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Let $A$ be a Noetherian domain and $K$ its field of quotients. I would like to prove that, given $x \in K$, we have $x \in A$ iff $x\in A_p$ for every $p$ prime ideal associated to an $y\in A$, $y\ne0$.

user26857
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Tommaso Scognamiglio
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  • Is $p$ an element of $A$ or a prime ideal? I think that the corresponding statement in $\mathbb{Z}$ would be that a fraction is an integer if and only if its denominator is not divisible by any prime, hence if $x \in \mathbb{Z}_{(p)}$ for all primes $p$, right? – 57Jimmy Apr 11 '18 at 13:23
  • Sry I edited now. – Tommaso Scognamiglio Apr 11 '18 at 13:31
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    https://math.stackexchange.com/questions/630752/an-integral-domain-a-is-exactly-the-intersection-of-the-localisations-of-a-a – ArtW Apr 11 '18 at 14:04
  • I knew this fact but I still can't see how to conclude – Tommaso Scognamiglio Apr 11 '18 at 14:41
  • By associated primes, I mean primes that appear as radical of primary ideal that appear in the primary decomposition of $(y)$ given $ y \in A$ – Tommaso Scognamiglio Apr 11 '18 at 20:38
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    Write $x=a/b$, and suppose $x\in A_p$ for every $p$ with the given property. Assume $x\notin A$. Then $I=(bA:a)$ is proper, and let $p$ be an associated prime of $I$. There is $c\in A$ such that $p=(bA:ac)$. This means that $p$ is associated to $b$. Now you can finish the proof. – user26857 Apr 11 '18 at 21:07
  • @Tommaso Scognamiglio - I apologize for my comment earlier. I don't know what I was thinking, but it wasn't mathematics. I am embarrassed to have posted such utter garbage. – Chris Leary Apr 13 '18 at 00:44
  • user26857 I'm sorry, but I still can't see how to end. I tried to write all things down, but I do not get how to use the hypothesis. Thank you a lot. – Tommaso Scognamiglio Apr 13 '18 at 17:07

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