Let $\Bbb F$ be a field with $k$ elements, where $k$ is a finite number. Prove that there exists a prime $p$ and a positive integer $n$ such that $$k=p^n$$
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1See Theorem 1.5 here. This is essentially the argument given by A Wertheim below. It might help to read the theorem in a context, and learn more cool stuff :) – Prism Aug 10 '13 at 00:50
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Thanks for posting that, @Prism. I love Dr. Conrad's notes (and fortunately, I do believe he's an active member of MSE)! – Alex Wertheim Aug 10 '13 at 00:56
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Related: http://math.stackexchange.com/questions/53877/is-there-anything-like-gf6 – Prism Aug 10 '13 at 00:58
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@AWertheim: Me too! And yes, always looking forward to reading his posts here :) – Prism Aug 10 '13 at 00:59
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This is equivalent to showing that $k$ is only divisible by one prime only. Suppose that $k$ is divisible by 2 distinct primes $p,q$ (with $p<q$) . Considering $F$ as a group under addition, we use Cauchy's theorem to deduce that there exists elements $x,y\in F$ such that $|x|=p,|y|=q$. Clearly, $x,y\not=0$.
Since $x+x+...+x=0\,$ (addition is done $p$-times), therefore by multiplying by $yx^{-1}$ we get $y+y+...+y=0$ (addition is done $p$-times). Therefore $|y|=q$ divides $p$, this contradicts the fact that $p<q$.
Amr
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1Wow. The way I've usually seen this proved is the one I posted, but I really like this answer! +1 :) – Alex Wertheim Aug 09 '13 at 23:54
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2New to me, too, and quite elegant. One remark: we don't need $p<q$ to know that $q$ cannot divide it. – Jonathan Y. Aug 10 '13 at 00:08
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@user142526 Sure. I thought that the answer would be clearer. I don't remember why I thought this way. – Amr Aug 10 '13 at 05:44
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This proof is not actually new. It is essentially the proof given by E.H. Moore in his paper on finite fields back in 1903. See http://en.wikipedia.org/wiki/Finite_field#Order. – KCd Dec 16 '13 at 09:16
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Hint: first show that the characteristic* of any field is prime. You'll want to show $F$ is a vector space over any subfield, and then use a dimension argument to count elements.
*If you don't know what the characteristic of a field is, it is the least integer $r$ such that
$$\underbrace{1_{f} + 1_{f} + \cdots 1_{f}}_{r \text{ times}} = 0$$
Alex Wertheim
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@Makkers certainly. Let $F'$ be a subfield of $F$. Suppose $F$ has dimension $n$ as a vector space over $F'$. What do elements of $F$ look like with respect to a basis in $F'$? Write every element of $F$ as a unique linear combination of basis elements in $F'$. Then we just count... – Alex Wertheim Aug 09 '13 at 23:58
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Well, there is an easy way to see this:
Consider the ring homomophism $\varphi: \mathbb{Z} \to F$ given by $\varphi(n)=n\cdot 1_F$ where $1_F$ is the multiplicative identity in $F$. What is the kernel of $\varphi$?
Well, From the first isomorphism theorem we know that $\mathbb{Z}/\ker(\varphi) \cong \operatorname{im}(\varphi)$. Since $\operatorname{im}(\varphi)$ is a subset of the field $F$, it must be an integral domain. Hence, $\ker(\varphi)$ must be a prime ideal of $\mathbb{Z}$ and because $|\operatorname{im}(\varphi)| < \infty$ since $F$ is finite we conclude that $\ker(\varphi)$ is a non-zero prime ideal of $\mathbb{Z}$. So, it's of the form $p\mathbb{Z}$ for a prime number p. But then, it would be a maximal ideal of $\mathbb{Z}$ and since $\mathbb{Z}/\ker(\varphi) \cong \operatorname{im}(\varphi)$ we conclude that $E = \operatorname{im}(\varphi)$ must be a subfield of $F$.
In other words, $F$ is a finite extension of $E=\operatorname{im}(\varphi)$ so we must have $\dim_E(F)=n$ for some n. So, everything in $F$ can be written as a linear combination of elements in $E$ and it's clear by a simple counting argument that we'll have $p^n$ such linear combinations. Therefore $|F| = p^n$
user66733
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