I'm currently looking for all the subsets $ S$ of $ \mathbb{Z}$ which are closed under integer multiplication. Such sets are called multiplicative. My first idea was that all of those subsets were of the form $n\mathbb{Z}$ but then I realized that $3^2\equiv_6 3 $ so $6\mathbb{Z}+3$ is also multiplicative. This led me to think that the multiplicative subsets of $\mathbb{Z} $ were of the form $ (n\mathbb{Z})+r$ where $ r$ is idempotent modulo $ n$. But then I realized that the complement of $p\mathbb{Z} $ (here $ p$ is a prime like always) is also multiplicative. Ok, after thinking a while I came to the conclusion that these are all the multiplicative subsets of the integers but I'm not absolutley sure nor know how to prove it. Any ideas?
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1Also ${2n\mid n\ge 0}$ is closed under multiplication, but different from $2\Bbb{Z}$. Isn't this question the same one? – Dietrich Burde Sep 15 '18 at 18:50
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You are right, we must add $ n\mathbb{N}+r$ where $ r$ is idempotent modulo $n $. – Natalio Sep 15 '18 at 18:52
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Closed under multiplication is different from multiplicatively closed, where we require in addition that $1\in S$. – Dietrich Burde Sep 15 '18 at 18:55
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4Natalio, you must also add the set of integers $S$, which can be represented as the sum of two squares, and many more things. So again, as in the duplicate, I am not sure whether or not a classification is possible. – Dietrich Burde Sep 15 '18 at 18:58
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2Also family of these sets is closed for intersection and product. – SMM Sep 15 '18 at 19:02
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@DietrichBurde thank you very much, probably a clasifiacation is not posible. – Natalio Sep 15 '18 at 19:26