Let $x^{14} – 16$ be an element of the polynomial ring $E =\mathbb{Z}[x]$ and let the bar notation to denote passage to the quotient ring $\mathbb{Z}[x]/(x^4 – 16)$.
a) Find a polynomial of degree less than or equal to 3 that is congruent to $7x^{13} -11x^9+5x^5 -2x^3+3 \pmod {x^4-16}$.
b) Prove that $\overline{x-2}$ and $\overline{x+2}$ are zero divisors in $\overline{E}$.
I tried it out, but not sure about notation and such. Can you please run through the problem solution? Thank you.
Best Answer
For (a), you need to do long division. In general, if $$p(x) = a(x)q(x) + r(x)$$ then $p(x) \equiv r(x)\pmod{q(x)}$, since $q(x)$ divides $p(x)-r(x)$. In particular, if you do it with long division so that $r(x)=0$ or $\mathrm{deg}(r)\lt \mathrm{deg}(q)$, you get a polynomial of degree smaller than the degree of $q(x)$ which is congruent to $p(x)$.
Here you have $q(x) = x^4-16$, so doing long division will give you a polynomial of degree at most $3$. So, divide $7x^{13} - 11x^9 + 5x^5 - 2x^3 + 3$ by $x^4-16$ to get the remainder, that's the polynomial you want.
For (b), simply show that neither of them is zero in $\overline{E}$ (easy: to be zero, they would have to be multiples of $x^4-16$). Then note that the product is $x^2-4$, and that $(x^2-4)(x^2+4) = x^4-16$. That gives you a witness to the fact that $\overline{x-2}$ and $\overline{x+2}$ are zero divisors, if you understand what it means to be a zero divisor and what $\overline{E}$ is.