Splitting field for the polynomial $x^3 + 1 \in \mathbb{Z}_2[x]$ is a subfield of the splitting field of $x^5 + 1 \in \mathbb{Z}_2[x]$

Question

Show that the splitting field for the polynomial $x^3 + 1 \in \mathbb{Z}_2[x]$ is a subfield of the splitting field of $x^5 + 1 \in \mathbb{Z}_2[x]$.

I'm not sure how to construct the splitting fields in the first place when we are over $\mathbb{Z}_2[x].$ I know they both have 1 as a root since we are in $\mathbb{Z}_2[x]$. I'm thinking that the degree of the splitting field $K_1$ over $x^3 + 1$ should be $3$ and the degree of the splitting field $K_2$ of $x^5 + 1$ is going to be $5,$ in which case $K_1$ could not be a subfield of $K_2.$ What am I missing here?

Answer 1

What you’re missing is that over the field $\Bbb F_2$ with two elements, you have the factorizations $X^3+1=(X+1)(X^2+X+1)$ and $X^5+1=(X+1)(X^4+X^3+X^2+X+1)$

Thus, the first splitting field is quadratic over $\Bbb F_2$, and so has four elements, while (*) the second is quartic over $\Bbb F_2$, with sixteen elements.

(*) That quartic is indeed $\Bbb F_2$-irreducible, not hard to show, since $X^2+X+1$ is the only irreducible quadratic over that little field.

Answer 2

$\mathbb F_{16}$ contains both splitting fields because its multiplicative group is cyclic of order $15$.

Since $\mathbb F_{16}$ has degree $4$ over $\mathbb F_{2}$, it can only have proper subfields of degree $2$, that is, of size $4$.

The splitting field of $x^{5}+1$ contains at least $6$ elements, and so must be $\mathbb F_{16}$.