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)
