TITLE: Finite fields as vector spaces
QUESTION [8 upvotes]: I'm having great difficulty understanding this topic.
Can someone concretely explain what it is meant by thinking of $GF(q^2)$ ($q$ a prime power) as a two-dimensional vector space over its subfield $GF(q)$ (fixed by the map $x \mapsto x^q$).
How would I construct a basis of this vector space?  If $\alpha$ was the primitive root in $GF(q^2)$ and $\beta$ of $GF(q)$.  Then $\alpha^{n(q+1)}=\beta^n$.  How would $\alpha^{n(q+1)+m}$ be represented in the vector space notation?

REPLY [3 votes]: I will attempt to understand the last part of the question. 
If $\alpha$ is a primitive element of GF$(q^2)$ then every non-zero element of GF$(q^2)$ can be written as a power of $\alpha$, and it's easy to do multiplication in the field when you represent its elements this way. 
Every element can be written uniquely as $a+b\alpha$ with $a$ and $b$ in GF$(q)$, and it's easy to do addition in the field when you represent its elements this way. 
I take it what you're asking about is how do the two different representations relate, e.g., how do you find the $a$ and $b$ that make $\alpha^2=a+b\alpha$. But to answer this, knowing $\alpha$ is primitive isn't enough. You have to know more about exactly which primitive element you've chosen - they all have the same properties multiplicatively, but different properties additively.