Factor Rings over Finite Fields Given a polynomial ring over a field $F[x]$, I can factor, for example, the ideal generated by an irreducible polynomial $ax^2 + bx + c$: $F[x]/\left<ax^2 + bx + c\right>$, and guarantee that this factor ring is also a field.
My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / \left<ax^2 + bx + c\right>$ for some irreducible polynomial $ax^2 + bx + c$, I can guarantee, for example, that this field has $p^2$ elements. 
I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by $ax^2 + bx + c$. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.
 A: The quotient you get is a two-dimensional vector space over your field. Can you show that any such vector space must have $p^2$ elements?
A: Indeed the elements of the factor ring $\Bbb{F}_p[x]/\langle ax^2+bx^2+c\rangle$ can be represented by the remainders by division by $ax^2+bx+c$. This is true because we can divide polynomials in $\Bbb{F}_p[x]$ by $ax^2+bx+c$ with remainder. What this means is that

For every polynomial $f\in\Bbb{F}_p[x]$ there exist unique $q,r\in\Bbb{F}_p[x]$ with $\deg r<2$ such that 
  $$f=q(ax^2+bx+c)+r.\tag{1}$$

This equality shows that $f$ and $r$ are in the same coset of $\langle ax^2+bx+c\rangle$, and hence they are mapped to the same element of the factor ring $\Bbb{F}_p[x]/\langle ax^2+bx+c\rangle$. Hence the image of
$f$ in the factor ring is represented by $r$, and so every element of the factor ring is represented by a linear polynomial.
To see that no two linear polynomials represent the same element of $\Bbb{F}_p[x]/\langle ax^2+bx+c\rangle$, it suffices to note that the remainder $r$ in $(1)$ is unique for every $f\in\Bbb{F}_p[x]$, meaning in particular that every linear polynomial is represented only by itself.
Alternatively, if two linear polynomials $r$ and $r'$ represent the same coset of $\langle ax^2+bx+c\rangle$ in the factor ring, then $r-r'$ is a multiple of $ax^2+bx+c$. Because $\deg r-r'<\deg(ax^2+bx+c)$ it follows that $r-r'=0$.
