$\newcommand{\scrO}{\mathscr{O}}$ $\newcommand{\A}{\mathbb{A}}$ Definitions: I denote $n$-dimensional affine space over a field $k$ by $\A^n$. Let $X$ be an algebraic subset of $\A^n$, a map $f:X \rightarrow k$ is regular if it is the restriction of some polynomial map, i.e if there exists some $g \in k[x_1,\dots, x_n]$ for which $g(x) = f(x), \; \forall x \in X$. We denote the collection of all regular maps defined on $X$ by $\scrO(X)$, and in particular $\scrO(\A^n) = k[x_1,\dots,x_n]$. We denote the vanishing set of a subset $S \subseteq \scrO(\A^n)$ by $V(S)$.
I'm trying to prove work through some basic algebraic geometry from Dummit and Foote. Specifically section 15.1. I know the question is basic, but I just want to make sure I'm understanding the essentials.
Problem: Let $S \subseteq \scrO(\A^n)$ and set $I=(S)$, i.e the ideal of $\scrO(\A^n)$ generated by $S$. Then $V(S) = V(I)$.
Attempt: Because $k$ is a field,i.e commutative, the ideal $(S)$ is essentially a span, specifically $$ I = \{a_1f_1+\dots +a_nf_n \mid n \geq 1, a_i \in \scrO(\A^n), f_i \in S\}. $$ Let $\alpha \in V(S)$. Then for any element $x = \sum_1^n a_if_i \in I$ it follows that $x(\alpha) = \sum_1^n a_i(\alpha)f_i(\alpha) = \sum_1^n a_i\cdot 0 = 0$ because every $f_i \in S$, so it follows that $V(S) \subseteq V(I)$. Conversely assume that $\beta \in V(I)$. This means that for any finite linear combination $x = \sum_1^n a_if_i$ we have that $x(\beta) = \sum_1^n a_if_i(\beta) = 0$. In particular, for every $f \in S$ we may write it trivially as $\sum_1^1 1 \cdot f$ and so $f(\beta) = 0$ for all $f \in S$. Therefore $\beta \in V(S)$ and so $V(S) = V(I)$.
Let me know if I need to clarify anything.
This is exactly right. You might simplify things by observing that the "vanishing set operator" trivially $V$ reverses inclusions, so whenever you have $S_1 \subset S_2$ then $V(S_1) \supset V(S_2)$. The not-so-obvious part is what you prove first, that if $S$ generates $I$, then $V(S) \subset V(I)$.