Dimension of the field let $F$ be a field with $7^6$ elements and let $K$ be a subfield of $F$ with $49$ elements then the dimension of $F$ as a vector space over $K$ is...I don't know how to proceed anyone please help me!!!
 A: Very simply, we have this formula, considering the prime subfield of $F$:
$$\underbrace{[F:\mathbf F_7]}_{=\,6}=[F:L]\cdot\underbrace{[L:\mathbf F_7]}_{=\,2}$$
A: Let's say the dimension of $F$ as a vector space over $K$ is $n$. This means there is a basis of $F$ with a cardinality of $n$ elements. We will call this basis $f_1, f_2, ..., f_n$. By the definition of a basis, every element of $F$ can be expressed in the form of $k_1f_1+k_2f_2+...+k_nf_n$ for some $k_1,k_2,...,k_n \in K$.
Now, since $K$ has $49$ elements, there are $49$ possible values of $k_i$ for any $1\leq i \leq n$. Thus, there are $49$ choices for each $k_i$ and $n$ different $k_i$ variables, meaning there are $49^n$ possible values of $k_1f_1+k_2f_2+...+k_nf_n$. Thus, there are $49^n$ elements in the field $F$.
Now, there are $7^6$ elements in the field $F$. Thus, we have the following equation:
$$49^n=7^6\rightarrow (7^2)^n=7^6\rightarrow 7^{2n}=7^6\rightarrow 2n=6\rightarrow n=3$$
Therefore, $F$ is a vector space of dimension $3$ over $K$.
