Let $L/K$ be a field extension and $\Gamma \subset L$ such that for all $\alpha \in \Gamma$, $\alpha$ is algebraic over K. Show that $K(\Gamma)/K$ is algebraic.
My attempt:
Since every element of $\Gamma$ is algebraic over $L$, we have that $K(\Gamma)=K[\Gamma]$.
Now, given $a \in K(\Gamma)$ we need to find a polinomyal $f \in K[x]- \lbrace 0 \rbrace $, such that $f(a)=0$.
Since $K(\Gamma)=K[\Gamma]$, $a$ is a polynomial, let's say $h$, which is evaluated on the elements of $\Gamma$.
From here I don't know how to build the polynomial $ f $. Any hint?
Proof.
Let $E=K(\Gamma)$ then $\Gamma \subset E$. Let $F= \lbrace \alpha \in E : \alpha$ is algebraic over $K \rbrace$.
We have that $F$ is a fiel such that $K\subset F \subset E$. Therefore, $E=K(\Gamma) \subset K(F)=KF=F$, hence $F=E$, i.e. $K(\Gamma)/K$ is algebraic.