Modulo polynomial in ring theory Let $x^4-16$ be an element of the polynomial ring $E= \mathbb{Z}[x]$ and use the bar notation to denote passage to the quotient ring $\mathbb{Z}[x]/(x^4-16)$. Find a polynomial of degree $\leq 3$ that is congruent to $7x^{13} -11x^9 + 5x^5-2x^3+3$ modulo $x^4-16$. Can anyone help understand how to mod out polynomials. 
 A: Note that modulo $x^4-16$, $x^4-16 = 0$, so $x^4 = 16$. Thus you can replace every copy of $x^4$ with 16. For one, you can write $7x^{13} = 7x\cdot(x^4)^3 = 7x\cdot (16)^3$, and similarly with the other terms in your polynomial.
...you get the idea.
A: A standard way to do this is by performing Euclidean division of $7x^{13} -11x^9 + 5x^5-2x^3+3$ by $x^4 - 16$. In this way you get a polynomial $r(x)$ of degree less than $x^4 - 16$, so of degree at most $3$, and a polynomial $q(x)$ such that 
$$7x^{13} -11x^9 + 5x^5-2x^3+3  = q(x)(x^4 -16) + r(x)$$
You see that $r(x)$ is congruent to $7x^{13} -11x^9 + 5x^5-2x^3+3$ modulo $x^4 -16$.
Note that this will always work as long as the polynomial that defines your ideal has leading coefficient $1$  (or more generally an element invertible in the base ring). 
A: It's the remainder of the division of $7x^{13} -11x^9 + 5x^5-2x^3+3$ by $x^4-16$.
A: By using the long division i was able to obtain $7x^{13} - 11x^9 + 5x^5-2x^3+3 = q(x)(x^4 -16) +r(x)$ where $q(x)= 7x^9+101x^5+1621x$ and $r(x) = -2x^3+25936x+3$ since i was looking for $r(x)$ then the polynomial is $-2x^3+25936x+3$ 
A: ${\rm mod}\,\ x^4-2^4\!:\,\ x^4\equiv 2^4\,\Rightarrow\, \color{#c00}{x^{4n}}\equiv 2^{4n}$
$\! \begin{align}{\rm therefore}\quad &\ \ 7\,\color{#c00}{x^{12}}x -11\,\color{#c00}{x^8}\,x + 5\,\color{#c00}{x^4}\,x-2x^3+3\\
 \equiv &\  \ 7\, 2^{12}\,x - 11\, 2^8\, x + 5\, 2^4\, x -2\,x^3+3\quad {\bf QED}\end{align}$
A: It is sometimes easier to see what is happening by looking at things a different way. Here, just as $p\equiv q \bmod r$ holds for integers precisely when $r \mid (p-q)$, modular arithmetic for polynomials has $p(x)\equiv q(x) \bmod r(x)$ precisely when $r(x)\mid (p(x)-q(x))$
The implication of this is (working modulo $r(x)$) we have $p(x)\equiv p(x)+r(x)s(x)$ for any polynomial $s(x)$ [working over the same ground ring]
The substitution rule here - that one can consistently replace $x^4$ by $16$ in the example in the question reflects the fact that $$x^4=16+(x^4-16)\equiv 16 \bmod (x^4-16)$$
