Let $E$ and $F$ be subfields of $GF(p^n)$. If $|E| = p^r$ and $|F| = p^s$, what is the order of $E\cap F$?

I read a corollary that "A finite field of order $p^n$ contains a unique subfield of order p^m for each $m$ | $n$ and no other subfields.

If that's the case, wouldn't that mean if two subfields have different orders, their intersection is $0$? So in this case, the order of $E \cap F$ = $0$?

  • 2
    $\begingroup$ The intersection can't be zero since it contains the prime subfield $GF(p)$! And anyway, $1 \in E, F$! $\endgroup$ – Robert Lewis Apr 6 '14 at 6:04
  • $\begingroup$ Very good point, thank you for the insight. $\endgroup$ – James Apr 6 '14 at 6:33
  • 1
    $\begingroup$ If you think about that corollary, you should be able to answer your question. $\endgroup$ – Gerry Myerson Apr 6 '14 at 6:35
  • $\begingroup$ Any source of this corollary in any text or notes will be helpful!. $\endgroup$ – BAYMAX Dec 23 '17 at 3:25

Extended hint:

So if $t\mid r$, and $t\mid s$, then, by the result you quoted, both $E$ and $F$ contain a unique copy of $GF(p^t)$. But also $GF(p^n)$ contains a unique copy of $GF(p^t)$, which means that the copies of $GF(p^t)$ inside $E$ (resp. $F$) must coincide with the one in $GF(p^n)$, and thus be contained in the intersection $E\cap F$.

In view of this the intersection $E\cap F$ is a copy of $GF(p^\ell)$, where $\ell=\gcd(r,s)$.

  • $\begingroup$ So if $E$ and $F$ contain a unique copy of $GF(p^t)$ and $GF(p^n)$ also contains a unique copy of $GF(p^t)$ then $|E \cap F|$ $\geq$ $p^t$, and the intersection also contains the prime subfield $GF(p)$ as well. So... $|E \cap F| = p^t + p$ in this case? $\endgroup$ – James Apr 6 '14 at 6:43
  • $\begingroup$ @James: The prime field is contained in all the copies of $GF(p^t)$. You should also realize that the intersection has to be a subfield. The question is: which subfield? Or what is the biggest common subfield? $\endgroup$ – Jyrki Lahtonen Apr 6 '14 at 6:51
  • $\begingroup$ so with that. Using your example $GF(p^t)$, wouldn't the biggest subfield ($E \cap F$) just be the biggest $t$ that divides $r$, $s$ and $n$? $\endgroup$ – James Apr 6 '14 at 7:08
  • $\begingroup$ Correct, @James! $\endgroup$ – Jyrki Lahtonen Apr 6 '14 at 7:41
  • $\begingroup$ Thank you for the hints, helped me out. :) $\endgroup$ – James Apr 6 '14 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.