I have a doubt about the proof of the following result:
Let $E/K$ be a finite field extension and $\{e_1,\ldots,e_n\}$ be a finite subset of $E$. Denote $K[e_1,\ldots,e_n]$ the smallest subring of $E$ that contains $K$ and $\{e_1,\ldots,e_n\}$ and $K(e_1,\ldots,e_n)$ the smallest subfield of $E$ that contains $K$ and $\{e_1,\ldots,e_n\}$. Then $$K[e_1,\ldots,e_n]=K(e_1,\ldots,e_n).$$
The proof goes something like this:
It suffices to show that $K[e_1,\ldots,e_n]$ contains every inverse of its elements. Let $a\in K[e_1,\ldots,e_n]$. We prove that the function $T_a:K[e_1,\ldots,e_n]\longrightarrow K[e_1,\ldots,e_n]$ given by $T_a(b)=ab$ is surjective. Since $K[e_1,\ldots,e_n]$ can be thought as a vector space of finite dimension and $T_a$ is an injective endomorphism, it follows that $T_a$ is also surjective.
My question is the following:
Over which field can $K[e_1,\ldots,e_n]$ be thought as a vector space?
I think $K$ is not the answer, since $T_a$ is not a $K$-linear mapping ($a$ may not be in $K$). Also, $K[e_1,\ldots,e_n]$ is not an $E$-vector space, since an element $K[e_1,\ldots,e_n]$ multiplied by an element in $E$ may not stay in $K[e_1,\ldots,e_n]$.
Any help would be appreciated!