Question: let $F$ be field of order $7^6$ and let $H$ be it's subfield of $F$ containing $49$ elements, then dimension of vector space form by $F$ over $H$ is?

I just know, every field form a vector space over its subfield. But from this we can't determine dimension. I had seen some familiar examples like, $dim(\mathbb{R}^3(\mathbb{R}))= 3$ etc. But here, it can't works, is there is any formula or any method? Please help me..

  • 2
    $\begingroup$ A vector space of dimension $n$ over a field of $q$ elements has size $q^n$. $\endgroup$ – Lord Shark the Unknown Feb 3 '18 at 16:41
  • $\begingroup$ @lord, I am not asking number of elements in vector space, I am asking about dimension $\endgroup$ – Akash Patalwanshi Feb 3 '18 at 16:42
  • 1
    $\begingroup$ @AkashPatalwanshi I'm almost certain Lord was hinting at you what the answer is. Check it carefully: it is a good hint. $\endgroup$ – DonAntonio Feb 3 '18 at 16:44


For any prime $\;p\;$ and natural numbers $\;n,\,m\;$ such that $\;m\,\mid \,n\;$ , we have that

$$\dim\left(\Bbb F_{p^n}/\Bbb F_{p^m}\right)=\frac nm$$

  • $\begingroup$ Sir, this done a job but, is there is any proof of this? $\endgroup$ – Akash Patalwanshi Feb 3 '18 at 16:44
  • $\begingroup$ @AkashPatalwanshi Of course there is. Check and think carefully about Lord's comment below your question! $\endgroup$ – DonAntonio Feb 3 '18 at 16:45
  • $\begingroup$ Sir, according to lord's sir comment, $|V(F)|=|F|^{dimV}$ but from this , how proof follows? $\endgroup$ – Akash Patalwanshi Feb 3 '18 at 16:49
  • $\begingroup$ @AkashPatalwanshi First, it must be clear that $\;\dim_{\Bbb F_p}\Bbb F_{p^n}=n\;$ , and from this $\;\dim_{\Bbb F_7}F=6\,,\,\,\dim_{\Bbb F_7}H=2\;$ ....well, what do you think now? $\endgroup$ – DonAntonio Feb 3 '18 at 19:06
  • $\begingroup$ sir, certainly, according to your hint, $dim(F/H) = 3$. But, still it's not proof I think. $\endgroup$ – Akash Patalwanshi Feb 4 '18 at 5:49

Let $\beta=\{x_1,x_2,x_3,...x_n\}$ be the basis we need to find $n$. Possible linear combinations $=7^2\times7^2...\times7^2=7^{2n}=7^6\implies n=3$


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.