1
$\begingroup$

I was trying to prove the following theorem:

A field $K$ of order $q=p^r$ contains a subfield $K'$ of order $q'=p^k$ if and only if $k\mid r$.

My attempt at the proof was:

$K'\setminus\{0\}$ is a subgroup of $K\setminus\{0\}$. So, $p^k-1\mid p^r-1$.

From the exponent gcd lemma [$\gcd(a^x-1,a^y-1)=a^{\gcd(x,y)}-1$ when $a,x,y\in\mathbb N,a\geq 2$.]

We have that $k\mid r$.

Is this proof correct? In fact, I'm worried about this particular statement:

$K'\setminus\{0\}$ is a subgroup of $K\setminus\{0\}$.

$\endgroup$
  • $\begingroup$ I don't understand the vote to close. This seems like a perfectly reasonable problem. $\endgroup$ – Robert Shore May 15 at 8:33
  • 2
    $\begingroup$ The proof seems correct to me as far as it goes. If you're worried about the initial step, it's easy enough for you to prove it directly. (I assume you mean that they're subgroups under multiplication.) But you also need to prove the other direction. $\endgroup$ – Robert Shore May 15 at 8:37
  • $\begingroup$ Yes, I meant they're subgroups under multiplication. $\endgroup$ – Thanic Nur Samin May 15 at 9:03
  • $\begingroup$ As for the other direction: I was sure about my proof of that part, which is why I didn't include it. Thanks a lot! $\endgroup$ – Thanic Nur Samin May 15 at 9:06
  • 1
    $\begingroup$ Then you still need to prove that if $k \vert r$ then an appropriate subfield exists. $\endgroup$ – Robert Shore May 15 at 9:06

Your Answer

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

Browse other questions tagged or ask your own question.