The splitting field over $Z/3\mathbb Z$

Question

What is the best method of finding the splitting field of $x^3+2x+1$ over $\mathbb Z/3\mathbb Z$? I believe I have the answer $\mathbb Z/3 \mathbb Z/ \langle x^3 +2x+1 \rangle$ but I'm unsure what the steps are leading up to it

Answer 1

Let $p(x) = x^3+2x+1$. Then:

• $p(0) = 1$
• $p(1) = 1$
• $p(2) = 1$

Therefore the polynomial is irreducible.

Let $F=(\Bbb Z/3\Bbb Z)[X]/\langle X^3+2X+1 \rangle$.

Let the generator be $a$.

Note that: $$ \begin{array}{rcl} p(x+1) &=& (x+1)^3 + 2(x+1) + 1 \\ &=& x^3+1+2x+2+1 \\ &=& x^3+2x+1 \\ &=& p(x) \end{array}$$

Therefore, $p(x) = p(x+1) = p(x+2)$.

Therefore, $a$, $a+1$, and $a+2$ are the roots of the polynomial, so the field $F$ is normal.

Note that the formal derivative of $p$ is $3x^2+2 = 2$, which cannot have any roots, so $F$ is separable.

Therefore, $F = (\Bbb Z/3\Bbb Z)[X]/\langle X^3+2X+1 \rangle$ is the splitting field of $x^3+2x+1$ over $\Bbb Z/3\Bbb Z$.