Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.17
Solution:
(1) Evidently, $$\overline{p(x)} = \overline{-x^2-11x+3},$$ $$\overline{q(x)} = \overline{8x^2 – 13x + 5},$$ $$\overline{p(x)+q(x)} = \overline{7x^2 – 24x + 8},$$ and $\overline{p(x)q(x)} = \overline{146x^2-236x+90}$.
(2) Evidently, $x^3 -2x+1 = (x-1)(x^2+x-1)$. Now neither $\overline{x-1}$ nor $\overline{x^2+x-1}$ is zero in $\overline{E}$, but their product is. Thus (for instance) $\overline{x-1}$ is a zero divisor in $\overline{E}$. We found this factorization by noting that $f(1)=0$, so that $x-1$ divides $f(x)$.
(3) Evidently, $$\overline{x}\overline{-x^2+2} = 1,$$ so that $\overline{x}$ is a unit. To find this inverse, recall that if $\overline{x}$ is a unit, its inverse is represented by a polynomial of degree at most 2- say $\overline{ax^2+bx+c}$. Suppose $$\overline{x}\overline{ax^2+bx+c} = \overline{bx^2 + (c+2a)x-a} = 1;$$ comparing coefficients, we find that $b=0$, $a=-1$, and $c=2$.
