Let $F$ be a field and suppose $I$ is an ideal of $F[x_1, \ldots , x_n]$ generated by a (possibly infinite) set $S$ of polynomials. Prove that a finite subset of polynomials in $S$ is sufficient to generate $I$.
So far I have this as my proof, but I'm not entirely sure if what I did is correct:
Since $F$ is a field, then it is noetherian and by Hilbert's Basis Theorem $F[x_1, \ldots, x_n]$ is also noetherian. Then $I$ is finitely generated. Let $I = (f_1, \ldots, f_m)$ where $f_i \in (s_1, s_2, \ldots , s_i)$ with $s_i \in S$, ie. $f_1 \in (s_1), f_2 \in (s_1, s_2), etc.$ Then we have that:
$(s_1) \subset (s_1, s_2) \subset (s_1, s_2, s_3) \subset \cdot \cdot \cdot \subset (s_1, \ldots , s_m) \subset (s_1, \ldots, s_{m+1}) \subset \cdot \cdot \cdot \subset I$.
But since $(f_1, \ldots, f_m)$ generates $I$, then the chain must stop:
$(s_1) \subset (s_1, s_2) \subset (s_1, s_2, s_3) \subset \cdot \cdot \cdot \subset (s_1, \ldots , s_m) = (s_1, \ldots, s_{m+1}) = \cdot \cdot \cdot = I$
So the finite set of elements in $S$, $\{s_1, s_2, \ldots, s_m\}$, is sufficient to generate $I$.
By Hilbert's Basis Theorem, as you said, $I$ has a finite number of generators $f_1, \dots, f_m$. Now each generator $f_i$ can be expressed with a finite subset $S_i\subset S \enspace (1\le i \le m$, so that the whole set of generators $f_1, \dots, f_m$ is expressible with the finite subset of $S$: $S'=S_1\cup\dots\cup S_m$. Hence $S'$ is a finite subset of generators of $I$ contained in $S$.