Identities of the spectrum of a ring

Question

Let $R$ be a commutative ring, then the spectrum of $R$ is a topological space:

$$Spec(R)=\{ P:P \;\text{is a prime ideal of R} \}$$

where the close sets are

$$V_{I}=\{P \in Spec(R) \;\text{such that} \; I \subset P\}$$

then we have that

$$V_I = \bigcap_{f\in I}V_f $$

So I want to prove that

1. $V_{\Sigma I_{\alpha}}=\bigcap _{\alpha}V_{I_{\alpha}}$
2. $V_{I_1 \cap I_2}=V_{I_1} \cup V_{I_2}$

But I don't know how to proceed in the first one. For the second one I used the distributive property of union and intersection to get:

$$V_{I_1} \cup V_{I_2}= \bigcap_{f \in I_1} V_{f} \cup \bigcap _{f_1 \in I_2} V_{f_1}=\bigcap_{f_1 \in I_2 } \bigcup_{f \in I_1} V_{f_1} \cap V_f$$

but I don't know how to compare it or get from this $V_{I_1 \cap I_2}$

So, can someone help me to prove this please?

Edition

For 1) I've got this

$\Rightarrow]$ Since we know that $\cup I_k \subset \Sigma I_k $ then the result follows.

$\Leftarrow]$ I don't know how to get the other contention because how can you embed a bigger set into a small one?

For 2) I've got this

$\Leftarrow]$ Since $I_1 \cap I_2 \subset I_1$ and $I_1 \cap I_2 \subset I_2$ then the result follows.

$\Rightarrow]$ The same problem as before.

Answer 1

Hint: Use the definition of $V_I$. Then for (1) you have to show that $P\in\mathrm{Spec}(A)$ contains $\sum I_k$ if, and only if it contains all the $I_k$ singularly, and for (2) you have to show that $P$ contains $I_1\cap I_2$ if, and only if it contains either $I_1$ or $I_2$.

Answer 2

I'll use $V(I)$ to denote what you mean as $V_I$ to clean up the notation.

1. $\sum I_\alpha$ is the smallest ideal containing all of the ideals $I_\alpha$. therefore, $P$ contains the sum if and only if $P$ contains each $I_\alpha$.

2. If $I$ or $J$ is contained in $P$, then $P$ certainly contains $I \cap J$. Conversely, suppose $I \cap J \subseteq P$. Suppose, WLOG, that $I$ is not contained in $P$ though. Then there is some $x \in I - P$. But for any $y \in J$, $xy \in I \cap J$ and so $xy \in P$. Thus, $y \in P$ so $J \subseteq P$ as desired.