Polynomials, finite fields and cardinality/dimension considerations Can someone please give me a hint how to prove that the quotient ring $\mathbb{F}_p[x]/{\langle f\rangle}$, where $f$ is a irreducible polynomial of degree $k$ and $p$ is a prime (and $\langle f\rangle$ the ideal generated by $f$), has cardinality $p^k$ and how to determine the dimension of $\mathbb{F}_p[x]/{\langle f\rangle}$ as a vector space ?
 A: Because $\mathbb{F}_p$ is a field, $\mathbb{F}_p[x]$ is a Euclidean domain.
So every element $q(x)\in \mathbb{F}_p[x]$ can be written as
$$q(x) = p(x)f(x) + r(x),$$
where $r(x)=0$ or $\deg(r)\lt \deg(f)$.
That means that in $\mathbb{F}_p[x]/\langle f\rangle$, every polynomial is equivalent (equal up to a multiple of $f(x)$) with $0$ or a polynomial of degree strictly smaller than $\deg(f)$. 
If $\deg(f)=k$, then the possible polynomials of degree smaller than $k$ are of the form
$$a_0 + a_1x + \cdots + a_{k-1}x^{k-1},\quad a_i\in\mathbb{F}_p.$$
Simple counting gives you that the quotient has size at most $p^k$.
Now, if $$a_0+a_1x+\cdots+a_{k-1}x^{k-1} +\langle f\rangle = b_0+b_1x+\cdots + b_{k-1}x^{k_1}+\langle f\rangle,$$
then $f(x)$ divides 
$$(a_0 + \cdots +a_{k-1}x^{k-1})-(b_0+\cdots+b_{k-1}x^{k_1}) = (a_0-b_0)+\cdots + (a_{k-1}-b_{k-1})x^{k-1}.$$
Since this is either $0$ or of degree strictly smaller than $f(x)$, the only way this can be a multiple of $f(x)$ is if it is equal to $0$. So two distinct polynomials of degree less than $\deg(f)$ represent the same coset if and only if they are identical, which shows that $\mathbb{F}_p[x]$ has at least $p^k$ elements. 
Putting the two together tells you the size is exactly $p^k$. 
Once you know the cardinality, the dimension follows, because an $n$-dimensional vector space over $\mathbb{F}_p$ has size $p^n$ (same counting argument as for polynomials of degree at most $k-1$ above; in fact, what we did above was to show that the classes $x^i+\langle f\rangle$, $i=0,\ldots,k-1$, are a basis). 
A: If a polynomial, e.g. $x^3 + 2x^2 - 5x + 7$, is to be identified with $0$, then $x^3$ gets reduced to $-(2x^2 - 5x + 7)$ and $x^4$ to $x\cdot x^3$, etc.---keep reducing until everything's a linear combination of $1$, $x$, and $x^2$.  And similarly for higher degrees. That gives you the dimension of the vector space.
A: Well, $\mathbb{F}_p/\langle f \rangle$ is a vector space over $\mathbb{F}_p$ (if that isn't immediately obvious, you should sit down and write down a proof to convince yourself).  If we write
$f=a_0 + a_1 x + \cdots + a_n x^n$
with $a_i\in \mathbb{F}_p$, then you can use $f$ to find a basis for $\mathbb{F}_p/\langle f \rangle$ as a vector space over $\mathbb{F}_p$.  (Hint:  a potential basis is staring you right in the face.)  That will tell you the dimension.
