## Compute in a field extension

Show that $p(x) = x^3 - 2x - 2$ is irreducible over $\mathbb{Q}$, and let $\theta$ be a root of $p(x)$. Compute $(1+\theta)(1+\theta+\theta^2)$ and $\dfrac{1+\theta}{1+\theta+\theta^2}$.

$p(x)$ is Eisenstein at 2, and so is irreducible over $\mathbb{Q}$.

In $\mathbb{Q}(\theta)$, we have $\theta^3 = 2\theta+2$. We can easily see that $(1+\theta)(1+\theta+\theta^2) = 3+4\theta + 2\theta^2$.

Next we compute $(1+\theta+\theta^2)^{-1}$; to this end, we use the extended Euclidean Algorithm to find polynomials $a(x)$ and $b(x)$ in $\mathbb{Q}[x]$ such that $a(x)(x^2+x+1) + b(x)(x^3-2x-2) = 1$; evidently, $a(x) = \frac{1}{3}(-2x^2+x+5)$ and $b(x) = \frac{1}{3}(2x+1)$ work. (WolframAlpha agrees.) Thus, mod $p(x)$, we have $(1+\theta+\theta^2)^{-1} = \frac{1}{3}(-2\theta^2 + \theta + 5)$. Now $(1+\theta)(1+\theta+\theta^2)^{-1} = \frac{1}{3}(-\theta^2 + 2\theta + 1)$.