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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?

Arctic Char
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homosapien
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  • But $p(\theta)=0$ thus $\theta^3=2 \theta+2$ where could i have gone wrong:?? @coudy – homosapien Mar 05 '23 at 21:29
  • We can compute inverses by the Extended Euclidean algorithms - just like for integers - see the linked dupe. – Bill Dubuque Mar 05 '23 at 21:51
  • that not the same question @BillDubuque .... – homosapien Mar 05 '23 at 21:53
  • Doesn't matter - the same method works here. – Bill Dubuque Mar 05 '23 at 21:55
  • Or by inverse reciprocity, let $,g!=!\theta^3-2\theta-2,\ f!=!\theta^2+\theta+1$. By Bezout $,af+bg = 1$ and we first compute $\bmod f!:\ b \equiv g^{-1},$ as follows, using $,\color{#0a0}{\theta^3 = 1}$ $$\bmod f!:,\ \color{#c00}b\equiv \dfrac{1}{g}=\dfrac{1}{\color{#0a0}{\theta^3}-2\theta-2} \equiv \dfrac{-1}{2\theta+1}\equiv\color{#c00}{\dfrac{2\theta+1}3},\ {\rm by}\ \ 2\theta+1 = \sqrt{-3}$$

    hence $,af+bg = 1\Rightarrow \bmod g!:\ f^{-1}\equiv a\equiv \dfrac{1-\color{#c00}b:!g}f\equiv (-2\theta^2+\theta+5)/3\ \ $

     – Bill Dubuque Mar 05 '23 at 22:33

2 Answers2

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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.

b00n heT
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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.

TheWildCat
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