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GF(p^n) are the only fields we need to worry about, so we just need to find the smallest field of this form that has exactly 6 subfields. Since GF(p^n) has exactly one subfield per divisor of n, we are looking at the smallest field GF(p^n) such that n has exactly 6 divisors. Just enumerate the divisors of n's and you should find the smallest n that has 6 divisors. See Theorem 22.3. -Josh | GF(p^n) are the only fields we need to worry about, so we just need to find the smallest field of this form that has exactly 6 subfields. Since GF(p^n) has exactly one subfield per divisor of n, we are looking at the smallest field GF(p^n) such that n has exactly 6 divisors. Just enumerate the divisors of n's and you should find the smallest n that has 6 divisors. See Theorem 22.3. -Josh | ||
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| + | So what did you get as an answer? if 1 and 12 count as divisors then I got n=12. | ||
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| + | Do 1 and 12 count? | ||
Latest revision as of 05:45, 11 December 2008
Do any of you have any helpful hints? That would be lovely.
GF(p^n) are the only fields we need to worry about, so we just need to find the smallest field of this form that has exactly 6 subfields. Since GF(p^n) has exactly one subfield per divisor of n, we are looking at the smallest field GF(p^n) such that n has exactly 6 divisors. Just enumerate the divisors of n's and you should find the smallest n that has 6 divisors. See Theorem 22.3. -Josh
So what did you get as an answer? if 1 and 12 count as divisors then I got n=12.
Do 1 and 12 count?
