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 \(Z_5[x]/<x^3+1>\)