# Splitting over some extension of $\mathbb Z_3$

I want to write $x^3 + 2x + 1$ as a product of linear polynomials over some extension field of $\mathbb Z_3$.

I know that the roots will lie in a field that will be isomorphic to the field $\mathbb Z_3[x]/<x^3 + 2x + 1>$ but not sure how to proceed further.

• By $\mathbb{Z}_3$, do you mean the integers modulo $3$ or the $3$-adic integers? – Viktor Vaughn Aug 6 '17 at 17:57
• The Frobenius automorphism $a \mapsto a^3$ gives as your polynomial is irreducible $X^3+2X+1 = (X-\alpha)(X-\alpha^3)(X-\alpha^{3^2})$ where $\alpha = x + (x^3+2x+1)$ is one of its roots in $\mathbb{Z}_3[x]/(x^3+2x+1)$. – reuns Aug 7 '17 at 2:21

Let $\omega$ the congruence class of $x$ in $\mathbf F_{27}=\mathbf Z/3\mathbf Z[x]/(x^3+2x+1)$. This polynomial splits completely over $\mathbf F_9$: $$x^3+2x+1=(x-\omega)(x-\omega-1)(x-\omega+1).$$

This means that $\omega\pm1$ are the other roots of the polynomial. Indeed $$(\omega+1)^3+2(\omega+1)+1=\omega^3+1+2\omega+2+1=\omega^3+2\omega+1=0,$$ and similarly for $\omega-1$.

• Edited the question, it was supposed to be Z and not Q. – The Doctor Aug 6 '17 at 18:09
• Thank you, this makes sense :) – The Doctor Aug 6 '17 at 18:15
• Just one cavil: this is the cubic extension of $\Bbb F_3$, and thus it’s an $\Bbb F_{27}$. – Lubin Aug 7 '17 at 2:35
• @Lubin: Oh! yes, of course. I really was absebt-minded. Thanks for pointing it! – Bernard Aug 7 '17 at 10:53

As your polynomial is irreducible, the Frobenius automorphism $a \mapsto a^3$ gives $$T^3+2T+1 = (T-\alpha)(T-\alpha^3)(T-\alpha^{3^2})$$ where $\alpha = x + (x^3+2x+1)$ is one of its roots in $\mathbb{Z}_3[x]/(x^3+2x+1)$.

Then you compute $\alpha^3 = x^3+(x^3+2x+1)= x+2 +(x^3+2x+1), \alpha^{3^2} = (x+2)^3 + (x^3+2x+1) = x^3+2^3 +(x^3+2x+1) = x+1+(x^3+2x+1)$ and hence

$\qquad\qquad T^3+2T+1 = (T-x)(T-x+1)(T-x+2)$ in $\mathbb{Z}_3[x]/(x^3+2x+1)$.