Basic problem about property of algebraic set

Question

We know there is a property about algebraic set:
$V(I\cap J) = V(I) \cup V(J)$

Where $I$ and $J$ are two ideals consisting polynomials contained in a polynomial ring (not shown). $V(I)$ is a algebraic set consisting of points which make polynomials in $I$ vanishing, i.e. $V(I) := \{x|f_i(x)=0, f_i\in I\}$.

Now, consider the following graph:

My question is

$V(I)$ should be the set of points making both 1 and 2 polynomials vanishing. So it should be points 1 and 2 (red pts). And $V(J)$ should be points 2 and 3 (red pts). The $V(I) \cup V(J)$ should be red points 1, 2 and 3.

However, I think $V(I\cap J)$ should be the green triangle, which is the intersected polynomial. My idea is $I\cap J$ = green triangle

I am confused where did I make a mistake?　

Answer 1

So it looks to me like you are doing something wrong when you are finding $I \cap J$ because you are claiming that any polynomial in it should vanish on your green triangle. Using your notation the polynomial $1*3$ is in $I \cap J$ and doesn't vanish on the entire green triangle (instead it vanishes if either $1$ or $3$ does). This means that, since $2$ is in the intersection of these ideals as well, the most places that could be in the vanishing set of $I \cap J$ are the three red dots in your picture.

Answer 2

The property $V(I\cap J) = V(I) \cup V(J)$ holds for ideals $I$, $J$ and not for just any sets of functions. The proof of it works like this

$$V(I) \cup V(J) \subset V(I\cap J) \subset V(I\cdot J) \subset\ V(I) \cup V(J)$$

The only nontrivial inclusion is $V(I\cdot J) \subset V(I) \cup V(J)$ and this uses in some way or another "$a\cdot b = 0 \Rightarrow a=0 \ \text{or}\ b=0$".

Let's look at your subsets of functions $I$, $J$. We have as sets $I \cdot J = \{ 1\cdot 2, 1\cdot 3, 2\cdot 2, 2\cdot 3\}$ and it's easy to check that

$$V(\{ 1\cdot 2, 1\cdot 3, 2\cdot 2, 2\cdot 3\}) = V(\{2,\, 1\cdot 3\}) = V(I) \cup V(J)$$

Why not for $I\cap J$? Well, $I \cap J \supset I\cdot J$ is a property of ideals. In general $I \cap J$ will just be too small so $V(I \cap J)$ will strictly include $V(I) \cup V(J)$.