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!

  • 2
    $\begingroup$ $E$ 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. $\endgroup$ – Qiaochu Yuan Aug 1 '12 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 $F$

  • $\begingroup$ 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? $\endgroup$ – roo Aug 1 '12 at 20:46
  • $\begingroup$ So the part in the edit is what you do not understand ? $\endgroup$ – Belgi Aug 1 '12 at 20:47
  • $\begingroup$ No please ignore the edit. It was due to temporary brain failure and I removed it. :) $\endgroup$ – roo Aug 1 '12 at 20:48
  • $\begingroup$ Oh, your edit. I just noticed it. Sorry I need to think about it for a moment. $\endgroup$ – roo Aug 1 '12 at 20:48
  • $\begingroup$ 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 $\endgroup$ – Belgi Aug 1 '12 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.