A basis of a field extension contained in a subring

Let $K\subseteq L$ be a finite extension of fields. Let $A\subseteq L$ be a subring such that the fraction field $\operatorname{Frac}(A)$ is equal to $L$.

Question. Is there a basis $\alpha_1,\dotsc,\alpha_n$ contained in $A$ of the $K$-vector space $L$?

My attempt. Since $K\subseteq L$ is finite, we have a basis $\alpha_1,\dotsc,\alpha_n\in L$ over $K$. I tried to multiply the $\alpha_i$ by some $x_i\in K$ and to get $x_i\alpha_i\in A$, but I could not proceed.

Thanks in advance.

Solution 1:

In fact, there is no need for the extension to be finite; we only need it to be algebraic. So suppose $L$ is algebraic over $K$. I claim first that the $K$-span of $A$ in $L$ is equal to $L$; to see this, fix an arbitrary $c\in L$. Since $\operatorname{Frac}A=L$, there exist $a,b\in A$ with $c=a/b$. Furthermore, since $L$ is algebraic over $K$, we have $b^{-1}=\lambda_0+\lambda_1b+\dots+\lambda_nb^n$ for some $\lambda_i\in K$. Hence $$c=\lambda_0a+\lambda_1ab+\dots+\lambda_nab^n,$$ which lies in the $K$-span of $A$, as desired.

So $A$ is a spanning set for the $K$-vector space $L$, and hence by Zorn's lemma there is a $K$-basis for $L$ contained within $A$, as desired.