Galois Theory - $\mathbb{Q}(\sqrt{2},\sqrt{3})$ I'm studying Galois Theory using Pinter's Book of Abstract Algebra. Quoting the book, questions are followed:
$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is an extension of degree 4 over $\mathbb{Q}$, so by Theorem 1, there are four automorphisms of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ which fix $\mathbb{Q}$: Now,  $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the root field of $(x^2 – 2)(x^2 − 3)$ over for it contains the roots of this polynomial, and any extension of containing the roots of $(x^2 – 2)(x^2 − 3)$ certainly contains $\sqrt{2}$ and $\sqrt{3}$.
Since every element of  $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is of the form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, these four automorphisms (shall call them ε, α, β, and γ) are the following:
$a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \xrightarrow{ε} a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$
$a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \xrightarrow{α} a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}$
$a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \xrightarrow{β} a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}$
$a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \xrightarrow{γ} a - b\sqrt{2} - c\sqrt{3} + d\sqrt{6}$
If $K$ is an extension of $F$, the automorphisms of $K$ which fix $F$ form a group.
Question:
1) How is $\sqrt{6}$ root of $(x^2 – 2)(x^2 − 3)$?
2) Wonder if someone could explain to me how does the four automorphisms ε, α, β, and γ mapping work? and how it is that $α \circ β = γ $?
I tried to do a little more reading on my own. Came across this pdf article, http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisappn.pdf
3) How is it that the minimal polynomial of $\sqrt{2} + \sqrt{3}$
is  $X^4 - 10X^2 +1$? 
Thank you!
 A: Answers to your questions...

*

*$\sqrt 6$ is not a root of $(X^2- 2)(X^2- 3)$, but it is an element of the field $\mathbb Q(\sqrt2, \sqrt3)$ since $\sqrt6 = \sqrt2\sqrt3$.


*An automorphism of $\mathbb Q(\sqrt2, \sqrt3)$ is completely determined by where $\sqrt2$ and $\sqrt3$ is sent. Moreover it must send $\sqrt2$ to a root of its minimal polynomial $X^2-2$, i.e., either $\sqrt2$ or $-\sqrt2$ (and similar for $\sqrt3$). For example, the automorphism of $\mathbb Q(\sqrt2, \sqrt3)$ defined by sending $\sqrt2\mapsto -\sqrt2$ and $\sqrt3 \mapsto \sqrt3$, ends up sending $\sqrt6 \mapsto -\sqrt6$ (since it's a ring homomorphism) and corresponds to your $\alpha$ above. Similar for the other maps...


*As noted in the comments, you can check that $\sqrt2 + \sqrt3$ is a root of that polynomial and that it is irreducible over $\mathbb Q$.
Where did it come from? One way: First note that the element $\sqrt2 + \sqrt3$ is not fixed by any nonidentity automorphism of $\mathbb Q(\sqrt2, \sqrt3)$ (check!), so $\mathbb Q(\sqrt2 + \sqrt3) = \mathbb Q(\sqrt2, \sqrt3)$ by Galois theory; since the latter is a degree 4 extension of $\mathbb Q$, it follows that the minimal polynomial of $\sqrt2 + \sqrt3$ over $\mathbb Q$ is degree 4 and its roots are the Galois conjugates of $\sqrt2 + \sqrt3$, i.e. $\epsilon(\sqrt2+\sqrt3), \alpha(\sqrt2 + \sqrt3), \beta(\sqrt2 + \sqrt3), \gamma(\sqrt2 + \sqrt3)$ in your notation. Thus the minimal polynomial factors as $$(X - \epsilon(\sqrt2+\sqrt3))(X-\alpha(\sqrt2 + \sqrt3))(X-\beta(\sqrt2 + \sqrt3))(X- \gamma(\sqrt2 + \sqrt3))$$ in $\mathbb Q(\sqrt2, \sqrt3)$ which you can check equals $X^4 - 10X + 1$.
