Galois group of a quartic which is also a quadratic in $x^2$

Question

A few weeks ago a professor of mine mentioned that the Galois group of a certain type of quartic polynomial is easy to calculate, and at the time it seemed obvious to me so I didn't ask why. Now i'm realizing that it's not entirely obvious.

Suppose we have an irreducible quartic polynomial, for instance $x^4 + 2x^2 + 2$, which is also a quadratic polynomial in the variable $y = x^2$. This becomes $y^2 + 2y + 2$, which is again irreducible. The splitting field of this polynomial is $\mathbb{Q}(\sqrt{-4})$, and the Galois group is $\mathbb{Z}/2\mathbb{Z}$. Does this information tell us anything about the Galois group of the original polynomial?

Answer 1

Seems to me the splitting field of $x^4+2x^2+2$ over the rationals is ${\bf Q}(i,\sqrt2)$, from which it's easy to work out that the Galois group is Klein-4.

Answer 2

If $f(x)=g(x^2)$, then the splitting fields satisfy $E_f\supseteq E_g\supseteq\mathbb Q$. Since $E_g/\mathbb Q$ is also Galois, you have that $H=\textrm{Gal}(E_f/E_g)$ is normal in $G_f$ and $G_f/H\cong G_g$.

In the situation you discuss this gives you that $G_f/\mathbb Z_2=\mathbb Z_2$, but that's as far as you can go without taking the specifics of the situation into account. In your example, $G_f=\mathbb Z_2\times\mathbb Z_2$, but the other possibility $G_f=\mathbb Z_4$ occurs, too.

An example for that is provided by $f(x)=x^4-8x^2+8$. It's easy to check that adjoining the zero $\alpha=2\sqrt{2}\sqrt{\sqrt{2}+1}$ produces a splitting field, and if we map $\alpha$ to one of the other two conjugates $\not=\pm\alpha$, then the corresponding automorphism doesn't have order $2$, which proves the claim on the Galois group.