# Let $f(x)=x^8-16$. Determine the Galois group of the splitting field of $f(x)$ over the fields $\Bbb{Q}$ and $\Bbb{Z}_{17}$.

Let $f(x)=x^8-16$. Determine the Galois group of the splitting field of $f(x)$ over the field $K$ in each case.

a) $K=\Bbb{Q}$

b) $K=\Bbb{Z}_{17}$.

a) The roots of $f(x)$ are $\alpha$, $\alpha\omega$, $\alpha\omega^2$, ... ,$\alpha\omega^7$ where $\alpha=16^{1/8} = (2^4)^{1/8} = 2^{1/2}$ and $\omega$ is the $8$th primitive root of unity.

First we need to find the degree of the extension since it is equal to the order of the Galois group.

$[\Bbb{Q}(\alpha,\omega):\Bbb{Q}]=[\Bbb{Q}(\alpha,\omega):\Bbb{Q}(\omega)][\Bbb{Q}(\omega):\Bbb{Q}]$.

$[\Bbb{Q}(\omega):\Bbb{Q}]=\Psi_8(x)$, where $\Psi_k(x)$ is the $k$th cyclotomic polynomial.

We know that

$$x^8-1 = \Psi_1(x)\Psi_2(x)\Psi_4(x)\Psi_8(x)$$

$$\implies x^8-1 = (x-1)(x+1)(x^2+1)\Psi_8(x)$$

$$\implies \Psi_8(x)=x^4+1$$

Since cyclotomic polynomials are irreducible over $\Bbb{Q}$, we know that $[\Bbb{Q}(\omega):\Bbb{Q}]=4$.

Now we need to find $[\Bbb{Q}(\omega,\alpha):\Bbb{Q}(\omega)]$.

Since $\alpha$ is a root of $x^8-16$, the minimal polynomial must be a divisor of it. We have $x^8-16 = (x^2-2)(x^2+2)(x^4+4)$. Here, we can see that $\alpha$ is a root of $x^2-2$, so we just need to check if $x^2-2$ is reducible. But $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$, and we know that $\sqrt{2} \not\in \Bbb{Q}(\omega)$ since it is irrational. So $[\Bbb{Q}(\omega,\alpha):\Bbb{Q}(\omega)]=2 \implies [K:\Bbb{Q}]=8$

There are five groups of order 8 up to isomorphism:

$\bullet \Bbb{Z}_8$

$\bullet \Bbb{Z}_4 \times \Bbb{Z}_2$

$\bullet \Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}_2$

$\bullet D_8$

$\bullet Q_8$

Since $x^8-16 = (x^2-2)(x^2+2)(x^4+4)$, the automorphisms are

$$\sigma_1: \alpha \rightarrow -\alpha$$

$$\sigma_2: \alpha\omega^2 \rightarrow \alpha\omega^6$$

$$\sigma_3: \alpha\omega \rightarrow \alpha\omega^3$$

$$\sigma_4: \alpha\omega \rightarrow \alpha\omega^5$$

$$\sigma_5: \alpha\omega \rightarrow \alpha\omega^7$$

$$\sigma_6: \alpha\omega^3 \rightarrow \alpha\omega^5$$

$$\sigma_7: \alpha\omega^3 \rightarrow \alpha\omega^7$$

$$\sigma_8: \alpha\omega^5 \rightarrow \alpha\omega^7$$

I guess I can tell which group of order 8 this is isomrophic to by direct computation, but I was kind of confused...for example, let's say I want to check if $\sigma_3\sigma_2(\sigma\omega) = \sigma_2\sigma_3(\alpha\omega)$. But $\sigma_2$ takes $\sigma\omega^2$ to $\alpha\omega^6$. But we don't have $\alpha\omega^2$, we have $\alpha\omega$. If we just raise it to the 3rd power, then what's the difference between $\sigma_2$ and $\sigma_3$? So that doesn't really make sense to me...

Also is there an easier way find out which group it's isomorphic to without actually having to directly go through all the elements and subgroups of the Galois group?

b) As in part a), $[\Bbb{Z_{17}}(\alpha,\omega):\Bbb{Z_{17}}]=[\Bbb{Z_{17}}(\alpha,\omega):\Bbb{Z_{17}}(\omega)][\Bbb{Z_{17}}(\omega):\Bbb{Z_{17}}]$. So, again, we need to check if $\Psi_8(x) = x^4+1$ is irreducible in $\Bbb{Z}_{17}$. In other words, since $x^4+1=0 \implies x^4=-1 \implies x^8=1$, we need to check if there are elements of order 8 in $\Bbb{Z}_{17}$.

We know that the multiplicative field of $\Bbb{Z}_{17}$ is $\Bbb{Z}^{\times}_{16}$. So we have a cyclic subgroup of order 8 $\implies$ the minimal polynomial of $\omega$ over $\Bbb{Z}_{17}$ is of degree 1 $\implies$ $[\Bbb{Z_{17}}(\omega):\Bbb{Z_{17}}]=1$.

For $[\Bbb{Z_{17}}(\alpha,\omega):\Bbb{Z_{17}}(\omega)]$, we get the same polynomial as we did for part a), which is not reducible since $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ and $\sqrt{2} \not\in \Bbb{Z}_{17}$. So $[\Bbb{Z_{17}}(\alpha,\omega):\Bbb{Z_{17}}]=2$, implying that the Galois group is isomorphic to a cyclic group of order 2.

Is my answer for part b), correct?

• In part b) you should also check, whether $2$ has a square root in $\mathbb{Z}_{17}.$ – Jyrki Lahtonen Aug 21 '13 at 9:36
• Another thing you need to be careful about is, as Dune pointed out, the eighth root of unity $\omega$. $$\omega=e^{2\pi i/8}=\cos\frac{\pi}4+i\sin\frac{\pi}4.$$ Remembering the $\sin(\pi/4)=1/\sqrt2$ might raise a flag. At least something worth thinking about for a while. – Jyrki Lahtonen Aug 21 '13 at 9:48
• Correct again, $\alpha\in\mathbb{Q}(\omega)$. Now which group is it? You can either look at the orders of the elements, or count a few intermediate fields... – Jyrki Lahtonen Aug 21 '13 at 10:02
• Well, no need to count intermediate fields any longer. I was hinting that your splitting field has $\mathbb{Q}(\sqrt2)$, $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-2})$ as intermediate fields, so the Galois group must have at least three non-trivial subgroups. The cyclic group of order four only has one non-trivial subgroup, so it would be ruled out this way also. Yours is an even better way of reaching that conclusion. – Jyrki Lahtonen Aug 21 '13 at 10:53
For part $b)$, you claim that $x^2-2$ is irreducible over $\mathbb{Z}/17\mathbb{Z}$. However, $$x^2-2=(x-11)(x-6),$$ from which it is immediate that the Galois group is trivial.
Another way to see this, is to note that $x^8-16\equiv x^8+1 (\operatorname{mod}17)$. The roots of $x^8+1$ in $\mathbb{Z}/17\mathbb{Z}$ are precisely the primitive roots of $(\mathbb{Z}/17\mathbb{Z})^{\times}$, of which there are precisely $\varphi(\varphi(17))=8$. Hence $x^8-16$ splits into distinct linear factors over $\mathbb{Z}/17\mathbb{Z}$, so the Galois group is trivial.