Let $k\subset F\subseteq k(X)$ be chains of field extension, prove that $k(X)/F$ is algebraic.
"Proof:" Let $y\in F\setminus k$ then $y=\frac{P(X)}{Q(X)}$ with $P\notin k$ or $Q\notin k$.
It sufficies to show that $k(X)/k(y)$ is algebraic, because this implies that $k(X)/F$ is algebraic.
Let $r(T)=P(T)-\frac{P(X)}{Q(X)}Q(T)\in k(y)[T]$, then $r(X)=0$, thus $k(X)$ is algebraic over $k(y)$.
Is correct?.
Thank you all.