18 at 21:55
  • $\begingroup$ OP it would help if you told us your definition of Galois extension $\endgroup$ – D_S Dec 29 '18 at 16:23

There are infinite algebraic Galois extensions of $\mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $p\in\mathbb{Z}$ is a prime. Remember that, since $\mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.

Now, if $K$ is a splitting field of a (only one) polynomial $p(x)\in\mathbb{Q}[x]$, then $K/\mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:\mathbb{Q}]\leq n!$, where $n=\deg p(x)$.

Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.


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