Prove that $q(x)$ does not divide $p(x)$?

Question

"Let $F$ be a field and suppose that $p(x),q(x) \in F[x]$ are the two polynomials $p(x) = x^5 - x^4 + x^3 - x^2 + x - 1$ and $q(x) = x^2-1$

(i) Prove that $q(x)$ does not divide $p(x)$ when $F = \mathbb{R}$

(ii) Prove that $q(x)$ does divide $p(x)$ when $F = \mathbb{Z}_3$"

I'm not sure how to approach this - is it enough for (i) to use long division to show that $p(x) = q(x) + 3x+3$?

I'm not sure what $\mathbb{Z}_3$ actually is - I know that $\mathbb{Z}$ is the set of integers, but I'm not sure what the 3 means.

Sure, long division is enough. Marginally easier is to show that $1$ and $-1$ are roots of our long polynomial in $\mathbb{Z}_3$. — André Nicolas, Jun 22 '16 at 04:17
The first question is, "What is $\mathbb Z_3$?" See http://math.stackexchange.com/questions/409133/definition-of-notation-mathbb-z-n — Jonas Meyer, Jun 22 '16 at 04:24

score 2 · Accepted Answer · answered Jun 22 '16 at 04:16

2

Hint $\ {\rm mod}\ x^2-1\!:\,\ x^2\equiv 1\,\Rightarrow\,p(x)\equiv x-1 + x-1 + x-1 \equiv 3(x-1)$

answered Jun 22 '16 at 04:16

Bill Dubuque

1

What a nice answer. It deals with both cases at once, and avoids long division entirely. – Alex Wertheim Jun 22 '16 at 04:38

score 0 · Answer 2 · answered Jun 22 '16 at 07:45

0

i. $$p(-1)=-6\ne 0$$

ii. If $F=\mathbb{Z_3}[x]$ then $\,-x=2x$ and $$ x^5 - x^4 + x^3 - x^2 + x - 1=(x^3-1)(x^2-x+1)=(x^3-1)(x^2+2x+1)=(x^3-1)(x+1)^2\quad$$

answered Jun 22 '16 at 07:45

Behrouz Maleki

2 Answers2