Find roots of primitive polynomial $x^2+4x+2$ in $\mathbb{Z}_{11}/(x^2+x+8)$

Question

I need to find the roots of primitive polynomial $x^2+4x+2$ in $\mathbb{Z}_{11}$ over field $\mathbb{Z}_{11}/(x^2+x+8)$ ($x^2+x+8$ is also primitive).

As far as I'm concerned the answer is the factor of this polynomial: $x^2+4x+2 = (x-\alpha_1)(x-\alpha_2)$ First of all I built the field $\mathbb{Z}_{11}/(x^2+x+8)$ and found generator $\alpha = x$ of a multiplicative group $\mathbb{F}^*_{121}$ and by raising $x^2 = 10x+3$ to a power of $2$, I can get all the elements of the field. Also I got an idea, that $x^2+4x+2$ will have two root in this field, so I need to find definite $\alpha^n$ so by raising it in power of $11$ I will get cyclic group of $2$ elements. The problem is, as far as I'm concerned, there are many cyclic groups that will be roots of other second order irreducible polymonials over $\mathbb{Z}_{11}$, and so I will need to check every of it, if it's this $x^2+4x+2$ polynomial or not. Are there ways to do it faster? How can I can use the fact that $x^2+4x+2$ is primitive?

Answer 1

A member of your extension field is $x = a + b \alpha$ where $\alpha^2 + \alpha + 8 = 0$, and $a, b \in \mathbb Z_{11}$. Expand $x^2 + 4 x + 2 $ and substitute $\alpha^2 = -\alpha - 8$ and you get an expression of the form $c_1 \alpha + c_0$: you want $c_1$ and $c_0$ to be $0$ (in $\mathbb Z_{11}$). Solve $c_1$ for $a$ and substitute in $c_0$: you should get an equation for $b$ that can be solved in $\mathbb Z_{11}$.