Ytukyg

I don't know how to begin to find this particular Galois group since I do not know the splitting field. Any hints will be appreciated.

NOTE: Whenever I say "the whole polynomial," I mean $(x^3+x^2+1)(x^2+x+1)(x^3+x+1)$.

Let $alpha$ be a root of $x^3+x^2+1$. We then have:

$$x^3+x^2+1=(x-alpha)(x^2+(alpha+1)x+alpha^2+alpha)$$

Now, we know that:

$$alpha^3+alpha^2+1=0rightarrow alpha^3=alpha^2+1$$
$$alpha^4=alpha^3+alpha=alpha^2+alpha+1$$
$$alpha^6=(alpha^3)^2=(alpha^2+1)^2=alpha^4+1=alpha^2+alpha$$

Using this info, let's guess and check values of $x$ to find more zeroes. Eventually, we find $x=alpha^2$:

$$x^2+(alpha+1)x+alpha^2+alpha=(x-alpha^2)(x-(alpha^2+alpha+1))$$

Thus, it took a single field extension of order $3$ to split $x^3+x^2+1$ into linear factors, so the Galois group of this splitting field is $C_3$.

Now, conveniently for us, this splitting field also splits the polynomial $x^3+x+1$. To verify this, again, using guess and check, we find $x=alpha^2+1$ is a root:

$$x^3+x+1=(x-(alpha^2+1))(x^2+(alpha^2+1)x+alpha^2+alpha+1)$$

Then, again using guess and check, we can factor that last quadratic by plugging in $x=alpha+1$:

$$x^2+(alpha^2+1)x+alpha^2+alpha+1=(x-(alpha+1))(x-(alpha^2+alpha))$$

Since the splitting field of $x^3+x^2+1$ also splits $x^3+x+1$, this means it took a single field extension of order $3$ to split both of these cubic polynomials. Thus, do not make the mistake of multiplying in two copies of $C_3$ into the Galois group of the only polynomial since we only made one field extension to split both cubics.

Now, let $beta$ be a root of $x^2+x+1$. We then have:

$$x^2+x+1=(x-beta)(x+beta+1)$$

Thus, it took a single field extension of order $2$ to split $x^2+x+1$ into linear factors, so the Galois group of this splitting field is $C_2$.

Now, to recap, the splitting field of the whole polynomial is $Bbb{F}_3(alpha)(beta)$, where $alpha$ has irreducible polynomial $x^3+x^2+1$ and $beta$ has irreducible polynomial $x^2+x+1$. Therefore, any automorphism of this splitting field will contain a 3-cycle between the roots of $x^3+x^2+1$ and a 2-cycle between the roots of $x^2+x+1$, so the Galois group of the splitting field is $C_3 times C_2$.

Since it is a finite field, you only need to find a degree of splitting field over $mathbb{F}_{2}$. We will show that the degree is 6 so that the splitting field is $mathbb{F}_{2^{6}}$ and the Galois group is $mathbb{Z}/6mathbb{Z}$.

In general, any degree $d$ irreducible polynomial over a finite field $mathbb{F}_{q}$ splits completely in $mathbb{F}_{q^{m}}$, since the splitting field of the polynomial is $mathbb{F}_{q^{d}}$, and it can be embedded into $mathbb{F}_{q^{m}}$ since $d|m$. Hence our polynomial splits completely in $mathbb{F}_{2^{6}}$, so the splitting field sits inside $mathbb{F}_{2^{6}}$.

By the way, any other proper subfields of $mathbb{F}^{2^{6}}$ can't be a splitting field. For example, we have 2 candidates: $mathbb{F}_{2^{2}}$ and $mathbb{F}_{2^{3}}$. In $mathbb{F}_{2^{2}}$, $x^{3} + x^{2} + 1$ doesn't split completely so it is impossible. Similarly, $x^{2} + x + 1$ doesn't split completely in $mathbb{F}_{2^{3}}$, so the only possibility is $mathbb{F}_{2^{6}}$.

Using this argument, you may show the following: splitting field of the polynomial $f(x) = p_{1}(x)cdots p_{r}(x)$ over $mathbb{F}_{q}$ where all $p_{i}(x)$'s are irreducible and $deg p_{i} = d_{i}$ is $mathbb{F}_{q^{d}}$, where $d = mathrm{lcm}{d_{1}, dots, d_{r}}$.