Zero Locus of Set is Zero Locus of Generated Ideal

Theorem
Let $k$ be a field.

Let $n\geq1$ be a natural number.

Let $A = k \left[{X_1, \ldots, X_n}\right]$ be the ring of polynomial functions in $n$ variables over $k$.

Let $T \subseteq A$ be a subset, and $V \left({T}\right)$ the zero locus of $T$.

Let $J = \left({T}\right)$ be the ideal generated by $T$.

Then:
 * $V \left({T}\right) = V \left({J}\right)$

Proof
Let $x \in V \left({T}\right)$, so $f \left({x}\right) = 0$ 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 $x \in V \left({J}\right)$.

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

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

So $x \in V \left({T}\right)$.