Are all Galois extensions simple? 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.
 A: It is a well-known result that: 

Every finite separable extension is simple. 

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.  
