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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?

Asaf Karagila
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Gil
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2 Answers2

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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
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Asaf Karagila
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  • 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. – k.stm May 23 '13 at 05:39
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    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. – Asaf Karagila May 23 '13 at 05:41
  • @K.Stm. What's NBG? – Gil May 23 '13 at 05:41
  • @GilYoungCheong: It's an extension of $\sf ZFC$ which allows proper classes to exist (in addition to sets). It carries the names of von Neumann-Godel-Bernays. – Asaf Karagila May 23 '13 at 05:42
  • AsafKaragila: Ah, of course. @GilYoung NBG. – k.stm May 23 '13 at 05:43
  • Then what is the "formula" for picking the ambient space? Do we look at what is written right next to the intersection symbol? – Gil May 23 '13 at 05:49
  • @GilYoungCheong: No, let me edit my answer to address this question. – Asaf Karagila May 23 '13 at 05:51
  • @AsafKaragila Sadly, I don't think this is the answer that I am looking for. My question is how do we know that we are supposed to agree on certain ambient space? I don't think my derivation of the contradiction depends on the choice or certain definition (if it does, that will also be an answer that I desire). That's why I chose this particular example. – Gil May 23 '13 at 06:25
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    @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. – Asaf Karagila May 23 '13 at 06:36
  • @AsafKaragila I did ask two questions and you answered one of them. I am still confused just because "the agreement on the ambient space" sounds too ambiguous to be a mathematical definition. – Gil May 23 '13 at 06:48
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    @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 $A$. – Asaf Karagila May 23 '13 at 06:50
  • @AsafKaragila Thanks for your patience. I think I understand what you are telling me, so I wrote it as a separate answer. Let me know if I distorted what you discern. – Gil May 23 '13 at 07:04
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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 $(1)$."

Gil
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