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François G. Dorais
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As observed by Rob, there is exactly one field for each prime power order. The exact formula for the number of fields is of coursethen $$F(n) = \pi(n) + \pi(n^{1/2}) + \pi(n^{1/3}) + \cdots$$ Therewhere $\pi(x)$ counts the number of primes up to $x$. There are $O(\log n)$ nonzero lower order terms each of which is $O(\sqrt{n})$. So the leading term $\pi(n)$ dominates and the Prime Number Theorem gives the asymptotic $F(n) \sim \mathrm{Li}(n) \sim n/\log(n)$.

The exact formula for the number of fields is of course $$F(n) = \pi(n) + \pi(n^{1/2}) + \pi(n^{1/3}) + \cdots$$ There are $O(\log n)$ nonzero lower order terms each of which is $O(\sqrt{n})$. So the leading term $\pi(n)$ dominates and the Prime Number Theorem gives the asymptotic $F(n) \sim \mathrm{Li}(n) \sim n/\log(n)$.

As observed by Rob, there is exactly one field for each prime power order. The exact formula for the number of fields is then $$F(n) = \pi(n) + \pi(n^{1/2}) + \pi(n^{1/3}) + \cdots$$ where $\pi(x)$ counts the number of primes up to $x$. There are $O(\log n)$ nonzero lower order terms each of which is $O(\sqrt{n})$. So the leading term $\pi(n)$ dominates and the Prime Number Theorem gives the asymptotic $F(n) \sim \mathrm{Li}(n) \sim n/\log(n)$.

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François G. Dorais
  • 44.4k
  • 6
  • 150
  • 233

The exact formula for the number of fields is of course $$F(n) = \pi(n) + \pi(n^{1/2}) + \pi(n^{1/3}) + \cdots$$ There are $O(\log n)$ nonzero lower order terms each of which is $O(\sqrt{n})$. So the leading term $\pi(n)$ dominates and the Prime Number Theorem gives the asymptotic $F(n) \sim \mathrm{Li}(n) \sim n/\log(n)$.