# Example of Galois extension over Q which is not cyclotomic

So, prof. introduced Galois extensions yesterday and I do apologise if I did not get something correctly. So, if I am right every finite extension of finite field is almost obviously Galois(using Artin lemma and theorem of prime element as your weapon), and every cyclotomic extension(extension $Q(\epsilon_n)/Q$, where $\epsilon_n$ is a prime n-th root of 1. Now, I am wondering are these two only examples of Galois field, or otherwise told, are there non-cyclotomic extensions of Q that are Galois and are there infinite extensions of finite fields that are Galois. Please provide me examples with proofs if yes, and proofs if no. Thanks for your help!

• You should be able to show that $\mathbb{Q}(\sqrt{2})$ is Galois. – Tobias Kildetoft May 26 '16 at 7:43
• Thanks a lot , Tobias! As I said, I got into expression right away, so I didn't see example! Thanks! You could make that to answer to get nice +10 reputation but your choice! – nikola May 26 '16 at 7:46
• Done (with some additional details). – Tobias Kildetoft May 26 '16 at 7:50

The extension $\mathbb{Q}(\sqrt{2})$ is easily seen to be Galois since it is the splitting field of the polynomial $x^2-2$.
Also (as you will probably see later on), the Galois groups of the examples mentioned here (finite extensions of finite fields and cyclotomic extensions of the rationals) are abelian. But not all Galois groups are abelian, for example one can show that $S_n$ and $A_n$ are obtainable as Galois groups of extensions of $\mathbb{Q}$ for all $n$ (and in fact it is conjectured that all finite groups can be so obtained, though this is still an open problem unless one allows extensions of extensions of $\mathbb{Q}$).
Whenever $F$ is a perfect field, its algebraic closure $\overline{F}$ is a Galois extension of $F$. Since $\mathbb{Q}$ and $\mathbb{F}_p$ for primes $p$ are perfect fields, we arrive at easy examples of infinite Galois extensions by considering their algebraic closures.