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I need to show that being an integral domain is a local property. That is, a commutative ring $A$ has no zero divisors iff $A_{\mathfrak p}$ has no zero divisors for every prime ideal $\mathfrak p$.

One way is obvious: if there are zero divisors in a localization, then there are zero divisors in $A$.

Proving the other direction I am running into the following difficulty: let $a\in A$ be a zero divisor. Then $Ann(a)\ne 0$. I can guarantee that either

  • for a certain prime ideal $\mathfrak p$ the fraction $a/1\ne 0$, in $A_{\mathfrak p}$ or

  • for a certain prime ideal $\mathfrak p$ there is a $b\in Ann(a)$ and an $s\in A-\mathfrak p$ such that $b/s\ne 0$ in $A_{\mathfrak p}$.

But I can't figure out how to get a $\mathfrak p$ for which both are true. If I had that, then the product $$\frac a1\cdot\frac bs$$ would equal zero and I would be done, but in both cases I only know for sure that one of the factors is nonzero.

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Artem
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    This isn't true--consider $\mathbb{R}\times\mathbb{R}$. – Alex Youcis Feb 22 '14 at 09:48
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    It is not a local property. Take the product of two integral domains, say, $\mathbb{Q} \times \mathbb{Q}$... – Zhen Lin Feb 22 '14 at 09:48
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    On the other hand, having no non-trivial nilpotent elements (i.e. being reduced) is a local property: $A$ is reduced, if and only if every $A_{\mathfrak{p}}$ is reduced. – Nils Matthes Feb 22 '14 at 11:11
  • Thank you! I probably wouldn't have thought of this myself despite how simple the answer is. – Artem Feb 23 '14 at 13:07
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    Can someone help me understand why if we take the localization of the direct product of two integral domains at a prime ideal we get an integral domain? For example the prime ideals of $A:=\mathbb{Q}\times \mathbb{Q}$ are $p:=\mathbb{Q}\times{0}$ and $q:={0}\times \mathbb{Q}$. But How does one compute $A_p,A_q$? – CWsl2 Mar 03 '14 at 05:51
  • @CWsl2: First, a ring is the finite product of domains if and only if its localization at every prime ideal is a domain. This is Proposition 2.20 in Eisenbud. Second, for the example you are referring, you compute the localization by definition. But here is the intuition of why the localization is a domain. Suppose in $k \times k$ that $(a,b)(a',b')=0$, where $(a,b), (a',b') \neq 0$. Then without loss of generality our elements are $(a,0),(0,b')$. Then notice either in the localization at $(k,0)$ or $(0,k)$, at least one of the elements must be zero. – Manos Mar 31 '16 at 14:10
  • Another class of counterexamples is $\mathbb{Z}/n\mathbb{Z}$ where $n$ is not prime and squarefree. See this post for an explanation. – Anakhand Jan 27 '24 at 15:16

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As is pointed out in the comments, this is not actually true. The product of any two fields is a counterexample.

The problem is, geometrically, your space isn't connected. Indeed, if you start with a Noetherian ring which isn't the product of non-trivial rings, and every localization is integral, then the ring is integral.

Geometrically, this is saying that if $X$ is a Noetherian scheme, then $X$ is integral if and only if $X$ is connected and $\mathcal{O}_{X,p}$ is a domain for all $p\in X$.

Try to prove this yourself. Here are some observations that might help:

1) If we can show that the irreducible components are disjoint then we're done, because there are finitely many of them (why?) and so we'd have a disconnection of $X$.

2) If two of the components intersected at $p$, you could descend this intersection to $\mathcal{O}_{X,p}$. Why is that bad?

Alex Youcis
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    The correct geometric statement is the following: A scheme $X$ is integral if and only if it is nonempty and connected, every stalk of its structure sheaf is integral, and its set of irreducible components is locally finite. So first, your answer is wrong in case of the empty scheme. And second, the parasitic noetherianness obfuscates as so often the real reasons. – Fred Rohrer Feb 23 '14 at 12:23
  • @FredRohrer, I'm curious as to your opinion about Noetherian rings. In what sense are they "parasitic"? (Note that I am not disagreeing.) – goblin GONE Apr 13 '14 at 10:22
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    @FredRohrer I just now saw this comment. I can't fathom why you would be so rude, and about as something as insignificant as the degenerate case of the empty scheme. Also, the whole "Noetherianess obfuscates the true nature of things" argument usually leaves me wanting. When one is sufficiently interested in scheme theory that they care about Noetherianess conditions, they probably don't need to be told that theorem ____ only requires qc or something. And, if they are not that advanced, then an extra Noetherian condition, if for simplification or just comfort, is not poison in baby's milk. – Alex Youcis Apr 13 '14 at 10:56
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    Dear @Alex, concerning the empty scheme: Truth and falseness are absolute in mathematics; this has nothing to do with rudeness. Concerning noetherianness: From a pedagogical point of view unnecessary hypotheses may be very poisonous. – Fred Rohrer Apr 14 '14 at 20:17
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    @user18921: I do not think noetherian rings are parasitic, but rather noetherian hypotheses. They are spread throughout the literature on commutative algebra and algebraic geometry in a way that makes it extremely difficult to get rid of them in case one wishes to do so. I cannot remember where I found the comparison with a parasite, but it seems very apt. (Noetherian rings are indeed rather nice, and thus they are often a good place to start some study. But after starting there usually should be some continuation...) – Fred Rohrer Apr 14 '14 at 20:27
  • Dear Alex, when you say "non-trivial rings" you mean "fields"? If so, then why not make it explicit? – Manos Mar 31 '16 at 14:15