Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $p$ is a prime, and $d$ is an integer greater than or equal to 1.

Let $n= p^d$, then there exists $K$ being a field of order $p^n$.

But how many different proper subfields does $K$ have?

The following property might be helpful, though I can't find a proof for it.

Say if $F$ is a subfield of $K$, then $F$ has order $p^d$ for some $d$ such that $d \ | \ n$.

Thanks in advance!

share|cite|improve this question
Well, that property pretty much answers your question, doesn't it? Well, together with one uniqueness property... – DonAntonio May 1 '14 at 20:41
@DonAntonio How will I be able to prove this property? – PandaMan May 1 '14 at 20:44
uniqueness because $d \ | \ n$? – PandaMan May 1 '14 at 20:46
No @PandaMan, uniqueness because we know (or we can prove depending on what you know), that there exists a field with $\;p^n\;$ elements, for any prime $\;p\;\;and\;\;n\in\Bbb N\;$, and it is unique up to isomorphism. – DonAntonio May 1 '14 at 20:48
In addition to what DonAntonio said you need the following bits: A) If $d\mid n$, then the field of order $p^n$ has a subfield of order $p^d$. B) And that subfield consists of zeros of $x^{p^d}-x$, and hence there is only one such subfield. In general it is possible for a field to have isomorphic but distinct subfields, so knowing them up to isomorphism is not enough. In the case of finite fields this never happens because of the above characterization. – Jyrki Lahtonen May 1 '14 at 21:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.