TITLE: Determine the Galois Group of $(x^2-2)(x^2-3)(x^2-5)$
QUESTION [7 upvotes]: Determine all the subfields of the splitting fields of this polynomial. 
I chose this problem because I think to complete it in great detail will be a great study tool for all of the last chapter, as well. (I'm using Dummit & Foote). 
Can I have help outlining the method? I need to first find the splitting field, correct? Is there a more elegant observation than noting that I can use difference of squares factoring and rewrite the polynomial as $(x- \sqrt{2})(x+\sqrt{2})(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})$, where none of $\sqrt{2}, \sqrt{3}, \sqrt{5}$ are in the field $\mathbb{Q}$? (The problem doesn't say this is the base field, but I'm assuming.) So the splitting field is $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$. Is this how you recommend finding the splitting field, i.e. explicitly calculating the roots and ensuring your extension contains them? I wish there were a more theoretical way of doing this that would apply to any polynomial. 
Now, to find the Galois group, do you exhaustively compute automorphisms and just observe which elements the automorphism fixes in order to find the corresponding fixed field? This seems to get complicated very quickly. Is there some theoretical machinery to aid in this, as well?
Once finding the fixed fields are accomplished, from the Fundamental Theorem of Galois Theory I know there exists a correspondence between the fixed fields and the subfields, so if I find all fixed fields of automorphisms, I will have found all subfields. 
Thank you for any help.

REPLY [2 votes]: To add to Mr. Becker's answer, an easier way of seeing that $\sqrt 5 \notin \mathbb Q(\sqrt2,\sqrt3)$ (once it has been determined that that $\sqrt 2, \sqrt 3$, and $\sqrt 6$ are linearly independent over $\mathbb Q$) is to notice that the Galois group of $\mathbb Q(\sqrt2,\sqrt3)$ is the Klein-4 group, which has 3 subgroups of order 2. By the fundamental theorem of Galois theory, these subgroups correspond to the quadratic extensions $\mathbb Q(\sqrt 2), \mathbb Q(\sqrt3)$ and $\mathbb Q(\sqrt 6)$. By following the same steps used to show that $\sqrt 2 \notin \mathbb Q(\sqrt 3)$, one can see that $\sqrt 5$ does not belong to any of the aforementioned quadratic extensions. So if $\sqrt 5 \in \mathbb Q(\sqrt2,\sqrt3)$, then $\mathbb Q(\sqrt 5)$ would constitute a fourth quadratic extension, corresponding to a fourth subgroup of degree 2 in the Klein-4 group, a contradiction.