[proofplan] The inclusion $V_+(I)\subseteq V_+(S)$ follows directly from the containment $S\subseteq I$. For the reverse inclusion, we take a projective point that annihilates every generator in $S$ and show that it annihilates every [homogeneous polynomial](/page/Homogeneous%20Polynomial) in the ideal generated by $S$. The key point is that every element of $(S)$ is a finite polynomial combination of elements of $S$, so evaluation at the chosen representative gives zero term by term. [/proofplan] [step:Use the containment of generators to prove $V_+(I)\subseteq V_+(S)$] Let $P \in V_+(I)$. Choose a representative $a=(a_0,\ldots,a_n)\in k^{n+1}\setminus\{0\}$ such that $P=[a_0:\cdots:a_n]$. Since $I=(S)$ is an ideal containing $S$, every $s\in S$ belongs to $I$. By the definition of $V_+(I)$, every homogeneous element of $I$ vanishes at $a$. Each $s\in S$ is homogeneous by hypothesis, so $s(a)=0$ for every $s\in S$. Hence $P\in V_+(S)$, and therefore \begin{align*} V_+(I)\subseteq V_+(S). \end{align*} [/step] [step:Express each element of the generated ideal as a finite sum of multiples of generators] Let $P\in V_+(S)$. Choose a representative $a=(a_0,\ldots,a_n)\in k^{n+1}\setminus\{0\}$ such that $P=[a_0:\cdots:a_n]$. Let $f\in I$ be homogeneous. Since $I=(S)$ is the ideal generated by $S$, there exist an integer $m\geq 0$, polynomials $h_1,\ldots,h_m\in R$, and elements $s_1,\ldots,s_m\in S$ such that \begin{align*} f=\sum_{j=1}^{m} h_j s_j. \end{align*} Here the sum is finite by the definition of an ideal generated by a subset. [guided] We want to prove that $P$ belongs to $V_+(I)$, so we must show that every homogeneous polynomial $f\in I$ vanishes at a representative of $P$. Fix such an $f\in I$. The phrase "$I$ is generated by $S$" means precisely that $f$ can be written as a finite $R$-linear combination of elements of $S$. Thus there are an integer $m\geq 0$, polynomials $h_1,\ldots,h_m\in R$, and generators $s_1,\ldots,s_m\in S$ with \begin{align*} f=\sum_{j=1}^{m} h_j s_j. \end{align*} The finiteness is essential: evaluation distributes over finite sums and products in the [polynomial ring](/page/Polynomial%20Ring). We do not need the coefficients $h_j$ to be homogeneous, because once $s_j(a)=0$, the product $h_j(a)s_j(a)$ is zero regardless of the value of $h_j(a)$. [/guided] [/step] [step:Evaluate the finite expression at a representative of the projective point] Because $P\in V_+(S)$, every $s\in S$ vanishes at $a$. In particular, \begin{align*} s_j(a)=0 \end{align*} for every $j\in\{1,\ldots,m\}$. Evaluation at $a$ is a ring homomorphism $\operatorname{ev}_a:R\to k$, so it respects finite sums and products. Therefore \begin{align*} f(a)=\sum_{j=1}^{m} h_j(a)s_j(a). \end{align*} Since $s_j(a)=0$ for every $j$, each summand is zero, and hence \begin{align*} f(a)=0. \end{align*} [/step] [step:Conclude that the two projective zero loci are equal] We have shown that every homogeneous $f\in I$ vanishes at $a$. Therefore $P\in V_+(I)$. Since $P\in V_+(S)$ was arbitrary, this proves \begin{align*} V_+(S)\subseteq V_+(I). \end{align*} Combining this inclusion with $V_+(I)\subseteq V_+(S)$ gives \begin{align*} V_+(S)=V_+(I). \end{align*} [/step]