Q: Find the smallest field that has exactly 6 subfields?
From our work and theorem 22.3 we know that a field of order p^n has a single subfield of order p^m for each m which divides n. So we must look for the smallest n with 6 divisors.
n Divisors
1 1
2 1,2
3 1,3
4 1,2,4
5 1,5
6 1,2,3,6
7 1,7
8 1,2,4,8
9 1,3,9
10 1,2,5,10
11 1,11
12 1,2,3,4,6,12
Eureka!
Thus choosing the smallest prime p=2 we see that the smallest field that has exactly 6 subfields has order 2^12.
--Bakey 09:11, 30 November 2012 (UTC)
