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!!!
Dimension of the field
linear algebra
Related Solutions
Let $\beta=\{x_1,x_2,x_3,...x_n\}$ be the basis we need to find $n$. Possible linear combinations $=7^2\times7^2...\times7^2=7^{2n}=7^6\implies n=3$
If the action is only on he basis elements then how would we find the operations like $σ(e_1+e_2)$?
This seems to be the key to your question.
You are right: at the outset, the action is explicitly only defined on the set $\{e_1, e_2, e_3\}$. Given a group element $\sigma$, and one of these basis elements $b$, we know what $\sigma\cdot b$ is, and that is all.
But every mapping defined on the basis of a vector space gives rise to a unique linear transformation extending that mapping to the entire vector space spanned by the elements. Furthermore, if the mapping is a bijection and the vector space is finite dimensional, the extension is an isomorphism.
So what is happening is that by definition the action on an arbitrary element of the vector space is defined to be the linear extension: $\sigma\cdot (\sum\lambda_ie_i):=\sum\lambda_i(\sigma\cdot e_i)$.
You can check all the axioms to confirm that indeed, the originally defined group action on the basis has now been promoted to a group action of $S_3$ on the entire vector space spanned by the basis.
This "lifting" of the group action is what you were missing, I think.
There's another way to see how this happens. You can rephrase a group action of $G$ on a set $X$ to be a homomorphism from $G\to Sym(X)$.
If $X$ is the finite basis of an $n$ dimensional $F$ vector space, we have another group homomorphism from $Sym(X)\to GL(n,F)$. Composing these two homomorphisms we get a group homomorphism from $G\to GL(n,F)$, and that defines an action of $G$ on $span(X)$.
It says, essentially, that every element of $G$ acts like a linear automorphism on $span(X)$ (very nice elements of $Sym(span(X))$!)
Best Answer
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 $K$.