$X^3-2$ splits completely in an extension of $\mathbb F_7$

Question

My question concerns the following problem:

Let $K=\mathbb F_7[T]/(T^3-2)$. Show that $X^3-2$ splits into linear factors in $K[X]$.

Write $K\simeq \mathbb F_7[\alpha]$ for a root $\alpha\in \overline{\mathbb F_7}$ of $X^3-2$. I found out through experimentation that $2\alpha$, $4\alpha$ are also roots of this polynomial (since $2^3\equiv 4^3 \equiv 1 \ \mathrm{mod}\ (7)$). Is there a more conceptual way to see this?

Answer 1

Notice that $X^3-2$ does not have any root in $\mathbb{F}_7$. Moreover polynomials of $\mathbb{F}_7[X]$ are invariant under the action of the Frobenius morphism with respect to $\mathbb{F}_7$. Therefore, $\alpha^7$ and $\alpha^{49}$ are roots of $X^3-2$. Besides, $\alpha,\alpha^7,\alpha^{49}$ are distincts, otherwise $\alpha\in\mathbb{F}_7$.

Remark. This is analogous to: if a real polynomial has a complex root, its conjuguate is also a root.

Answer 2

$X^3-2$ is irreducible mod $7$. This implies that $K$ has $7^3$ elements.

Since the group $K^\times$ is cyclic of order $7^3-1= 2 \cdot 3^2 \cdot 19$, there is an element $\omega \in K^\times$ of order $3$.

Then $X^3-2=(X-\alpha)(X-\omega\alpha)(X-\omega^2\alpha)$.

As you have discovered, you can take $\omega=2$.