Let $(I_a)_{a \in A}$ be a family of ideals of $K[x_1,x_2, \dots, x_n]$. I want to prove that:
$$V \left ( \sum_{a \in A} I_a\right )=\bigcap_{a \in A} V(I_a)$$
Do we have to use the definition:
$$V(S)=\{ (a_1,a_2, \dots, a_n) \in K^n| f_a(a_1,a_2, \dots, a_n)=0 \forall a \in A\}$$
If so, how could use it?
EDIT: In order to show that $V \left( \sum_{a \in I_a}\right) \subseteq \bigcap_{a \in A} V(I_a)$ I tried the following:
Let $\overline{x} \in V\left( \sum_{a \in I_a}\right)$. That means that $f(\overline{x})=0 \forall f \in \sum_{a \in A} I_a$.
We want to show that $\overline{x} \in \bigcap_{a \in A} V(I_a)$, i.e. that $f(\overline{x})=0 \forall f \in I_a \subseteq \sum_{a \in A} I_A$. And thus, it holds.
Therefore $V \left( \sum_{a \in I_a}\right) \subseteq \bigcap_{a \in A} V(I_a)$.
For the other direction I tried the following:
Let $\overline{x} \in \bigcap_{a \in A} V(I_a)$. That means that $f(\overline{x})=0, \forall f \in I_a, \forall a \in A$.
$$I_a \subseteq \sum_{a \in A} I_a$$
$$\Rightarrow f(\overline{x})=0 \forall f \in \sum_{a \in A} I_a \Rightarrow \overline{x} \in V \left (\sum_{a \in A} I_a \right )$$
Could you tell me if it is right?
This is straightforward but confusing.
$\subseteq$:
Suppose $x \in V(\sum I_a)$. By definition $f(x)=0$ for all $f \in \sum I_a$. But the generators of $\sum I_a$ is the union of the generators of each $I_a$, so in particular $f(x)=0$ for each generator of $I_a$. This means that $x \in V(I_a)$ for all $a$. So $ x \in \cap_a V(I_a)$.
$\supseteq$:
This is similar: Assume that $x \in \cap_a V(I_a)$. By definition of intersection, this means that $x \in V(I_a)$ for all $a$. Do you see the rest of the proof?