## Compute in a quotient of a polynomial ring

Consider $f(x) = x^4 - 16$ as an element of $E = \mathbb{Z}[x]$ and use bars to denote the natural projection $E \rightarrow E/(f(x))$.

1. Fine a polynomial of degree at most 3 which is congruent to $g(x) = 7x^{13} - 11x^9 + 5x^5 - 2x^3 + 3$ modulo $x^4-16$.
2. Prove that $\overline{x+2}$ and $\overline{x-2}$ are zero divisors in $\overline{E}$.

1. Evidently, $\overline{g(x)} = \overline{-2x^3 + 25936x + 3}$.
2. Note that $x^4-16 = (x+2)(x-2)(x^2+4)$. By §7.4 #14, $\overline{x+2}$ and $\overline{x-2}$ are zero divisors in $\overline{E}$.

• murphy  On November 29, 2011 at 1:32 am

can you tell me how you’re doing these? there’s nothing like this in the book

• nbloomf  On November 29, 2011 at 2:55 am

Remember that two polynomials $p(x)$ and $q(x)$ are equivalent mod $f(x)$ if $f$ divides $p-q$. Using the Division Algorithm in the Euclidean domain $\mathbb{Q}[x]$, we can find $b$ and $r$ such that $p = bf + r$, and the degree of $r$ is less than that of $f$; this $r$ is the (in this case unique) representative of the coset $p(f)$ with this property.
Sometimes, it is easier to reduce powers of $x$ directly. For example, in this ring we have $x^4 \equiv 16$; so $x^13 \equiv (x^4)(x^4)(x^4)(x)$ $\equiv 16^3x$. In this fashion, we can eliminate powers of $x$ larger than the degree of $f$.