I have been looking at Galois extensions and Galois groups and have been wondering if all Galois extensions are simple. I don't think this is true.
For example with $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$ would be a simple extension of $\mathbb{Q}$.
With something like $\mathbb{Q}(i,\sqrt{3})$ over $\mathbb{Q}$ this is Galois, but I don't think it is simple. Is that correct? My initial guess was that $\mathbb{Q}(i,\sqrt{3}) = \mathbb{Q}(i\sqrt{3})$, but I think the right side is just a subfield.
It is a well-known result that:
See the Primitive element theorem.
Thus, yes, all finite Galois extensions are simple.
Infinite Galois extension cannot be simple as every simple algebraic extension is of course finite; basically this is the definition of an element being algebraic.
Let me add that for the specific example, as mentioned in comments, one can take for example $ i + \sqrt{3}$. You are correct that $i \sqrt{3}$ will not work; as it is a root of $X^2+3$ and thus its degree is only two.
It is the third subfield of degree $2$; the other two are the ones generated by $i$ and $\sqrt{3}$ respectively.
Under a Galois theory angle one can say that those elements $\alpha$ work as generators that are not fixed by any of the three non-identity elements of the Galois group, and thus have four conjugates.
Differently the only ones that do not work are those fixed by some non-identity element of the Galois group. These are $i$ and its rational multiples, $\sqrt{3}$ and its rational multiples, and $i\sqrt{3}$ and its rational multiples.