The cardinal of finite field

Question

I know that the cardinal of a finite field is $p^n$ for $p$ a prime. The proof what I found is: if $K$ is a finite field then there is a sub-field $\Bbb F_p$ with cardinal $p$ and then we can see $K$ as a vector space over $\Bbb F_p$ so its cardinal is $p^n$ for some $n$. The problem with this proof is that I can't use the vector spaces so my question: is there another proof?

Answer 1

Suppose the field has an order divisible by two primes, $p\neq q$. The additive group of the field has elements $a$ of order $p$ and $b$ of order $q$ (this is group theory). The field also, therefore, contains the non-zero elements $pb=b+b+\dots +b\neq 0$ where $b$ is taken $p$ times, and this is non-zero because $b$ has order $q$, and also $qa\neq 0$.

Now $pb\cdot qa=pa\cdot qb=0$, contradicting the fact that we have a field.

So there can only be one prime dividing the order of the field, which must therefore be a power of that prime.

Answer 2

The fact that $K$ is a vector space over $\mathbb{F}_p$ is the core of the result so you're not going to find a truly different proof. If you don't want to use the language of vector spaces, you can imitate the proof as follows. Since the field $K$ is finite, it contains $\mathbb{F}_p$ for some prime $p$. Let $K_0 = \mathbb{F}_p$. If $K_0 = K$ we are done. If $K \ne K_0$, choose some element $x \in K - K_0$ and let $K_1 = K_0[x]$. Now prove by hand that $K_1$ is a field and its number of elements is a power of $p$. If $K_1 = K$ we are done. If not, choose another element and add it to $K_1$ to get a field $K_2$. Prove that $|K_2|$ is still a power of $p$. If $K_2 = K$ we are done, else repeat again to $K_3$, etc. Since $K$ is finite, this algorithm eventually terminates and proves that $|K|$ is a power of $p$.