Number of elements in a finite field extension for finite fields Given an arbitrary finite field $K$ (not necessarily $\mathbb{F}_p$ with $p \in \mathbb{P}$) with $|K| = q$ and an irreducible polynomial $f$ with $\alpha$ as root and degree of $n$. Is $|K(\alpha)| = q^n$ and why? Its clear to me for $K$ isomorphic to $\mathbb{Z}_p$ with $p$ is a prime number.
 A: The division algorithm holds in $k[x]$ for any field $k$. In fact, it holds in $R[x]$ for any commutative ring $R$ provided that the divisor has leading term a unit.
Theorem. Let $R$ be a commutative ring, and let $a(x)$ be a polynomial in $R[x]$ whose leading coefficient is a unit in $R$. Then for every $b(x)\in R[x]$ there exist unique $q(x),r(x)\in R[x]$ such that $b(x) = q(x)a(x)+r(x)$, where $r(x)=0$ or $\deg(r)\lt\deg(a)$.
The proof is exactly the same as the proof in any field, and can be done by induction on $\deg(b)$. 
With the division algorithm in hand, it is immediate that if $f(x)$ is of degree $n$, then $k[x]/\langle f\rangle$ is a vector space over $k$ of dimension $n$, with basis $1+\langle f\rangle$, $x+\langle f\rangle,\ldots,x^{n-1}+\langle f\rangle$. 
In particular, for $k=\mathbb{F}_q$, the resulting quotient is dimension $n$ over $\mathbb{F}_q$, and so has $q^n$ elements. This holds whether or not $f(x)$ is irreducible, though we need $f(x)$ irreducible in order to know that the quotient is a field.
