TITLE: Why are finite unions of algebraic sets algebraic?
QUESTION [8 upvotes]: Suppose $F,G$ are polynomials in $k[x_1,\cdots, x_n]$ (k is a field). Let $$V(F) = \{ (a_1,\cdots,a_n)\in k^n : F(a_1,\cdots,a_n)=0 \}.$$
Then $V(F)\cup V(G) = V(FG),$ essentially because integral domains don't have zero divisors. What I'm having trouble seeing is 
$$ V(I)\cup V(J) =\{ (a_1,\cdots, a_n)\in k^n: F(a_1,\cdots,a_n)G(a_1,\cdots,a_n)=0 \ \ \ \forall F\in I, \ \ G\in J\}.$$ The same argument as in the simpler case doesn't seem to apply since if $x$ is a point in the RHS, then it could make $FG$ always vanish by sometimes making $F$ vanish but not $G$ for some $F,G$, and other times making $G$ vanish but not $F$ for some other $F,G.$ 
Can someone help me see why the equation is true?
EDIT: More specifically, I can see why the LHS is a subset of the RHS, but not the reverse.

REPLY [11 votes]: Suppose that $x\in V(I)\cup V(J)$. Then it is either in $V(I)$ or $V(J)$; let's say it is in $V(I)$. Thus, $F(x)=0$ for all $F\in I$, and hence, for all $F\in I$ and $G\in J$, we have that $F(x)G(x)=0$. The same reasoning applies if we assumed $x$ was in $V(J)$. Thus,
$$V(I)\cup V(J)\subseteq\{x\in k^n\mid F(x)G(x)=0\text{ for all }F\in I,G\in J\}.$$
Conversely, let's suppose that the point $x$ satisfies $F(x)G(x)=0$ for all $F\in I$ and $G\in J$. If it is the case that $F(x)=0$ for all $F\in I$, then by definition, we have $x\in V(I)$, and hence $x\in V(I)\cup V(J)$. So, now let's suppose that there were some $F_0\in I$ such that $F_0(x)\neq 0$. Then for any $G\in J$, since we assumed that $F_0(x)G(x)=0$ and we are working over a field, we must have $G(x)=0$. Thus, $x\in V(J)$, and hence $x\in V(I)\cup V(J)$.