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S R Wicker in his book states ... $GF(p^m)$ has all sub-fields of order $p^b$ provided $b$ divides $m$.

Qn: In order to have a sub-filed of order $p^b$ shouldn't we have also the constraint: $p^b - 1$ divides $p^m -1$ ? (that is to say, the multiplicative order of a sub-field should divide the multiplicative order of the entire GF)

Jyrki Lahtonen
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user290776
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    That's right. But if $b\mid m$ then automatically also $p^b-1\mid p^m-1$. Do you know why the polynomial $x^b-1$ is a factor of $x^m-1$. Or even more obviously, that $x-1$ is a factor of $x^{m/b}-1$? – Jyrki Lahtonen Nov 16 '15 at 22:45
  • Note that in some sense this is in fact why one needs to restrict to divisors. It would be true for any $b \le m $ that $p^b \mid p^m$. – quid Nov 16 '15 at 23:01
  • Re: "That's right. But if b∣mb∣m then automatically also (p^b − 1)∣(p^m − 1)" – user290776 Nov 16 '15 at 23:10
  • Dear Jyrki, the case (x-1) divides x^(m/b)-1 is easy (x^n-1/x-1) is the sum of a finite geometric series. I suppose the more general case (x^(nm)-1)/(x^m -1) implies a ratio q = x^m ? – user290776 Nov 16 '15 at 23:30

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