Root field of $x^3+2x+1$ over $\mathbb{Z}_3$

Question

Find the root field of $a(x)=x^3+2x+1$ over $\mathbb{Z}_3$.

Suppose $u$ is a root of $x^3+2x+1$. Then $a(u+1)=(u+1)^3+2(u+1)+1=u^3+2u+1=0$, so $u+1$ is also a root, and similarly $u+2$ is also a root.

Now, the root field should contain $u,u+1,u+2$, so we should find $\mathbb{Z}_3(u,u+1,u+2)$. How can we find it?

Answer 1

Adjoin a root $u.\,$ Let $\,\Bbb Z_3[u] = \Bbb Z_3[t]/(t^3\!-t+1).\,$ Since $t^3-t+1$ has no roots in $\Bbb Z_3$ it is irreducible over $\Bbb Z_3,\,$ so the quotient is a field. As you noted, $u, u\pm1$ are roots in $\Bbb Z_3[u].$ Generally

$$(x\!-\!y)(x\!-\!y\!+\!1)(x\!-\!y\!-\!1)\, =\, x^3\!-x+1 - \color{#c00}{(y^3\!-y+1) - 3xy(x\!-\!y)} $$

so one explicitly sees the factorization by letting $y = u,$ modulo $3$, so final two summands $\, = \rm\color{#c00}0.$