A question about the Zariski topology

Question

Let $(\mathfrak a_i)$ be an infinite family of ideals in commutative ring $R$. Is $\bigcap\limits_{i=1}^\infty \mathfrak a_i$ not defined?

I am trying to understand Zariski topology. Here, $V(\bigcap_i \mathfrak a_i)= \bigcup\limits_{i} V(\mathfrak a_i)$.

If $\bigcap\limits_{i=1}^\infty \mathfrak a_i$ is defined, which I think it should be, then we would have an infinite union of sets of the form $V(X)$, all of which are closed sets. This is a contradiction as the infinite union of closed sets need not be closed.

Answer 1

In general, $V(\bigcap\mathfrak a_i)= \bigcup\limits_{i} V(\mathfrak a_i)$ is wrong!

Take $\mathfrak a_i=p_i\mathbb Z$, where $p_i$ is the $i$th prime number. Then $\bigcap_{i=1}^\infty \mathfrak a_i=(0)$, so $V(\bigcap_{i=1}^\infty \mathfrak a_i)=V((0))=\operatorname{Spec}\mathbb Z$ while $\bigcup_{i=1}^\infty V(\mathfrak a_i)=\bigcup_{i=1}^\infty p_i\mathbb Z$.

However the equality holds true for finitely many ideals.

Answer 2

First of all, the question of whether or not "$\bigcap_i\mathfrak a_i$ is defined" doesn't quite makes sense. Each $\mathfrak a_i$ is a set, and intersections of (arbitrary sets of) sets always makes sense as a set, and since each $\mathfrak a_i$ is a subset of $R$, so is the intersection. The right question to ask is:

If each $\mathfrak a_i$ is an ideal of $R$, is the intersection an ideal of $R$?

It's a subset of $R$, so it has a chance to be an ideal. You just need to check the conditions in the definition of an ideal. This is not difficult, using that each $\mathfrak a_i$ is an ideal, and that an element of $R$ lies in the intersection if and only if it lies in each $\mathfrak a_i$.

Now we accept that $\mathfrak a=\bigcap_i\mathfrak a_i$ is an ideal. Great. So we can look at $V(\mathfrak a)=V(\bigcap_i\mathfrak a_i)$, as well as $\bigcup_i V(\mathfrak a_i)$. Now, why should your intuition be that these aren't equal? By definition, the Zariski topology on $X=\mathrm{Spec}(R)$ has as its closed sets the sets $V(\mathfrak b)$ where $\mathfrak b$ is an ideal of $R$. In particular, $V(\mathfrak a)$ above is definitely a closed set. Each $V(\mathfrak a_i)$ is also a closed set... does that mean $\bigcup_i V(\mathfrak a_i)$ is closed? No! The axioms for a topology on a set dictate that a finite union of closed sets is closed, but there is no reason an infinite union should be. If $V(\mathfrak a)=\bigcup_i V(\mathfrak a_i)$, then every union of closed sets in the Zariski topology on $X$ would be closed. This isn't always true.

However, since finite intersections of closed sets are closed, it is reasonable (and correct!) to guess that $V(\bigcap_{i=1}^n \mathfrak a_i)=\bigcup_{i=1}^n V(\mathfrak a_i)$. You should prove this and see why the argument won't work for an infinite intersection. My suggestion for proving this is to first prove that $V(\prod_{i=1}^n\mathfrak a_i)=V(\bigcap_{i=1}^n\mathfrak a_i)$, and then that $V(\prod_{i=1}^n \mathfrak a_i)=\bigcup_{i=1}^n V(\mathfrak a_i)$ using the defining property of a prime ideal of $R$ (note that this argument definitely doesn't obviously translate to an infinite intersection, because infinite products of ideals don't generally make sense).

Answer 3

$V(\cap_i \mathfrak{a}_i)$ is the closure of $\cup_i V(\mathfrak{a}_i)$. In general, $\cup_i V(\mathfrak{a_i})$ might be not closed. For example, take $\mathbb{Z}$ so that $\mathrm{Spec}(\mathbb{Z})$ carries the cofinite topology except for the generic point $\eta$ which is contained in every non-empty open subet. We consider the ideals $(p)$ for prime numbers $p$. Then $\cup_p V(p) = \mathrm{Spec}(\mathbb{Z}) \setminus \{\eta\}$ is not closed (but it is dense in $\mathrm{Spec}(\mathbb{Z})$).