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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) Please 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) Please 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.

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I love the way you added "please" to both items. – Bruno Stonek Apr 11 '11 at 0:13
After the edit, what is the question? – lhf Apr 20 '11 at 10:14
per @lhf 's comment, I rolled back the edit. It is not polite to edit the question so that the answers no longer makes sense. – Willie Wong Apr 20 '11 at 11:43

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.

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a) Much simpler than using the polynomial long division algorithm is to exploit Horner form, i.e. rewrite the polynomial in nested form pulling out factors of $\rm\:x^4\:$ as follows.

$\rm\quad\quad\quad\quad\quad 3 - 2\ x^3\ \ \ +\ \ \ 5\ x^5\ \ \ -\ \ \ 11\ x^9\ \ \ +\ \ \ 7\ x^{13} $

$\rm\quad\quad\quad =\ 3 - 2\ x^3 + x^4\ (5\ x + x^4\ (-11\ x + x^4\ (7\ x)))$

Substituting $\rm\ x^4 \to 16\ $ above yields the sought polynomial of degree $< 4\:.$

b) $\rm\ \ x\pm2\ |\ x^4-16\ $ since $\ (\pm2)^4 = 16$

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Arturo Magidin's answer and Bill Dubuque's answer give nice systematic ways to find the representative of $\overline{7x^{13}-11x^9+5x^5-2x^3+3}$ with degree less than or equal to 3. Here's another quick way to work out the answer in your case:

You have $\overline{x^4-16}= \overline{0}$ in $\overline{E}$, which has a ring structure; so $\overline{x^4-16} = \overline{x^4} - \overline{16}= \overline{0}$. We can rewrite this as $\overline{x^4}=\overline{16}$. So every place you can find $\overline{x^4}$, you can replace it with $\overline{16}$, which means we can simplify any polynomial in $\overline{E}$ to one with degree less than or equal to 3.

You should think about all 3 suggestions to see that they give you the same answer.

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