I would like to show that $V(S)=V(<S>)$. I know that $S$ is a collection of polynomials from $\mathbb{P}^n$. $V(S)=\{x\in \mathbb{R}^n : p(x)=0, \forall p\in S\}$
$\langle S\rangle=\{f_1g_1h_1+\dots+f_kg_kh_k: k\in \mathbb{N}, f_i,h_i\in \mathbb{P}^n, g\in S\}$
I have tried to take a $V(S)\subseteq V(\langle S\rangle)$ and $V(\langle S\rangle)\subseteq V(S)$ approach, but I am getting stuck. I found a proof of the proposition Here: Proof of the proposition $V(S)=V(\langle S \rangle )$
But, I am not following it well especially since I have defined things slightly different. Any help on this would be greatly appreciated; thank you!
It is clear that, if $S\subseteq T$, then $V(T)\subseteq V(S)$. Therefore $$ V(\langle S\rangle)\subseteq V(S) $$ Suppose now $x\in V(S)$ and let $p\in\langle S\rangle$. By definition of $\langle S\rangle$, $$ p=f_1g_1h_1+\dots+f_kg_kh_k $$ where $f_1,\dots,f_k,h_1,\dots,h_k$ are suitable polynomials and $g_1,\dots,g_k\in S$. Then $$ p(x)=f_1(x)g_1(x)h_1(x)+f_2(x)g_2(x)h_2(x)+\dots+f_k(x)g_k(x)h_k(x)=0 $$ and so $p\in V(\langle S\rangle)$.
The definition of $\langle S\rangle$ can actually be simplified, because the set you describe is the same as $$ \{f_1g_1+\dots+f_kg_k: k\in \mathbb{N}, f_i\in \mathbb{P}^n, g\in S\} $$ as multiplication in $\mathbb{P}^n$ is commutative.