Let $A$ be a commutative ring with $1 \neq 0$. Then writing $V(1) = V((1))$, we have $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \sqrt{(1)} = (1)$.

But then $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \{x : x \in \mathfrak{p} \text{ for all prime ideals } \mathfrak{p} \supseteq (1)\} \supseteq \text{Set}$, where the last containment is justified since there is no prime ideal that contains $(1) = A$, so $x$ can be anything by vacuous truth.

I had a similar confusion like this in my topology course, but the answer that I heard from my teacher was that "of course, $A$ is the universe that we consider," so $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \{x \in A : x \in \mathfrak{p} \text{ for all prime ideals } \mathfrak{p} \supseteq (1)\} = (1)$.

But then why "pure" definition from (naive) set theory fails and forces us to consider $A$ as the ambient space?


Consider the defining formula for the class $\bigcap\cal S$, where $\cal S$ is a collection of sets:

$$\bigcap\mathcal S=\{x\mid\forall S\in\mathcal S, x\in S\}$$

When $\cal S=\varnothing$, the assumption is satisfied by every object in the universe. And as we know, the collection of all objects in the universe is not a set. Therefore $\bigcap\varnothing$ is not a set, it doesn't "exist" in the sense that we want it to exist.

But if we agree that the ambient space is $A$ itself, our ring, then this is just all the objects which are in $A$, that is to say $A$ itself. Since $A$ exists, it is a set, it's fine.

When we agree on an ambient space, we actually say that whatever formula is used to pick whatever elements, they will always be members of that ambient space. So if we agree that $A$ is our ambient space, and $\cal S$ is a collection sets, then we actually write:

$$\bigcap\mathcal S=\{x\in A\mid\forall S\in\mathcal S, x\in S\}$$

Now the empty intersection would be $A$ itself, which we assume is a set.

Also see:

  1. intersection of the empty set and vacuous truth
  2. Unary intersection of the empty set
  • $\begingroup$ In NBG, you may consider it to be a class, and then it exists and the paradox remains, right? I think, one has then to define $\bigcap \mathcal{A} = \{x ∈ \text{Set};\; ∀A ∈ \mathcal{A}: x ∈ A\}$, right? Hm, this does not resolve the paradox. $\endgroup$ – k.stm May 23 '13 at 5:39
  • 2
    $\begingroup$ Even in $\sf NBG$, only sets are allowed to be members of other classes. So by writing $x\in ...$ you immediately have that $x$ is a set. $\endgroup$ – Asaf Karagila May 23 '13 at 5:41
  • $\begingroup$ @K.Stm. What's NBG? $\endgroup$ – Gil May 23 '13 at 5:41
  • 1
    $\begingroup$ @Gil: You asked why the naive definition fails. It fails because otherwise you end up with a proper class, which is certainly not a subset of your ring, $A$. In fact the only way to assure that you end up with a subset of $A$, which is what you want, is to bound the ambient space by a subset of $A$. Since any proper subset will definitely lose information, you have to use $A$ itself. $\endgroup$ – Asaf Karagila May 23 '13 at 6:36
  • 1
    $\begingroup$ @Gil: It's not mentioned anywhere. This is one of these things that should be "clear from context". We are interested in ideals of $A$, so they are subsets of $A$. Therefore we take the ambient universe as $A$. $\endgroup$ – Asaf Karagila May 23 '13 at 6:50

I think the following is what Asaf Karagila means (of course, I am not going to select my answer).

Definition. Let $\{X_{\alpha}\}_{\alpha \in I}$ be a collection of subsets of $A$. We define $\bigcap_{\alpha \in I}X_{\alpha} := \{x \in A : x \in X_{\beta} \text{ for all } \beta \in I\}$.

Now, the problem I had can be resolved by reading "$\mathfrak{p} \in V(1)$" as "prime ideal $\mathfrak{p}$ of $A$ that contains $(1)$," which again can be paraphrased as "a subset $\mathfrak{p}$ of $A$ that is a prime ideal and contains $(1)$."


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.