Relation between subvarieties corresponding to union and intersection

Question

This may be a rather trivial question, but here goes:

I have been attempting to prove (for myself) that the Zariski topology is indeed a topology, i.e showing that the varieties of the affine space $A_\mathbb{K}^n$ do indeed behave as closed sets:

1) $A_\mathbb{K}^n$ and $\emptyset$ are varieties (which is easy)

2) If $V_1$ and $V_2$ are varieties then $V_1\cup V_2$ is a variety.

3)If $\{ V_\alpha \}_{\alpha\in \Lambda}$ is a collection of varieties, then $\underset{\alpha \in \Lambda}{\bigcap}V_\alpha$ is also a variety.

Now since $V=Z(S)$ for some $S\subseteq \mathbb{K}[x_1,...,x_n]$ in order for $V$ to be a variety, the question is regarding how the zero-locus of a union and intersection should be. It seems rather simple to show that:

$Z(S_1)\cap Z(S_2)=Z(S_1\cdot S_2)$ and $Z(S_1)\cup Z(S_2)=Z(S_1+S_2)$

But is it also true that $\underset{\alpha \in \Lambda}{\bigcap}Z(S_\alpha)= Z \Big( \underset{\alpha \in \Lambda}{\bigcup} S_\alpha \Big)$ and $\underset{\alpha \in \Lambda}{\bigcup}Z(S_\alpha)= Z \Big( \underset{\alpha \in \Lambda}{\bigcap} S_\alpha \Big)$ ?

A second question I'm having is whether:

$Z\Big( (S_1)\Big) = Z\Big( (S_2)\Big) $ if and only if $(S_1)=(S_2)$?

I would appreciate any and all help.

Answer 1

$Z(S_1)\cap Z(S_2)=Z(S_1\cdot S_2)$ and $Z(S_1)\cup Z(S_2)=Z(S_1+S_2)$

You got this wrong. If $f \in S_1 \cdot S_2$, this means that $f$ will vanish on both sets $Z(S_1)$ and $Z(S_2)$, because $f$ is in both ideals generated by $S_1$ and $S_2$. Thus $f$ will vanish on $Z(S_1) \cup Z(S_2)$. From this it is easy to see that $Z(S_1) \cup Z(S_2) = Z(S_1 \cdot S_2)$.

The same goes for the other case, if $f \in S_1 + S_2$, you only know that it will vanish, where both $S_1$ and $S_2$ vanish, i.e. $f$ vanishes on $Z(S_1)\cap Z(S_2)$. Now it should be easy to show that in fact, $Z(S_1) \cap Z(S_2) = Z(S_1 + S_2)$.

For the general case show that $\bigcap_\alpha Z(S_\alpha) = Z(\sum_\alpha S_\alpha)$.

A second question I'm having is whether: $Z\Big( (S_1)\Big) = Z\Big( (S_2)\Big) $ if and only if $(S_1)=(S_2)$?

That is not true. In general we have $Z(S) = Z(\sqrt{S})$, where $\sqrt{S} = \{f \in k[x_1,\dots,x_n] \mid \exists n: f^n \in S \}$ is the so called radical of $S$. Try proving this identity, it's not hard and a good exercise.

So we see immediatly, that there might be different ideals which have the same zero locus. If $S \neq \sqrt{S}$ this is the case, which happens for example if $S = (x_1^2)$. Then $\sqrt{S} = (x_1)$.