Find the product of 2x^3+3x^2+1 and 2x^2+4 in z_5[x]/<x^3+1>

Yulia 2021-09-16 Answered

Find the product of \(\displaystyle{2}{x}^{{3}}+{3}{x}^{{2}}+{1}\) and \(\displaystyle{2}{x}^{{2}}+{4}\) in \(z_5[x]/<x^3+1>\)

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Expert Answer

Talisha
Answered 2021-09-17 Author has 4321 answers

Group of integers modulo 5
\(\displaystyle{Z}_{{5}}={\left\lbrace{0},{1},{2},{3},{4}\right\rbrace}\)
Now,
\(\displaystyle{Z}_{{5}}{\left[{x}\right]}\) denotes the ring of polynomials with coefficients from \(\displaystyle{Z}_{{5}}\) and \(\displaystyle{<}{x}^{{3}}+{1}{>}\) denotes the principal ideal generated by \(\displaystyle{\left({x}^{{3}}+{1}\right)}\) that is
\(\displaystyle{<}{x}^{{3}}+{1}>={\left\lbrace{f{{\left({x}\right)}}}{\left({x}^{{3}}+{1}\right)}{\left|{f{{\left({x}\right)}}}\in{Z}_{{5}}{\left[{x}\right]}\right|}\right\rbrace}\)
\(Z_5[x]/<x^3+1>={\left\lbrace{g{{\left({x}\right)}}}+{<}{x}^{{3}}+{1}{>}{\mid}{g{{\left({x}\right)}}}\in{Z}_{{5}}{\left[{x}\right]}\right\rbrace}\)
\(\displaystyle={\left\lbrace{a}{x}^{{2}}+{b}{x}+{c}+{<}{x}^{{3}}+{1}{>}{\mid}{a},{b},{c}\in{Z}_{{5}}\right\rbrace}\)
We know that
\(\displaystyle{x}^{{3}}+{1}+{<}{x}^{{3}}+{1}\ge{0}+{<}{x}^{{3}}+{1}{>}\)
So, \(\displaystyle{x}^{{3}}+{1}\) is equivalent 0
Equivalently, \(\displaystyle{x}^{{3}}=-{1}\)
Now \(\displaystyle{2}{x}^{{3}}+{3}{x}^{{2}}+{1}={2}{\left(-{1}\right)}+{3}{x}^{{2}}+{1}+{<}{x}^{{3}}+{1}{>}\)
\(\displaystyle={3}{x}^{{2}}-{1}+{<}{x}^{{3}}+{1}{>}\)
\(\displaystyle={3}{x}^{{2}}+{4}+{<}{x}^{{3}}+{1}{>}\in Z_5[x]/<x^3+1>\)
since \(\displaystyle-{1}\equiv{4}\) in \(\displaystyle{Z}_{{5}}\)
\(\displaystyle{2}{x}^{{2}}+{4}={2}{x}^{{2}}+{4}+{<}{x}^{{3}}+{1}{>}\in Z_5[x]/<x^3+1>\)
Product of given two polynomials
\(\displaystyle{\left({2}{x}^{{3}}+{3}{x}^{{2}}+{1}\right)}{\left({2}{x}^{{2}}+{4}\right)}={\left({3}{x}^{{2}}+{4}+{<}{x}^{{3}}+{1}{>}\right)}\cdot{\left({2}{x}^{{2}}+{4}+{<}{x}^{{3}}+{1}{>}\right)}\)
\(\displaystyle={6}{x}^{{4}}+{20}{x}^{{2}}+{16}+{<}{x}^{{3}}+{1}{>}\)
Since \(\displaystyle{x}^{{3}}=-{1}\) implies \(\displaystyle{x}^{{4}}=-{x}\)
So we have
\(\displaystyle{\left({2}{x}^{{3}}+{3}{x}^{{2}}+{1}\right)}{\left({2}{x}^{{2}}+{4}\right)}={6}{\left(-{x}\right)}+{20}{x}^{{2}}+{16}+{<}{x}^{{3}}+{1}{>}\)
\(\displaystyle=-{6}{x}+{20}{x}^{{2}}+{16}+{<}{x}^{{3}}+{1}{>}\)
As \(\displaystyle-{6}\equiv{4},{20}\equiv{0},{16}\equiv{1}\) in \(\displaystyle{Z}_{{5}}\)
\(\displaystyle\therefore{\left({2}{x}^{{3}}+{3}{x}^{{2}}+{1}\right)}{\left({2}{x}^{{2}}+{4}\right)}={4}{x}+{1}+{<}{x}^{{3}}+{1}{>}\) in \(Z_5[x]/<x^3+1>\)

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