Union of two Zero Loci is Zero Locus

Theorem

Let $k$ be a field.

Let $n \in \N_{>0}$.

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

Let $S,T \subseteq A$.

Then:

$\ds {\map V S} \cup \map V T = \map V {S \cdot T}$

where:

$\map V \cdot$ denotes the zero locus

$S \cdot T := \set { fg : f \in S, g \in T }$

Let $x \in {\map V S} \cup \map V T$.

That is, either $x \in \map V S$ or $x \in \map V T$.

Then, if $x \in \map V S$:

$\forall \paren {f,g} \in S \times T : \map f x \map g x = 0 \map g x = 0$

else:

$\forall \paren {f,g} \in S \times T : \map f x \map g x = \map f x 0 = 0$

Therefore:

$x \in \map V {S \cdot T}$

On the other hand:

$\ds x$

$\not \in$

$\ds {\map V S} \cup \map V T$

$\ds \leadsto \ \ $

$\ds x$

$\in$

$\ds \paren {k^n \setminus \map V S} \cap \paren {k^n \setminus \map V T}$

$\ds \leadsto \ \ $

$\ds \exists f \in S : \map f x \ne 0$

$\land$

$\ds \exists g \in T : \map g x \ne 0$

$\ds \leadsto \ \ $

$\ds \exists \paren {f,g} \in S \times T: \, $

$\ds \map f x \map g x$

$\ne$

$\ds 0$

$\ds \leadsto \ \ $

$\ds x$

$\not \in$

$\ds \map V {S \cdot T}$

$\blacksquare$