Dimension of vector space generated by finite field over its subfield

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..

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

Hint:

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

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

• Sir, this done a job but, is there is any proof of this? – Akash Patalwanshi Feb 3 '18 at 16:44
• @AkashPatalwanshi Of course there is. Check and think carefully about Lord's comment below your question! – DonAntonio Feb 3 '18 at 16:45
• Sir, according to lord's sir comment, $|V(F)|=|F|^{dimV}$ but from this , how proof follows? – Akash Patalwanshi Feb 3 '18 at 16:49
• @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? – DonAntonio Feb 3 '18 at 19:06
• sir, certainly, according to your hint, $dim(F/H) = 3$. But, still it's not proof I think. – 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 $=7^2\times7^2...\times7^2=7^{2n}=7^6\implies n=3$