Show that $V(I \cap J)=V(I) \cup V(J)$.

Question

Let $I$, $J$ ideals of $K[x_1, x_2, \dots , x_n]$.

I want to show that $$V(I \cap J)=V(I) \cup V(J)$$

I tried the following:

$$\subseteq: $$

Let $x \in V(I \cap J)$.

From the definition of $V$:

$V(S)=\{ (a_1,a_2, \dots, a_n) \in K^n| f_a(a_1,a_2, \dots, a_n)=0 \forall a \in A\}$

we have that $f(x)=0, \forall f \in I \cap J$.

So, $f(x)=0, \forall f \in I \text{ AND } f \in J$.

So, $f(x)=0, \forall f \in I \text{ AND } f(x)=0, \forall f \in J$.

So, $x \in V(I) \text{ AND } x \in V(J)$.

So, $x \in V(I) \cap V(J)$.

But... I have to show that $x \in V(I) \cup V(J)$.

What have I done wrong?

Answer 1

Assume $x$ is neither in $V(I)$ nor $V(J)$ (i.e. $x \notin V(I)\cup V(J)$). That means that there is an $f\in I$ and a $g\in J$ such that $f(x) \neq 0$ and $g(x) \neq 0$.

We have $fg \in I\cap J$, but $fg(x) \neq 0$, and therefore $x \notin V(I\cap J)$.

This shows that $x \notin V(I)\cup V(J)$ implies $x \notin V(I\cap J)$, which is the contrapositive of what you want to show.

As for the direct approach, how about something along these lines? $$ x \in V(I\cap J) \implies\\ \forall f\in I\cap J: f(x) = 0 \implies\\ \forall f \in I, g\in J: fg (x) = 0\implies\\ \forall f \in I, g\in J: f(x) = 0 \text{ OR } g(x) = 0 \implies\\ x \in V(I) \text{ OR } x\in V(J)\implies\\ x \in V(I)\cup V(J) $$

Answer 2

Let $x \in V(I \cap J) \Rightarrow f(x) = 0, \forall f \in I \cap J.$ Suppose there exists $h \in I$ such that $h(x) \neq 0.$ The for any $g \in J, g(x) = 0,$ because $hg \in I \cap J.$