What is the union of two varieties I am confused about the union of two varieties. I get two options:
$$
\mathbb{V}(I)\cup\mathbb{V}(J)=\mathbb{V}(IJ)
$$
and 
$$
\mathbb{V}(I)\cup\mathbb{V}(J)=\mathbb{V}(I\cap J)
$$
I can prove the first one is right. But I also find the second on in the note. I don't think we always have $IJ=I\cap J$. So could anyone confirm which is correct? And provide an example to illustrate the incorrectness of the other?
 A: If you're talking about the union of two subvarieties of some ambient variety $V$, purely in a geometric sense, then you'd like $V(I)\cup V(J)$ to be $V(K)$, where $K$ is some ideal containing all the functions which vanish on both $V(I)$ and $V(J)$, hence you'd want $K = I\cap J$.
It's educational to consider the example given by rghthndsd. If $I = J$, then do you want the union $V(I)\cup V(I) = V(I\cap I) = V(I)$ or $V(I^2)$?
Geometrically they're the same space, but scheme theoretically $V(I^2)$ has "more functions" on it.
Usually $IJ\ne I\cap J$. When this is true is discussed at great length in https://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection
A simple situation where $IJ = I\cap J$ is when $I,J$ are comaximal, that is, when $I+J$ is the entire ring.
A: You always have $IJ \subset I \cap J$ by definition of $IJ$. You do not need $IJ = I \cap J$ for $V(IJ) = V(I \cap J)$ though.
To see that $IJ \neq I \cap J$ for some $I, J$ take $I = J = <x>$.  Then $I \cap J = <x>$ but $IJ = <x^2>$.
Some more stuff to read: The first equation is shown in Theorem 7 and the second in Theorem 15 in Chapter 4.3 (page 185 and 190) of the classic book "Ideals, Varieties and Algorithms".
A: I'm ressurecting this thread because I the answers didn't provide explicit answers.
It's easy to see that $V(I)\cup V(J) = V(IJ)$, for if $p\in V(I)\cup V(J)$ then for any $fg\in IJ$ we have $(fg)(p) = 0$ because either $f(p) = 0$ or $g(p) = 0$. On the other hand, $p\in V(IJ)$ but $p\notin V(I)$ then $f(p)\neq 0$ for some $f\in V(I)$, but $(fg)(p) =0 \implies g(p) = 0$ for all $g\in V(J)$, so $p\in V(J)$
Also, if $I\subset J$ then $V(I)\supset V(J)$, because if $p\in V(J)$ then $f(p) = 0$ for all $f\in J$, in particular for all $f\in I$.
Finally, notice that $(I\cap J)^2\subset IJ\subset I\cap J$, the last inclusion because $I,J$ are ideals.
Then
$$
V((I\cap J)^2)\supset V(IJ)\supset V(I\cap J)\implies V(IJ) = V(I\cap J)_\blacksquare
$$
A: This is a question of definitions, and so perhaps what you mean to ask is what is standard. Many I've seen use the fact that one should have $X \cup X = X$. Do you see which one to take if this is desired?
