# Quick way to show a finite field always has prime power order? [duplicate]

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Order of finite fields is $p^n$

Is there a quick way to prove that finite fields always have cardinality that is the power of a prime p?

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## marked as duplicate by Jyrki Lahtonen♦, Hagen von Eitzen, Stefan Hansen, draks ..., Davide GiraudoJan 24 '13 at 14:11

Pretty funny: three nearly identical answers at virtually the same time. – rschwieb Jan 23 '13 at 19:48
This is not only a duplicate, but an $n$-plicate. See the list of related questions. – Jyrki Lahtonen Jan 24 '13 at 12:02

Here is a quick way to derive the characteristic of a finite field with pretty low level concepts. Consider the ring homomorphism $$f\colon\mathbb{Z}\to F:n\mapsto n\cdot 1_F$$ where $1_F$ is the multiplicative identity in $F$. Now $$\mathbb{Z}/\ker(f)\simeq\operatorname{im}(f)$$ but $\operatorname{im}(f)$ is a subring of a field, hence an integral domain. So necessarily $\ker(f)$ is a prime ideal, thus of form $p\mathbb{Z}$ for some prime $p$. Then $\operatorname{im}(f)$ is a subfield of order $p$ of $F$, so viewing $F$ as a vector space over this field immediately implies $|F|=p^n$ for some $n$.

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If $F$ is a finite field, then the prime subfield $P$ (that is, the smallest subfield) has prime order $p$. $F$ is a vector space over $P$ of dimension $n$, say, and hence $F$ has order $p^n$.

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A finite field has prime characteristic. Hence, it contains a field isomorphic to $\mathbb{F}_p$ for some prime $p$. Thus, it is a finite dimensional vector space over $\mathbb{F}_p$ (finite dimensional because it is finite). What is the order of a finite dimensional vector space over a finite field?

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Yes. First, note that the characteristic of any integral domain (if not $0$), will be a prime, by noting that the subring generated by $1$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ where $n$ is the characteristic (and any subring of an integral domain is again an integral domain).

Now we know that for our finite field (which canot have characteristic $0$), there is some prime $p$ such that adding any element to itself $p$ times gives $0$. This means that the underlying abelian group (for the addition) is elementary abelian (meaning that all non-identity elements have prime order for a fixed prime), and it is well-known from group theory that such groups have order a power of $p$ (if some other prime divided the order, then the group would have an element of that order).

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