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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?

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If the minimal polynomial is of degree $n$, then the Galois group is a subgroup of $S_n$ so it must divide $n!$. That's about all that can be said in general. Your question basically asks, "How do you compute a Galois group?" The answer, "In general, it's complicated." –  Bill Cook Apr 5 '12 at 1:54
    
In this example it's not that complicated, it's just tedious! Whereas if from the get-go I was able to say, I know there will be 4 elements in the Galois group, then if I find one, the others can follow smoothly, and I know when to stop. But what you are saying is that there is no cut and paste method for situations like this. I.e., we know that for cyclic groups we can apply Euler's totient function to determine the size of the group. –  daniel Apr 5 '12 at 2:31
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If your base field is the rational numbers, there is a beautiful statistical method which allows you to guess the size of the Galois group without computing the Galois group directly. By a change of variables a polynomial $f(x)$ in ${\mathbf Q}[x]$ can be made monic in ${\mathbf Z}[x]$ without changing its splitting field over ${\mathbf Q}$. As $N \rightarrow \infty$, the ratio $\#\{p \leq N : f(x) \bmod p \text{ splits completely}\}/\#\{p \leq N\}$ tends to $1/d$, where $d$ is the size of the Galois group of the splitting field of $f(x)$ over ${\mathbf Q}$. –  KCd Apr 5 '12 at 3:11
    
Since an automorphism of your splitting field must send a root of an irreducible polynomial in $\mathbb{Q}[X]$ to another root of that polynomial, there are really only $4$ permutations of the roots which might give rise to automorphisms. It's not too hard to prove that the extension has degree $4$, and it is Galois because it is the splitting field of a separable polynomial, so the Galois group has $4$ elements, and therefore each admissible permutation of the roots must give rise to an automorphism. –  Keenan Kidwell Apr 5 '12 at 3:13
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For example, let $f(x) = (x^2-2)(x^2-3)$. The ratio in my previous comment, for $N = 100, 1000, 10000$, and 100000 is .2, .2142, .2400, and .2470. If this is supposed to be tending towards a reciprocal of an integer, the likely limit is $1/4 = .25$, so this suggests the splitting field has degree 4 and hence the Galois group has order 4. But this is not a proof since I didn't give error estimates on that limit. This interesting result is a consequence of the Chebotarev density theorem. –  KCd Apr 5 '12 at 3:15

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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.

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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.

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