If a set $S$ generates an ideal $I\subset F[x_1,x_2,\ldots,x_n]$, then there is a finite subset $S_0 \subseteq S$ which generates $I$ The question:

If $I$ is an ideal in $F[x_1,x_2,\ldots,x_n]$ generated by a set of polynomials $S$, then there is a finite subset $S_0 \subseteq S$ which generates $I$.

By the Hilbert Basis Theorem, I know that any ideal in $F[x_1,x_2,\ldots,x_n]$ is finitely generated.
So, I can suppose $I$ has a finite basis $i_1, i_2 ,\ldots, i_n$.
I also know that $I = \langle S\rangle$.
This seems almost self evident. There has to be something obvious I am missing. Can someone offer a hint or some assistance?
 A: It's not quite "self-evident" from what you wrote, because the finite basis of $i_j$'s you found are not necessarily a subset of $S$. (Also, you should be careful with your notation; it's not true that an ideal of $F[x_1,\ldots,x_n]$ necessarily can be generated by $n$ elements – take a look at this math.SE thread – but you wrote your finite basis as $i_1,\ldots,i_n$.)
However, how can we convert from $S$ to the $i_j$'s?
We know that $S$ generates $I$, in the sense that for any $f\in I$, there are some $s_1,\ldots,s_t\in S$ and some $g_1,\ldots,g_t\in F[x_1,\ldots,x_n]$ such that $f=g_1s_1+\cdots +g_ts_t$.
Thus, finding elements of $S$ and elements of $F[x_1,\ldots,x_n]$ with
$$\begin{align*}
i_1 & = g_{11}s_{11}+\cdots+g_{1t_1}s_{1t_1}\\
&\;\;\vdots\\
i_r & = g_{r1}s_{r1}+\cdots+g_{rt_r}s_{rt_r}\\
\end{align*}$$
you know that the set of elements
$$S_0=\{s_{11},s_{12},\ldots,s_{1t_1},s_{21},s_{22},\ldots,s_{2t_2},\ldots,s_{rt_r}\}$$
are, together, enough to generate all of the $i_j$'s, and hence all of $I$.
