Is every element of $E$ algebraic over $F(S)$?

Question

We have the set $E/F$.

Suppose that $S\subseteq T\subseteq E$ and that every element of $E$ is algebraic over $F(T)$ and that $S$ is algebraically independent over $F$.

Does it follow that every element of $E$ is algebraic also over $F(S)$ ?

Answer 1

Not necessarily.

Let $X$ and $Y$ be indeterminates, and let $E=F(X,Y)$. Furthermore, let $S=\{X\}$ and $T=\{X, Y\}$.

Then $E=F(X,Y)$ is an algebraic extension of $F(T) = F(X,Y) = E$, but it is not algebraic over $F(S) = F(X)$. Moreover, $S$ is algebraically independent over $F$.