# Elements in a finite field

• Nov 16th 2009, 03:04 PM
lisa
Elements in a finite field
Hello.

I have to proof the following proposition:

Let $K$ be a finite field. Then $K$ has $q=p^n$ elements, whereas $p$ is the characteristic of $K$ and $n=[K:\mathbb{F}_p]$.
Furthermore I have to proof, that $x^q = x$ for all $x \in K$.

Could anybody of you explain, how to proof the proposition?

Bye,
Lisa
• Nov 16th 2009, 04:15 PM
aliceinwonderland
Quote:

Originally Posted by lisa
Hello.

I have to proof the following proposition:

Let $K$ be a finite field. Then $K$ has $q=p^n$ elements, whereas $p$ is the characteristic of $K$ and $n=[K:\mathbb{F}_p]$.
Furthermore I have to proof, that $x^q = x$ for all $x \in K$.

Could anybody of you explain, how to proof the proposition?

Bye,
Lisa

I'll assume p is a prime number. A finite field K of order $q=p^n$ is the splitting field of $x^q - x$ over $F_p$ (see here). Since a finite field is perfect, K is a separable extension of $F_p$. It follows that K is a Galois extension of a field $F_p$. By the fundamental theorem of Galois theory, $[K:F] = |Gal(K/F)|=n$, where $F=F_p$ and Gal(K/F) is a cyclic group of order n generated by a Frobenius automorphism.
• Nov 16th 2009, 04:29 PM
tonio
Quote:

Originally Posted by lisa
Hello.

I have to proof the following proposition:

Let $K$ be a finite field. Then $K$ has $q=p^n$ elements, whereas $p$ is the characteristic of $K$ and $n=[K:\mathbb{F}_p]$.
Furthermore I have to proof, that $x^q = x$ for all $x \in K$.

Could anybody of you explain, how to proof the proposition?

Bye,
Lisa

One of the easiest proofs is, perhaps, to note that any field is a vector space over any of its subfields, and then: a finite field $K$ of characteristic p is a v.s. over the prime field of char. p, $\mathbb{F}_p\cong \mathbb{Z}\slash p\mathbb{Z}$, obviously of finite dimension n, and choosing any basis for this v.s. a simple combinatoric reasoning gives that the number of elements in $K$ must be $p^n$

Take the polynomial $f(x)=x^{p^n}-x\,\,\,over\,\,\mathbb{F}_p\left[x\right]$ . Now you can go with the argument from fields extensions and Galois theory that was given to you by Aliceinwonderland or else use a little group theory: the multiplicative group $K^{*}$ has order $p^n-1\,\Longrightarrow \omega^{p^n-1}=1\,\,\forall\,\,\omega\,\in\,K\,\Longrightarrow \,\omega^{p^n}=\omega$ , and we're done.

Tonio
• Nov 17th 2009, 04:15 AM
lisa
Thank you. Your explanations were very helpful for me.

Best greets,
Lisa