How to show characteristic of a finite field that has $p^k$ elements is $p$

Question

I'm studying linear algebra recently and I started K. Hoffman and R. Kunze book about linear algebra. I faced this problem as my teacher asked:

Show that a field of order $p^k$ where $p$ is a prime number, has a characteristic of $p$.

I've searched the internet and found some results. But they're using groups, rings, etc. I want a proof that just uses field axioms and field characteristic definition. Not groups and so on. Some proofs were using subfields and their characteristic, but I couldn't understand. There was no proof to what they said. Actually I know a proof exists because my teacher asked this after teaching field, their axioms, characteristic of a field, some properties about fields, etc. I know when $k=1$, there exists a set $\mathbb{Z}_p = \{0, 1, 2, ..., p-1\}$ which has $p$ elements and addition and multiplication are in hang of $p$. Or we can say $\mathbb{Z}_p = \{[0], [1], [2], ..., [p-1]\}$ which its elements are equivalent classes. For example $[1]$ means all the numbers in this form: $ap+1$. Its characteristic is obviously $p$. How can I prove it for fields that have $p^k$ elements.

Note that I'm studying linear algebra and I haven't studied abstract algebra.

Answer 1

Let $p^r$ be the cardinality of $K$. For integers $a\geq 1$, write $[a]$ for $1+\cdots+1\in K$ ($a$ times).

Lemma. $[p^r]=0_K$.

Proof. Notice that the map $K\ni y\mapsto 1+y\in K$ is a bijection, hence we have $$\sum_{y\in K}y=\sum_{y\in K}(1+y)=[p^r]+\sum_{y\in K}y.$$ Subtracting $\sum_{y\in K} y$ from both sides yields the result.$\,\tiny\blacksquare$

Lemma. The characteristic of $K$ divides $p^r$.

Proof. Euclidean division of $p^r$ by the characteristic $n$ gives $p^r=an+b$ with $0\leq b<n$. In $K$, we then have $$0=[p^r]=[a]\cdot[n]+[b]=[a]\cdot 0+[b]=[b].$$ We must thus have $b=0$ since $b<n$.$\,\tiny\blacksquare$

Proposition. The characteristic of $K$ is $p$.

Proof. By the previous Lemma, the characteristic is $p^s$ for some $1\leq s\leq r$. If we had $s\geq 2$, then $$[p^{s-1}]\cdot[p]=[p^s]=0.$$ Thus, a product of two non zero elements in $K$ gives zero, contradiction. We must therefore have $s=1$.$\,\tiny\blacksquare$

Answer 2

By definition of field, the field must have a multiplicative identity; call this element 1.

Note that there must be some $n$ such that $$\underbrace{1+1+\dots+1}_{n \text{ times}} = 0 \quad (\star)$$ because there are only finitely many elements in your field. (Do you buy this part?). From now on, let $n$ be the smallest positive integer such that the $(\star)$ is satisfied. We seek to show that $n = p$.

Another fact about fields is that every element must have a multiplicative inverse. If you have not done so already, prove that this implies that if $ab = 0$, then $a=0$ or $b=0$. In other words, fields have no nonzero zerodivisors.

We want to show that $n=p$, and $p$ is a prime number. Our first step will be to show that $n$ is also prime. So we will have to use some facts about prime numbers; probably that the only factors of $p$ are $\pm 1$ and $\pm p$. So let's write $n=\ell m $ where $\ell, m \in \mathbb{Z}$. Calculate

$$(\underbrace{1+1+\dots+1}_{m \text{ times}})\underbrace{1+1+\dots+1}_{\ell \text{ times}}$$ using distributivity, and use the fact you just proved about nonzero zerodivisors to determine that either $\ell = n$ or $m = n$.

You have now shown that $n$ is a prime number; but does it have to be $p$?

At this point, we will break our field up into distinct subsets. The first subset is things that can be written as a sum of the multiplicative identity, i.e., $$\{ 1, 1+1, \dots, \underbrace{1+1+\dots+1}_{n-1 \text{ times}}, 0 \}.$$ For simplicity of notation, we'll denote $\underbrace{1+1+\dots+1}_{m \text{ times}}$ by $m$, (but this isn't necessarily the 'same' $m$ as the $m$ in the natural numbers is).

If this is the whole field, great. Otherwise, there is some $a_1$ that is not in this set: consider $$\{ a_1+1, a_1+2, \dots, a_1+n-1, a_1+0 \}.$$ Prove that all of these elements are distinct, i.e., if $a_1 + i = a_1+j$ with $1\leq i \leq n$ and $1 \leq j \leq n$, then $i=j$. Thus each set has $n$ elements.

Again, if we have not captured the whole field between these two subsets, there is some $a_2$ that is not in either set, so consider $$\{ a_2+1, a_3+2, \dots, a_2+n-1, a_2+0 \}.$$ Carry on until you have written down every element in the field. Since your field has finitely many elements, this process will eventually terminate, maybe you end up with $r$ many sets, all of which have $n$ elements.

Since every element in the field shows up in some subset, by construction, you know that $p^k \leq nr$.

There could, a priori, be repeats, so maybe we have overcounted. Prove that $a_{\iota} + i \neq a_{\kappa} + j$ whenever $\iota \neq \kappa$ or $i \neq j$. Then you will have shown that $p^k = nr$. Since we've already shown that $n$ must be prime, the only prime divisors of $p^k$ are $p$, and we are done!