Zero Locus of Set is Zero Locus of Generated Ideal

Theorem
Let $k$ be a field.

Let $A = k[X_1,\ldots,X_n]$ be the Ring of Polynomial Functions in $n$ variables over $k$.

Let $T \subseteq A$ be a set, and $V(T)$ the zero locus of $T$.

Let $J = (T)$ be the ideal generated by $T$.

Then $V(T) = V(J)$.

Proof
Let $x \in V(T)$, so $f(x) = 0$ for all $f \in T$.

By definition, $J$ is the set of linear combinations of elements of $T$ over $k$.

So any $g \in J$ is of the form


 * $g = k_1 t_1 + \cdots + k_r t_r$

with $k_i \in k$ and $t_i \in T$.

Therefore

Therefore $x \in V(J)$.

Conversely, if $x \in V(J)$, then $f(x) = 0$ for all $f \in J$.

But $T \subseteq J$, so in particular $f(x) = 0$ for all $f \in T$.

So $x \in V(T)$.