Compute $\frac{1+\theta}{1+\theta+\theta^2}$ in $\Bbb{Q}(\theta)$

Question

So im told to compute $(1+\theta)(1+\theta+\theta^2)$ and $\frac{1+\theta}{1+\theta+\theta^2}$ where $\theta$ is a root of $p(x)=x^3-2x-2$ in some extension of $\Bbb{Q}$. I got for the first part:

\begin{align} (1+\theta)(1+\theta+\theta^2)&= \theta^3 + 2 \theta^2 + 2 \theta +1\\\\ &= 2 \theta^2 + 4 \theta + 3 \end{align}

But for the second part, do I write it as $(1+ \theta)(1+\theta+\theta^2)^{-1}$ and compute the inverse? If so, I did the long division and have

$$\theta^3-2 \theta -2=(\theta^2+\theta+1)(\theta-1)-2 \theta - 1$$
where do I go from here?

Answer 1

If you multiply the denominator by $a\theta^2+b\theta+c$ then you get

\begin{align*} (1+\theta+\theta^2)(a\theta^2+b\theta+c)=a\theta^4+(a+b)\theta^3+(a+b+c)\theta^2+(b+c)\theta+c \end{align*} but since $\theta^3=2\theta+2$ this can be rewritten as $$(3 a+b+c)\theta^2 + (4 a+3b+c)\theta + 2 a+ 2 b + c$$ now you want a constant/rational denominator, thus you want the $\theta,\theta^2$ terms gone.

By solving the linear system you deduce that can pick $a=2,b=-1,c=-5$ and obtain a denominator equal to $-3$ and you are done.

Answer 2

You can try the following direct but frustrated way: assume that $(1+\theta )(1+\theta +\theta^2)^{-1} =p(\theta)$ for some polynomial $p(\theta)$ of degree not greater than or equal to $2$. Multiply both sides by $(1+\theta +\theta^2)$ gives us an identity $1+\theta =p(\theta)(1+\theta +\theta^2)$. Let $p(\theta )=a+b\theta +c\theta^2$, by plugging in this $p(\theta)$ and using the given relation to reduce to polynomials of degree at most $2$, you get a linear system, whose solution tells you the conclusion.