Zero Locus of Set is Zero Locus of Generated Ideal

Theorem

Let $k$ be a field.

Let $n \ge 1$ be a natural number.

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

Let $T \subseteq A$ be a subset, and $\map V T$ the zero locus of $T$.

Let $J = \ideal T$ be the ideal generated by $T$.

Then:

$\map V T = \map V J$

Let $x \in \map V T$, so $\map f x = 0$ 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:

$\ds \map g x$

$=$

$\ds k_1 \map {t_1} x + \cdots + k_r \map {t_r} x$

$\ds $

$=$

$\ds 0$

because $\map f x = 0$ for all $f \in T$

Therefore:

$x \in \map V J$

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

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

So:

$x \in \map V t$

$\blacksquare$