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If

$$ A \subset \Bbb{Z} $$

is such that $xy \in A \implies x \in A, $ or $y \in A$. Then $A$ is either a prime ideal or ?

Can we describe all "prime subsets" of $\Bbb{Z}$ that aren't prime ideals in one fell swoop?

It's not just a subset closed under taking divisors, though those are counted, and so it's also not just a subset of negative prime numbers together with $\{0, -1\}$ either.

Daniel Donnelly
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  • Just checking: you only mandate $xy\in A\implies x\in A\lor y\in A$ and you are not even mandating the opposite implication? –  Dec 06 '20 at 01:05
  • @StinkingBishop yes, that is correct – Daniel Donnelly Dec 06 '20 at 01:16
  • To see the relationship with primes, make the inferences bidirectional, then the complement of $A$ Is a saturated monoid, hence $A$ is a union of prime ideals. This is best viewed in terms of localization, e.g. see this answer in the dupe. See also this answer on the complementary view of a subgroup. – Bill Dubuque Dec 06 '20 at 04:29

1 Answers1

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Notice that the implication $$ xy \in A \implies \big( x \in A \text{ or } y \in A \big) $$ is equivalent to its contrapositive $$ \big( x \notin A \text{ and } y \notin A \big) \implies xy \notin A. $$ In other words, $A$ has the given property if and only if $\Bbb Z\setminus A$ is closed under multiplication.

In particular, there is a wide variety of examples of such sets $A$, such as:

  • $A = \{-m,\dots,m\}$ for any $m\in\Bbb N$;
  • $A = \{n\in\Bbb Z\colon n$ cannot be written in the form $n = 13^a2020^b\}$;
  • $A = \{n\in\Bbb Z\colon n$ is not a perfect square$\}$;
  • $A = \{n\in\Bbb Z\colon \text{there exists } p\mid n \text{ such that } p^p\nmid n\}$....
Greg Martin
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