Let $E/F$ is a field extension, and $\alpha_1,\ldots, \alpha_n\in E$. I was thinking that $F[\alpha_1,\ldots,\alpha_n] = F(\alpha_1,\ldots, \alpha_n)$ if and only if $\alpha_i$ are all algebraic over $F$ (Maybe this statement is false).
The converse direction is trivial. For the forward direction, we can prove that $F[\alpha_1]\subsetneq F(\alpha_1)$ if $\alpha_1$ is not algebraic over $F$. This is because $\alpha_1$ is not invertible in $F[\alpha_1]$ or else $\alpha_1$ would be algebraic over $F$. Hence $\alpha_1^{-1}\in F(\alpha_1)\notin F[\alpha_1]$.
However, I have trouble generalizing this statement for $n>1$. I was thinking if we can assume without loss of generality that $\alpha_1$ is not algebraic over $F$. Then we have $F[\alpha_1]$ is an integral domain which is not a field. I feel that this would imply $F[\alpha_1][\alpha_2,\ldots, \alpha_n]$ is not a field either and this would finish the proof. However, I don't know how to prove this statement.
So my question is: if $D$ is an integral domain which is not a field. If $D\subset E$ for some field $E$ and $\alpha\in E$, is it possible that $D[\alpha]$ is a field.
I am also interested in proofs or counterexamples to my initial question.