Question

Find the inverse of x+1 in QQ[x]/(x^3-2). Explain why this is the same as finding the inverse of root(3)(2) in RR.

Abstract algebra
Find the inverse of $$\displaystyle{x}+{1}\in\mathbb{Q}\frac{{{x}}}{{{x}^{{3}}-{2}}}$$. Explain why this is the same as finding the inverse of $$\displaystyle{\sqrt[{{3}}]{{{2}}}}\in\mathbb{R}$$.

2021-01-09

Let tthe polynominal $$\displaystyle{f{{\left({x}\right)}}}={x}+{1}{\quad\text{and}\quad}{p}{\left({x}\right)}={x}^{{3}}-{2}.$$
Using the Educlidean algorithm theorem, divide $$\displaystyle{x}^{{3}}-{2}\ {b}{y}\ {x}+{1}.$$
$$\displaystyle{x}^{{3}}-{2}={\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}-{3}$$
$$\displaystyle{x}^{{3}}-{2}-{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}=-{3}$$
Divide by -3 on the both the sides of the above equation.
$$\displaystyle{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{3}}-{2}\right)}-{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}={\left(-\frac{{3}}{{-{{3}}}}\right)}$$
$$\displaystyle{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{3}}-{2}\right)}+{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}={1}$$
That is $$\displaystyle{\left\langle{p}{\left({x}\right)}\right\rangle}+{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({f{{\left({x}\right)}}}\right)}={1}$$
Therefore, the inverse of the element $$x+1$$ is $$\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}$$
By the theorem, "Let F be a field and let $$\displaystyle{p}{\left({x}\right)}\in{F}{\left({x}\right)}$$ be an irreducible polynominal. Suppose K is an extension field of F containing the root $$\displaystyle\alpha$$ of p(x), $$\displaystyle{p}{\left(\alpha\right)}={0}$$. Let $$\displaystyle{F}{\left(\alpha\right)}$$ denote the subfield of K generated by $$\displaystyle\alpha$$ over F. Then $$\displaystyle{F}{\left(\alpha\right)}\stackrel{\sim}{=}{F}\frac{{{x}}}{{{p}{\left({x}\right)}}}'$$
Here $$F=Q$$ and $$\displaystyle{p}{\left({x}\right)}={x}^{{3}}-{2}$$
Note that, the root of the polynominal p(x) is $$\displaystyle{\left({2}\right)}^{{\frac{{1}}{{3}}}}$$ and p(x) is ireeducible over Q
By the theorem $$\displaystyle{Q}{\left({\left({2}\right)}^{{\frac{{1}}{{3}}}}\right)}\stackrel{\sim}{=}{Q}\frac{{{x}}}{{{x}^{{3}}-{2}}}$$
Therefore, the inverse of $$x+1$$ over $$\displaystyle{Q}\frac{{{x}}}{{{x}^{{3}}-{2}}}$$ sameas inverse $$\displaystyle{o}{f}{2}^{{\frac{{1}}{{3}}}}+{1}$$ over $$\displaystyle\mathbb{R}$$
The inverse of the element $$x+1$$ is $$\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}.$$