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
ANSWERED
asked 2021-01-08
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 \(\displaystyle{\sqrt[{{3}}]{{{2}}}}\in\mathbb{R}\).

Answers (1)

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 \(\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}.\)

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