Hey!!

We have field extension $E/F$. We have that $S\subseteq E$ is algebraically independent over $F$. Let $t\in E$, then $S\cup \{t\}$ is algebraically independent over $F$.

I want to show that $t$ is algebraic over $F(S)$.

Since $S\subseteq E$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n\in S$ (different from each other) and non-zero $f\in F[x_1, \ldots , x_n]$ such that $f(s_1, \ldots , s_n)\neq 0$.

Since $S\cup \{t\}$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n, t\in S\cup \{t\}$ (different from each other) and non-zero $g\in F[x_1, \ldots , x_n, y]$ such that $g(s_1, \ldots , s_n, t)=0$.

To show that $t$ is algebraic over $F(S)$, we have to show that there exist $s_1, \ldots , s_n\in S$ and non-zero $h\in F[x_1, \ldots , x_n]$ such that $t=h(s_1, \ldots , s_n)$.

But how exactly can we show the existence of such a polynomial?