## Count the elements of F[x]/(p(x)) where F is a finite field

Let $F$ be a finite field of order $q$ and let $p(x)$ be a polynomial in $F[x]$ of degree $n \geq 1$. Show that $F[x]/(p(x))$ has order $q^n$.

In this previous exercise, we showed that every coset in $F[x]/(p(x))$ is represented by a polynomial of degree less than $n$, and that two such polynomials represent distinct cosets. Evidently then the number of elements of $F[x]/(p(x))$ is precisely the number of polynomials in $F[x]$ having degree less than $n$. Each such polynomial is uniquely determined by its coefficients, of which there are $n$, each taking one of the $q$ values in $F$. Thus $F[x]/(p(x))$ contains $q^n$ elements.