(New page: 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...) |
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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. | 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 | + | n Divisors |
| + | |||
1 1 | 1 1 | ||
| + | |||
2 1,2 | 2 1,2 | ||
| + | |||
3 1,3 | 3 1,3 | ||
| + | |||
4 1,2,4 | 4 1,2,4 | ||
| + | |||
5 1,5 | 5 1,5 | ||
| + | |||
6 1,2,3,6 | 6 1,2,3,6 | ||
| + | |||
7 1,7 | 7 1,7 | ||
| + | |||
8 1,2,4,8 | 8 1,2,4,8 | ||
| + | |||
9 1,3,9 | 9 1,3,9 | ||
| + | |||
10 1,2,5,10 | 10 1,2,5,10 | ||
| + | |||
11 1,11 | 11 1,11 | ||
| + | |||
12 1,2,3,4,6,12 | 12 1,2,3,4,6,12 | ||
| + | |||
Eureka! | Eureka! | ||
Latest revision as of 04:11, 30 November 2012
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)
