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A field $F$ has at most a finite number of elements of order $\leq n$ for any $n$ in integers.

How can I prove this? I thought it's related to the fact that a polynomial of degree $n$ would have at most $n$ roots over $F$. But I am not sure. Would appreciate hints.

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An element of order $n$ will be a root of $x^n-1$. So, straight away, there can be only finitely many elements of order dividing $n$. But I don't see immediately how that tells you something about elements of orders $\leq n$ not dividng $n$. –  hoyland Feb 18 '12 at 14:54
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If $x\in F$ has order $k\geq 1$ then $x^k=1$ so either $x=1$ or $1+x+\ldots+x^{k-1}=0$. –  Davide Giraudo Feb 18 '12 at 14:55
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@hoyland But you could also look at $x^k - 1$ for $k < n$, and there are only finitely many of these polynomials. –  Dylan Moreland Feb 18 '12 at 15:09
    
@Dylan: simple + brutal= perfect! –  Georges Elencwajg Feb 18 '12 at 17:29

2 Answers 2

Your idea works. There are various ways to carry it out. For example, let $N$ be the least common multiple of the numbers $1, 2, \dots, n$. Then every element of order $\le n$ is a zero of the polynomial $x^N-1=0$. Or else, more wastefully, you can use the polynomial $\prod_{k=1}^n (x^k-1)$.

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Hint: by pigeonholing, if there are infinitely many elements of order $\le n$ then for some $k\le n$ there are infinitely many elements of order $k,\:$ hence $x^k - 1$ has infinitely many roots $\:\Rightarrow\Leftarrow$

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