# Galois Group of some polynomials of low degree

• May 2nd 2008, 10:04 PM
mpetnuch
Galois Group of some polynomials of low degree
Okay, I still having some trouble with finding Galois group.

Let's say that I wish to find the Galois group of $f(x)=x^4+1$ over $\mathbb{Q}$. Now the roots of $f(x)$ are $\omega, \omega^3, \omega^5, \omega^7$ where $\omega = \sqrt{i}$ and so the splitting fields of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\omega)$. Also the minimal polynomial of $\omega$ is $f(x)$ hence we have $[\mathbb{Q}(\omega) : \mathbb{Q}]=4$.

Since the characteristics of $\mathbb{Q} = 0$ we know that $|Gal(\mathbb{Q}(\omega)/\mathbb{Q})| = [\mathbb{Q}(\omega) : \mathbb{Q}]=4$. There are only two unique groups (up to isomorphism) of order 4, hence either $Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_4$ or $Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2}$.

I feel like it should be $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$. This is my reason (which I know is not right) but I think I am on the right track. Since we have $\omega^2 = i$ we can write $f(x)=(x^2+\omega^2)(x^2-\omega^2)$. Now if $x^2+\omega^2, x^2-\omega^2 \in \mathbb{Q}[x]$ I would be able to assert that for $\phi \in Gal(\mathbb{Q}(\omega)/\mathbb{Q})$ we must have $\phi(\omega^2) = \pm \omega^2$ and $\phi(\omega^3) = \pm \omega^3$.

So in essence I need some like the above that will enable me to eliminate some of the 24 permuatations (unless my argument is totally wrong and it's acutally isomorphic to $S_{4}$ (Crying) ).

Also, the second problem I was doing is this: find the Galois group of $f(x) = x^8 -1$. Since $f(x)=(x^4-1)(x^4+1)=(x^2-1)(x^2+1)(x^4+1)=(x-1)(x+1)(x^2+1)(x^4+1)$ the spltting field of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\omega)$ (because $i = \omega^2 \in \mathbb{Q}(\omega)$). Hence once I solve the above problem I have solved this one as well since they have the same splitting fields, right?

Let $\phi \in Gal(\mathbb{Q}(\omega)/\mathbb{Q})$, then $\phi(\omega) = \omega, \omega^3, \omega^5,$ or $\omega^7$. Once we choose the value for $\phi(\omega)$, we have defined $\phi$ entirely because it is an automorphism! That is $\phi(\omega^{n}) = \phi^{n}(\omega)$. Thus we must have $Gal(\mathbb{Q}(\omega)/\mathbb{Q}) \cong \mathbb{Z}_{4}$.
And for the second problem, since the splitting field of $f(x)=x^8-1$ is $\mathbb{Q}(\omega)$ as well, the Galois group of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Z}_{4}$ as well.