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I would like to show that if $K$ is a field with $p^n$ elements then its characteristic has to be $p$, $p$ prime. I'm not sure where to start. It's clear to me that I can construct a field of order $p^n$ if I have a ring of characteristic $p$ by taking the ring of polynomials and quotienting it by an irreducible polynomial of degree $n$. But this doesn't help here.

Many thanks for your help!

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7  
A field is a vector space over the subfield generated by 1. –  Qiaochu Yuan Feb 13 '11 at 11:37

6 Answers 6

There is a unique map of rings with unity ${\Bbb Z}\rightarrow K$. Since the target is finite, the kernel must be a non trivial ideal $I$ of ${\Bbb Z}$. Moreover, $I$ must be a prime ideal, since the target is a domain (a field, in fact).

On the other hand the map maps $p^n=|K|$ to $0$, so that $p^n\in I$ and $p^n{\Bbb Z}\subset I$. But the only prime ideal containing $p^n{\Bbb Z}$ is $p{\Bbb Z}$. Therefore $I=p{\Bbb Z}$ and ${\Bbb F}_p={\Bbb Z}/p{\Bbb Z}\hookrightarrow K$ so that $K$ has characteristic $p$.

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Since the additive group has order $p^n$ it is obivious that the (additive) order of every element is a power of $p$.

Now assume $py \neq 0$ for some $y$ (with $p = 1 + 1 + ... + 1$, p times) then $p\neq 0$. Now $px = 0$ implies that $x = 0$, so there are no elements of order $p$; a contradiction.

Therefore $py = 0$ and every element has additive order p.

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With any field, either $char(K) = m \neq 0$, or $char(K) = 0$ ($1$ has either finite additive order, or infinite additive order).

Clearly, in a finite field, we cannot have characteristic $0$, since we only have finitely many elements to obtain by considering $1,1+1,1+1+1$, etc.

Suppose $char(K) = m$. if m is composite, then $m = st$ for some $1 < s,t < m$.

But a field is an integral domain, so $m1 = 0 \implies (st)1 = (s1)(t1) = 0$, so either $s1 = 0$ or $t1 = 0$, contradicting that $char(K) = m$.

The only two options left then, is that either $m$ is prime, or $m = 1$. but the latter option means $1 = 0$, which is not allowed in a field. So we are forced to conclude that $m$ is prime.

However, Lagrange's theorem applied to the additive group of $K$, tells us the order of $1$ in this group is a divisor of $|K|$, that is: $char(K)|p^n$, hence $char(K) = p$ (this being the only prime that divides $p^n$).

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up vote 2 down vote accepted

First of all, thanks for your help. I used your helpful posts to come up with this:

claim: $|K|=p^n \Rightarrow char(K)=p$

proof:

Consider the additive group of $K$. Then $\forall g \in K: \hspace{2mm} |g| \Big | p^n$. In particular, $|1| \Big | p^n$. Now let $k$ be the smallest integer such that

$$ 0 = \sum_{i=1}^{p^k}1 = (1 + \dots + 1)\cdot \dots \cdot(1 + \dots +1) = \bar{p} \cdot \dots \cdot \bar{p}$$

$k$ times, where we define $\bar{p} = \sum_{i=1}^p1$. By definition, this is the characteristic of $K$. But if $$ 0 = \bar{p}^{k} = \bar{p}\bar{p}^{k-1}$$ then either $\bar{p} = 0$ or $\bar{p}^{k-1} = 0$. If $\bar{p} = 0$ then we're done since then the characteristic is $p$. If $\bar{p}^{k-1} = 0 = \bar{p}\bar{p}^{k-2}$ then we repeat the argument until we reach $\bar{p} = 0$ and hence $char(K) = p$.

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As a Corollary of Lagrange's Theorem, we know that $p^nx = 0$ for all $x$. Therefore, the characteristic of $K$, which is the smallest positive $k$ such that $kx = 0$ (if such exists) must be a divisor of $p^n$. Hence, it is of the form $p^i$, $0\leq i\leq n$.

It cannot be $p^0$, since $1\neq 0$ in a field.

Now, suppose that the characteristic is $p^i$; then there exists an $x\in K$ such that $p^{i-1}x\neq 0$. Therefore $$0 = p^ix = (p^{i-1}p)x = (px)(p^{i-1}x).$$ Since we are in a field, and $p^{i-1}x \neq 0$, then $px=0$. It follows that $i=1$, so the characteristic is $p$, as desired.

Note that this works in any integral domain whose underlying additive group is of exponent a power of $p$.

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Thank you! That's much nicer than what I wrote! –  Rudy the Reindeer Mar 18 '12 at 0:15
    
It would appear, Arturo, that you and I are on the same page. By the definition of the characteristic, we lose no generality if we take $x = 1$. However, I like the way you directly get to p without using primality per se. Cheers :) –  David Wheeler Mar 18 '12 at 0:34
    
@David: I disagree: I am using primality, but it is perhaps hidden a bit. Primality is used when going from "characteristic is a divisor of $p^n$" to "characteristic is $p^i$ for some $i$". That would not be true if $p$ were not prime. (-: –  Arturo Magidin Mar 18 '12 at 0:53
    
hence the "per se" in my comment. it's sneaky...ninja math ftw... –  David Wheeler Mar 18 '12 at 8:56

Consider the prime field of $K$, $P$ ,and then $K$ is a vector space over $P$ so that $x^{d}=p^{n}$ for some natural integer d in which x is the order of $P$.
Moreover, since the order of the prime field must be a prime, it is equal to p in fact, hence we obtain the result.

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