A finite field of order $p^n$ has characteristic $p$.

Question

Let $K$ be a finite field with $|K|=p^n$ for a prime $p$ and $n\in\mathbb{N}$.

Now I have to show that the characteristic of $K$ is $p.$

So I know that $(K,+)$ is a $p$-Group and therefore there exists a $m\in\mathbb{N}$ such that $p^m\cdot 1_K = 0_K$. This means that char$(K)=p^m$ because $ker(\varphi)=(p^m)$, where $\varphi: \mathbb{Z}\rightarrow K, n\mapsto n\cdot 1_K$.

Now I don't know how to show $m=1$.

Help is very much appreciated!

Answer 1

If the characteristic of $K$ is $p^m$ for some $m\neq 1$, then $p\cdot 1\neq 0$. However, $(p\cdot 1)^m = 0$, contradicting that $K$ is a field.

Answer 2

The characteristic of a field $K$ is the minimum number of $1_K$s which, when added together, yield $0$:

$\underset{\text{char} K \text{times}} {\displaystyle \sum 1_K} = 0; \tag 1$

by virtue of the fact that $K$ is a field,

$\text{char} K = q \tag 2$

for some prime

$q \in \Bbb P; \tag 3$

for if not, we may write

$q = rs, \; 2 \le r, s < q, \tag 4$

and by grouping the $q = \text{char} K$ $1_K$s under the summation sign in (1) into $r$ subgroups of $s$ $1K$s each, we find that

$rs = \text{char} K = q = 0 \tag 5$

in $K$; but neither

$r, s = 0 \tag 6$

in $K$, whence, since $K$ is a field

$0 \ne rs \in K; \tag 7$

the contradiction 'twixt (5) and (7) forces the conclusions that $q$ is indeed prime, and thus that $1_K$ generates via sucessive addition a $q$-element subfield $F_q$ of $K$; now $K$ being finite it is a vector space over $F_q$ with

$\exists m \in \Bbb N, \; \dim_{F_q} K = m, \tag 8$

whence

$\vert K \vert = q^m; \tag 9$

but we are given that

$\vert K \vert = p^n, \tag{10}$

from which

$q^m = p^n; \tag{11}$

from this we infer that

$q = p, \; m = n; \tag{12}$

that is,

$\text{char} K = p. \tag{13}$

$OE\Delta$.