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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!

amWhy
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TwoStones
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    Hint: Show $(p^m\cdot 1)=(p\cdot 1)^m$ for $m\geq 1.$ Then use that $K$ is a field. – Thomas Andrews Jun 11 '19 at 17:21
  • @ThomasAndrews Why are you answering in a comment? – Arthur Jun 11 '19 at 17:41
  • @Arthur A question like this is guaranteed to be a duplicate or a near duplicate. Answering it does not add to the repository of knowledge this site strives to be. A comment should steer the asker though, may be even prompt them to answer themselves, or at least make them see the light. At which point the question can either be deleted or self-answered (or possibly answered by another noob). May be this discussion should take place in meta? Possibly your "complaint" should also be there? – Jyrki Lahtonen Jun 12 '19 at 05:18

2 Answers2

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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.

Arthur
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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,

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

$OE\Delta$.

Robert Lewis
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