From Galois Theory (Rotman):
Let E be the splitting field of $f(x)=x^4-10x^2+1$ over $\Bbb{Q}$. find $Gal(E/\Bbb{Q})$. The roots of $f(x)$ are $\sqrt{2}+\sqrt{3}$, $\sqrt{2}-\sqrt{3}$, $-\sqrt{2}+\sqrt{3}$, $-\sqrt{2}-\sqrt{3}$.
We know that the splitting field of $f(x)=x^4-10x^2+1$ over $\Bbb{Q}$ is $\Bbb{Q}(\sqrt{2}, \sqrt{3})$.
There are two theorems in the textbook:
Theorem 1: If $f(x) \in F[x]$ has $n$ distinct roots in its splitting field $E$, then $Gal(E/F)$ is isomorphic to a subgroup of the symmetric group $S_n$, and so its order is a divisor of $n!$.
Theorem 2: If $f(x) \in F[x]$ is a separable polynomial and if $E/F$ is its splitting field , then $$|Gal(E/F)|=[E:F].$$
From the first theorem, we know that the order of the Galois group divides 4!. There is an example in the textbook that shows that [E:F]=4, so the second theorem says that $|Gal(E/F)|=4$ as well. However, I find this a bit confusing. Since the Galois group is the set of all automorphisms that fix F…we need to look at the possible automorphisms that permute the four distinct roots of $f(x)$. Since there are four distinct roots, there must be 4! ways to permute them (while fixing all the elements of F), right? Doesn't that imply that we are supposed to have 4! elements in the Galois group? Why is the order 4 then?
Thanks in advance.
Best Answer
Not every permutation of the roots is an automorphism of $E/\mathbb Q$. For example, any permutation $\sigma$ which sends $\sqrt2$ to $\sqrt3$ is not an automorphism, since $\sigma(\sqrt2^2)=\sigma(2)=2$ but $\sigma(\sqrt2)^2=3$.
More generally any automorphism of $E/\mathbb Q$ must preserve the roots of polynomials with rational coefficients, in particular the roots of $x^2-2$ and $x^2-3$. So the only possible automorphisms are given by $$\sqrt2\mapsto \sqrt2,\ \sqrt3\mapsto \sqrt3$$ $$\sqrt2\mapsto -\sqrt2,\ \sqrt3\mapsto \sqrt3$$ $$\sqrt2\mapsto \sqrt2,\ \sqrt3\mapsto -\sqrt3$$ $$\sqrt2\mapsto -\sqrt2,\ \sqrt3\mapsto -\sqrt3$$ and since there are exactly $4$ automorphisms, you know that all of these are automorphisms.