# Question about the Characteristic of $\mathbb{F}_{p^n}$

We can prove that any finite field of prime characteristic $p$ must have $p^n$ elements.

Conversely, let $F$ be a finite field with $p^n$ elements, where $p$ is a prime number. Is the following argument a correct way to prove that $F$ has characteristic $p$ ?

We can show, without assuming that $F$ has characteristic $p$, that every finite field with $p^n$ elements is the splitting field of $x^{p^n}-x\in\mathbb{F}_p[x]$. Therefore $E/\mathbb{F}_p$ is a field extension and the characteristic of $E$ must equal the characteristic of $\mathbb{F}_p$ which is $p$.

• You say that $F$ is a splitting field of some polynomial in $\mathbb{F}_p[x]$ but this already includes the assumption that $\mathbb{F}_p$ is a subfield of $F$, so I do not think this argument should work. – Matthias Klupsch Oct 16 '14 at 6:08

Like @Martín-Blas Pérez Pinilla: said, $$\underbrace{1+ \cdots + 1}_{p^n}=0$$ and so

$$\underbrace{p \times \cdots \times p}_{n}=0$$

in a field, so $p=0$.

• Would this also be correct? Since the characteristic of $\mathbb{F}_{p^n}$ is the order of $1$ in the additive group of $\mathbb{F}_{p^n}$, and we know it must be a prime that divides $p^n$ (by Lagrange's Theorem), the only possible choice is $p$ ? – Guest Oct 16 '14 at 15:00
• @Guest: Yes, it's also correct! – orangeskid Oct 16 '14 at 15:46

The additive group of $\Bbb F_{p^n}$ has order $p^n$, that is...

• @orangeskid, thanks. And thanks for the correction of my typo. – Martín-Blas Pérez Pinilla Oct 16 '14 at 10:06

If $$\mathbb{F}_q \subseteq \mathbb{F}$$ is a subfield, then $\mathbb{F}$ is a finite-dimensional $\mathbb{F}_q$ vectorspace, so it must have $q^m$ elements. But $p^n=q^m$ implies $p=q$

• And to prove that $\mathbb{F}$ does contain $\mathbb{F}_q$ for some prime $q$, would this be correct: let $q:=\text{Char}(\mathbb{F})$, then we know $q$ is a prime, and $\{1,1+1,\ldots,1+1+\cdots+1\}$ (with $q$ elements) is a cyclic subgroup of $\mathbb{F}$ that is isomorphic to $\mathbb{F}_q$. – Guest Oct 16 '14 at 6:22
• For any Ring $R$ there is a Homomorphism $\mathbb Z \rightarrow R$, which cannot be injective if $R$ is finite. If $R$ is a domain, then the kernel must be a prime ideal $(p)$. – Blah Oct 17 '14 at 6:09