## Compute the inverse of a given element in a field extension

Show that $p(x) = x^3 + 9x + 6$ is irreducible over $\mathbb{Q}$. Let $\theta$ be a root of $p(x)$. Find the inverse of $1+\theta$ in $\mathbb{Q}(\theta)$.

$p(x)$ is Eisenstein at 3, and so is irreducible, and thus $\mathbb{Q}(\theta) \cong \mathbb{Q}[x]/(p(x))$ is a field.

Note that every element of $\mathbb{Q}(\theta)$ has the form $a + b\theta + c\theta^2$ for some rational numbers $a,b,c$. Suppose $(1+\theta)(a+b\theta+c\theta^2) = 1$; comparing coefficients, we see that $a-6c = 1$, $a+b-9c = 0$, and $b+c = 0$. Solving this system, we can see then that $(1+\theta)^{-1} = \frac{1}{4}(\theta^2 - \theta + 10)$.