Compute $[(2+x^2)+k(x)(x^3-2)]^{-1}$

Question

I have this exercise:

Let $I=<x^3-2>$ be ideal of $\mathbb{Q}[x]$. Consider the element $a=(x^2-2)+I\in\mathbb{Q}[x]/I$, compute $a^{-1}$.

I know that $\mathbb{Q}[x]/I$ is a field, and $a=(x^2-2)+k(x)(x^3-2)$ for some $k(x)\in\mathbb{Q}[x]$, then i want to have some $b\in\mathbb{Q}[x]$ such that $ab=ba=1$, but i haven't managed to know how to continue, can you help me please?

Answer 1

Well, we do know that all elements can be reduced to the form $ax^2 + bx + c$, so that reduces our search space.

Now, we can actually just compute this directly, as we want \begin{align*} 1 &= (ax^2 + bx + c)(x^2 - 2)\\ &= ax^4 + bx^3 + (c-2a)x^2 - 2bx - 2c\\ &= (c-2a)x^2 + (2a-2b)x + 2b-2c \end{align*} Thus we just have to solve \begin{align*} c-2a &= 0\\ a-b &= 0\\ 2b-2c &= 1 \end{align*} So we get $b=a=-1/2$ and $c=-1$

Answer 2

Subtract below equations, or use the extended Euclidean algorithm as in the $\rm\color{#0a0}{Example}$ below.

$\, \ \quad\qquad\qquad\qquad\begin{align} (x^2-2)(x^2+2) &\,=\, \color{#c00}2x-4\ \ \ {\rm by}\ \ (\color{#c00}{x^3})x = \color{#c00}2x\\ {\bf -}\qquad\qquad\quad\ x\,(x^2+2) &\,=\, \color{#c00}2+2x\ \ \ {\rm by}\ \ \ \color{#c00}{x^3 = 2}\\ \hline \Rightarrow\ (x^2-x-2)(x^2+2) &\,=\, -6 \end{align}$

$\begin{align}{\rm generally}\ \ \ [\![1]\!]\,\qquad\qquad\quad (x^2-a)(x^2+a) &\,=\, \color{#c00}bx-a^2\ \ \, {\rm by}\ \ (\color{#c00}{x^3})x = \color{#c00}bx\\ [\![2]\!]\ \ \ \qquad\qquad\qquad\quad\ x\,(x^2+a) &\,=\, \color{#c00}b+ax\ \:\!\ \ \ {\rm by}\ \ \ \color{#c00}{x^3 = b}\\ a[\![1]\!]-b[\![2]\!]\Rightarrow\ (ax^2-bx-a^2)(x^2+a) &\,=\, -a^3-b^2 \end{align}$

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$\rm\color{#0a0}{Example}$ Euclidean calculation from a different question. Exactly the same method works in the OP. Generally it's easiest to use said augmented-matrix form of the extended Euclidean algorithm, e.g. below we compute $\,1/g \pmod{\!f} = 1/(x^2\!+\!1) \pmod{\!x^3\!+\!2x\!+\!1}\,$ over $\,\Bbb Q,\,$ as in this answer.

$\,\begin{eqnarray} [\![1]\!]&& &&f = x^3\!+2x+1 &\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}0\,\right>\quad\ \ \, {\rm i.e.}\ \qquad f\, =\ \color{#c00}1\cdot f\, +\, \color{#0a0}0\cdot g\\ [\![2]\!]&& &&\qquad\ \, g =x^2\!+1 &\!\!=&\, \left<\,\color{#c00}0,\,\color{#0a0}1\,\right>\quad\ \ \,{\rm i.e.}\ \qquad g\, =\ \color{#c00}0\cdot f\, +\, \color{#0a0}1\cdot g\\ [\![3]\!]&=&[\![1]\!]-x[\![2]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\ \ x+1 \,&\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}{-x}\,\right>\ \ \ {\rm i.e.}\quad x\!+\!1\, =\, \color{#c00}1\cdot f\color{#0c0}{-\,x}\cdot g\\ [\![4]\!]&=&[\![2]\!]+(1\!-\!x)[\![3]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\qquad\ 2 \,&\!\!=&\, \left<\,\color{#c00}{1\!-\!x},\,\ \color{#0a0}{1\!-\!x+x^2}\,\right>\\ \end{eqnarray}$

Hence the prior line implies: $\ \ \ 2\ =\ (\color{#c00}{1\!-\!x})f + (\color{#0a0}{1\!-\!x\!+\!x^2})g $

$\!\!\bmod\! f\,$ this yields in $\Bbb Q[x]/f\!:\,\ 2\, =\, (\color{#0a0}{1\!-\!x\!+\!x^2})g\ \Rightarrow\ {1/g= (\color{#0a0}{1\!-\!x\!+\!x^2})/2}$