# Finite fields as vector spaces

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?

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You construct a basis in the same way you do for any finite-dimensional vector space: repeatedly pick a vector not in the span of what you've already got until you're done. I don't know what you mean by "the vector space notation". – Chris Eagle Jun 1 '11 at 11:41

Let $E$ be a finite field of order $p^n$ where $p$ is some prime number. Let $F=\{0,1,2,\cdots,p-1\}$ be the subfield of $E$ generated by $1$. We view $E$ as an $F$-vector space in the following manner:

(1) $E$ is an abelian group under addition.

(2) If $\alpha\in F$ and if $x\in E$, we define $\alpha\cdot x$ to be the product of $\alpha$ and $x$ in the field $E$. Note that this defines "scalar multiplication" of elements of $F$ by elements of $E$.

You should now check that $E$ is an $F$-vector space. For example,

(1) If $\alpha\in F$ and if $x,y\in E$, then $\alpha\cdot(x+y)=\alpha\cdot x + \alpha\cdot y$ since multiplication is distributive over addition in the field $E$.

(2) If $\alpha,\beta\in F$ and if $x\in E$, then $(\alpha+\beta)\cdot x=\alpha\cdot x + \beta\cdot x$ by similar reasoning to that of (1).

(3) If $\alpha,\beta\in F$ and if $x\in E$, then $\alpha\cdot(\beta\cdot x)=(\alpha\beta)\cdot x$, where $\alpha\beta$ is the product of $\alpha$ and $\beta$ in the field $F$. This follows from associativity of the multiplication in $E$.

(4) Finally, $1\cdot x=x$ for all $x\in E$ since the identity $1$ of $F$ is also the identity of $E$.

Therefore, $E$ is indeed an $F$-vector space as claimed.

Vector spaces occur in unexpected ways throughout mathematics and it is therefore important to get used to this particular example. For example, if you have studied the theory of field extensions, you will know that if $F\subseteq E$ are fields, then $E$ can be viewed as an $F$-vector space. A basic (but important) result in this situation is that if $E$ is a finite-dimensional vector space over $F$, then $E$ is algebraic over $F$. (Of course, the converse fails as you should check.)

We know that $GF(q^2)$ is the splitting field of a polynomial of degree $2$ over $GF(q)$. If $\alpha$ is a root of this polynomial in $GF(q^2)$, then $[GF(q)][\alpha]=GF(q^2)$ as you should check. In this case, we know by elementary field theory that $\{1,\alpha\}$ is a basis for $GF(q^2)$ over $GF(q)$. (Intuitively, since $\alpha$ satisfies a polynomial of degree $2$ over $GF(q)$, we can write $\alpha^2$ as a $GF(q)$-linear combination of $1$ and $\alpha$. Repeat this procedure for higher powers of $\alpha$.)

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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.

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