Relation between Variety of $(I\cap J)$ and Variety of $(I)$ $\cap$ Variety of $(J)$

Question

I was wondering whether a relationship exists between $V(I\cap J)$ and $V(I)\cap V(J)$. Where $I$ and $J$ are ideals of the ring $R=K[x_1,...,x_n]$.

Answer 1

The correct correspondence is

$$V(I) \cap V(J) = V(I+J)$$

The intuition here is that the algebra and geometry are dual to one another. If I add more polynomials to my ideal there are fewer points that vanish on the whole ideal. If I take away polynomials, there are more points which vanish on the whole ideal. With that in mind, you're not going to get an identity where both the algebraic and geometric objects are getting smaller.

As an example, take $I = (x)$ and $J = (y)$ in $K[x,y]$. The intersection of $V(x)$ and $V(y)$ is the origin. But the intersection of $I$ and $J$ is $(xy)$, where $V(xy)$ is the union of the coordinate axes.

If we take $I = (x-3y)$ and $J = (y)$ then $V(I) \cap V(J)$ is still the origin. But $V(I \cap J)$ will be neither the origin nor the union of the coordinate axes. This shows that we cannot deduce the identity of $V(I \cap J)$ from our knowledge of $V(I) \cap V(J)$.

Answer 2

We know $V(I) \cap V(J) \subseteq V(I\cap J)$.