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.
Best Answer
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 $s=1$.$\,\tiny\blacksquare$