How to prove that the order of $F[x]/(g(x))$ is $|F|^{\deg(g)}$, where $F$ is a finite field?

Question

In the textbook that I am currently learning from, it says that:

Given $F[x]/g(x)$ for some fixed $g(x) \in F[x]$ of degree $d ≥ 1$,

$$F[x]/g(x) = {r(x) + g(x):\ \deg(r) ≤ d − 1}$$

If $F$ is finite, $r(x)$ has $d$ coefficients, each of which has $|F|$ choices. Hence $$|F[x]/g(x)| = |F|^{\deg(g)}$$

My question is: how did we get the "$r(x)$ has $d$ coefficients" statement? There are $d-1$ different $\deg(r)$, and the maximum $\deg$ for $r(x)$ is $d-1$, so what do the $d$ coefficients imply?

Thank you!

Answer 1

Think about the standard isomorphism $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$. After long division, every coset has the form $r(x) + (x^2+1)$ where the degree of $r$ is $1$ or less. Hence $r(x)=ax+b$ for some $a, b \in \mathbb{R}$---either of which could be $0$---and this $r(x)$ is unique. Thus every element of $\mathbb{R}[x]/(x^2+1)$ is "parameterized" by the choice of $a$ and $b$, so it takes $2$ "coordinates" to specify the element. In this case, if we exchange $ax+b$ for $ai+b$, we get the standard isomorphism with $\mathbb{C}$.

The same thing would happen if you pondered $\mathbb{Z}_3[x]/(x^2+1)$. Here $x^2+1$ is irreducible and so $\mathbb{Z}_3[x]/(x^2+1)$ is a field. How large is it? Using the exact same analysis as above you get that every coset may be written $ax+b + (x^2+1)$, where $a, b \in \mathbb{Z}_3$. Hence you have $3$ choices for $a$ and $3$ choices for $b$, so there are $3 \cdot 3=9$ elements in the field $\mathbb{Z}_3[x]/(x^2+1)$. Yes, the remainder is linear, but a linear term requires $2$ (not $1$!) coefficients.