If $F$ is a finite field and $K$ is a subfield, then $F$ is isomorphic to $V_m(K) = \{(\lambda_1,…,\lambda_m)|\lambda_i \in K, i=1,2,…,m\}$

Question

Suppose we have the finite field $F$ with characteristic $p$. Suppose $K$ is a subfield in $F$. Then $(\forall \lambda \in K, \forall \alpha \in F)\,\, \lambda \alpha \in F$. Also $(\forall \lambda_1, \lambda_2 \in K)\,\, \lambda_1 + \lambda_2 \in F$ where the addition is the operation of addition of $F$. Then the book says that $F$ is a linear space over $K$. I understand it this way: as $K$ is closed for any operation in $F$ then $F$ can be considered a linear space and $K$ is a subfield of $F$, am I getting it right? Then as $F$ is finite then it is finite dimensional and has a finite basis. Suppose $F$ is $m$ dimensional. Then the book states that $F$ is isomorphic to the $V_m(K) = \{(\lambda_1,…,\lambda_m)|\lambda_i \in K, i=1,2,…,m\}$ $m$-dimensional vector space and the number of elements in $F$ is equal to the number of elements in $V_m(K)$. But I don't get the reason. Why are they isomorphic? I suppose it means that $V_m(K)$ contains the basis of $F$, but why? Also why is $V_m(K)$ m dimensional? Why isn't it $m-1$ dimensional? If we have a linear space and a subspace of it, the dimension of the subspace can be smaller. I think the reason is in the fact that we are considering a finite linear space, but I can't see precisely why. Could you tell me what I am missing? Thanks in advance.

Answer 1

An important result of linear algebra tells us that two $K$-vector spaces are isomorphic if and only if they have the same dimension.

As $m = \dim_K(L)$ by definition and $m = \dim_K(V_m(K))$ (for example by considering the standard basis), you get $L \cong V_m(K)$ as $K$-vector spaces.