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

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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Talisha

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 $$Z_5[x]/<x^3+1>$$