Maybe somebody knows how to proove the following algebraic theorem:
$C \subset U$ is a field extension and $N \subset U$ so, that all $x \in N$ are algebraic over $C$ and $C[N]=\left\lbrace f(x_1,\dotsc,x_n); n \in \mathbb{N}, f \in C[X_1,\dotsc,X_n], x_1,\dotsc,x_n \in N \right\rbrace$. Then:
(A) $C[N]$ is a field extension of $C$.
(B) All $x \in C[N]$ are algebraic over $C$.
First prove (B): every element of $C[N]$ is a polynomial expression $f(x_1,\dots,x_n)$ in elements $x_1, \dots, x_n$ that are algebraic over $C$ and hence is itself algebraic over $C$.
As already noted in the comments, for (A), it is clear that $C[N]$ is closed under multiplication, addition, subtraction; what is left is to show that it is closed under taking inverses.
Start with $0 \neq a \in C[N]$ and look at the subring $C[a]$ of $C[N]$. By (B), $a$ is algebraic over $C$, which makes $C[a]$ into an algebraic field extension of $C$. So $a^{-1} \in C[a] \subseteq C[N]$.
(Note that the fact that $C[x]$ is an algebraic field extension doesn't directly - or with induction - prove both (A) and (B), as $N$ could be infinite; the point of this exercise seems to be that you have to restrict attention to a finite subset of $N$.)
If you don't (yet) know that $C[a]$ is actually a field, I'll show why $a^{-1} \in C[a]$, which is enough here.
Say $g(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_0 \in C[x]$ is the minimum polynomial of $a$ over $C$. Then $$a(c_n a^{n-1} + c_{n-1} a^{n-1} + \dots + c_1) = -c_0.$$ Note that $c_0 \neq 0$ because $g$ is the minimum polynomial and hence $$a^{-1} = -c_0^{-1}(c_n a^{n-1} + c_{n-1} a^{n-1} + \dots + c_1) \in C[a].$$