# Is a Galois extension over $\mathbb{Q}$ always finite?

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.

• There are infinite Galois extensions, for instance the field of all algebraic numbers. – Lord Shark the Unknown Dec 27 '18 at 20:38
• No, the whole algebraic closure is an example of an infinite Galois extension – Wojowu Dec 27 '18 at 20:38
• Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials? – PerelMan Dec 27 '18 at 20:40
• @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 – Hagen von Eitzen Dec 27 '18 at 21:55
• OP it would help if you told us your definition of Galois extension – 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 $$n=\deg p(x)$$.