Size of Galois Group based on roots and minimal polynomial If we know the minimal polynomial of a field extension, how can we determine the number of elements in the Galois Group? 
For example, take $\mathbb{Q}(\sqrt{2},\sqrt{3})$, with minimal polynomial $ (x^{2} - 2)(x^{2} - 3)$ and roots $ \pm \sqrt2, \pm \sqrt3$.  Of the $4!$ bijections possible between the roots, is there a method for determining the number that are automorphisms?  
 A: As explained in my comment, to conclude that all $4$ admissible permutations of the roots give rise to automorphisms, it suffices to show that the degree of the splitting field $K$ of $(x^2-2)(x^2-3)$ over $\mathbb{Q}$ has degree $4$. Clearly $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. You have a tower $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2})\subseteq K$. The first extension is degree $2$ because $X^2-2$ is irreducible by Eisenstein. Note that $K$ is obtained from $L:=\mathbb{Q}(\sqrt{2})$ by adjoining a root of $X^2-3$. Thus the degree of $L$ over $K$ is at most $2$. If it is equal to $1$, then $X^2-3$ splits over $L$. Thus you can write $\sqrt{3}=a+b\sqrt{2}$ for rational numbers $a$ and $b$. Squaring gives $3=a^2+2ab\sqrt{2}+2b^2$. Since $\{1,\sqrt{2}\}$ is a $\mathbb{Q}$-basis for $L$, this forces $a^2+2b^2=3$ and $2ab=0$, so $a=0$ or $b=0$. If $a=0$, then $2b^2=3$ so $3/2$ is a rational square, which is false. Similarly, $b=0$ leads to $a^2=3$ with $a$ rational, another contradiction. Thus $X^2-3$ can't split in $L$, so it's irreducible over $L$, and $K/L$ has degree $2$, giving $[K:\mathbb{Q}]=2\cdot 2=4$.
If for some reason you don't want to use the fact that $3/2$ is not a rational square, you could use the (non-trivial) fact that $a$ and $b$ must actually be integers, and then $a^2+2b^2=3$ forces $a$ and $b$ to each be $\pm 1$, while $2ab=0$ forces one of them to be zero, again a contradiction.
A: I think the question is badly stated. Let's work over the rationals. Let $K$ be a finite extension. What do you mean by the minimal polynomial of $K$? Presumably, you mean the minimal polynomial $p$ for some $\alpha$ such that $K={\bf Q}(\alpha)$. Of course, there are many such $\alpha$, so there are many minimal polynomials. 
Now, what do you mean by "the Galois group"? You might mean the group of automorphisms of $K$, but many sources don't call that a Galois group unless $K$ is a normal extension. You might mean the Galois group of the polynomial $p$, in which case you are referring not to the automorphisms of $K$ but of the splitting field of $p$, which field might be $K$ but might be considerably larger. 
Now let's look at your example. That field is the splitting field of $(x^2-2)(x^2-3)$ (it contains all the roots of that polynomial, and no smaller field does), so it is a normal extension. When $K$ is normal, the number of elements in its Galois group equals its degree which, in this case, is clearly 4. The hard cases are the ones where the field is not a splitting field. 
