Definition: A field extension $E$ of $F$ is of degree $n$ (and is called a finite field extension) if $E$ is an $n$-dimensional vector space over $F$.

Theorem: Let $E$ be a degree $n$ finite extension of a field $F$. If $F$ has $q$ elements, then $E$ has $q^n$ elements.

Definition: Let $E$ be a field. Suppose there exists $n\geq 1$ such that $n\cdot x = 0$ for all $x\in E$. Then the smallest such choice of $n$ is the characteristic of $E$. If no such $n$ exists, then $E$ is of characteristic $0$.

Corollary: Let $E$ be a finite field with characteristic $p$. Then $E$ contains exactly $p^n$ elements for some choice of $n\geq 1$.

Proof: (Taken from Fraleigh - A First Course in Algebra, 7Ed) Every finite field $E$ is a finite field extension of a prime field isomorphic to the field $\mathbb{Z}_{p}$, where $p$ is the characteristic of $E$. The result follows from the theorem using $F = \mathbb{Z}_{p}$.

This proof is probably very simple, but I'm having problems with showing that the degree of $E$ over $\mathbb{Z}_{p}$ is finite. It seems intuitively obvious since $E$ itself is finite, but I cannot see how to conclude from this that $E$ is a finite-dimensional vector space over $\mathbb{Z}_{p}$. Obviously it is not an infinite-dimensional vector space over $F$ since it is not infinite.

This may seem too simple to answer, but what am I missing here?

Thanks for any assistance!

• $E$ is finite, so it has a finite generating set as a vector space (namely take every element). Standard linear algebra shows that this generating set can be reduced to a (finite) basis by throwing out enough elements. Aug 1, 2012 at 20:47

If the degree was not finite then this means any basis of $E$ over $\mathbb{Z}_p$ must containg infinite number of elements, $E$ is finite hence $E$ spans $E$ over $\mathbb{Z}_p$ hecne a basis must be also finite.

Edit: if $K/F$ is a field extension then $K$ is a vector space over $F$

• My question goes a bit deeper. Why must $E$ be a vector space over $F$ at all? As I mentioned at the end, clearly it cannot be infinite-dimensional, but I do not see how this implies that it is finite dimensional. That is, why are we guaranteed a dimension?
– roo
Aug 1, 2012 at 20:46
• So the part in the edit is what you do not understand ? Aug 1, 2012 at 20:47
• No please ignore the edit. It was due to temporary brain failure and I removed it. :)
– roo
Aug 1, 2012 at 20:48
• Oh, your edit. I just noticed it. Sorry I need to think about it for a moment.
– roo
Aug 1, 2012 at 20:48
• You may also need to know that every vecotr space have a basis and that the cardinality of every such basis is the same and is called the dimension of the vector space Aug 1, 2012 at 20:49