# Multiplicative system in the ring

Let $R$ be a commutative ring with unit element. A nonempty subset $S$ of $R$ is called multiplicative system if

$1. \ 0\notin S$

$2. \ s_1,s_2\in S$ implies that $s_1s_2\in S$.

Let $n=pq$ where $p,q$ are prime numbers. We know that $\mathbb{Z}_n$ is a ring with addition and multiplication modulo $n$. Consider the $\mathbb{Z}_n^{\times}=\{x: 1\leq x<n, \ \text{gcd}(x,n)=1\}$.

It's easy to check that $\mathbb{Z}_n^{\times}$ is multiplicative system.

What if we consider $\mathbb{Z}_n^{\times}\cup \{p\}$? I have checked that this is not multiplicative system.

Am I right? If yes, can anyone suggest some other example of multiplicative system in $\mathbb{Z}_n$ distinct from $\mathbb{Z}_n^{\times}$?

• Well, that is not a multiplicative system because $\;pq=0\;$ and $\;0\notin\Bbb Z_n^*\;$ ...And you can take, in a general ring $\;R\;$ , any prime ideal $\;I\;$, and then $\;R\setminus I\;$ is a multiplicative system. This is probably the most important example in algebraic number theory, commutative algebra, algebraic geometry and stuff... – DonAntonio May 20 '18 at 19:32
• @DonAntonio, Thanks for reply! – ZFR May 20 '18 at 19:40
• @DonAntonio Are you assuming that also $q\in S$? – Hagen von Eitzen May 20 '18 at 19:44
• @HagenvonEitzen I was until your comment, so the first part of the first line of my comment is incorrect. Thanks. – DonAntonio May 20 '18 at 19:46

The multiplicative closure of $\{p\}\cup \Bbb{Z}_n^\times$ , namely $\{x: 1\le x< n, \gcd(x,q)=1\}$ is a multiplicative system distinct from $\Bbb{Z}_n^\times$.
• Why the multiplicative closure of $\{p\}\cup \mathbb{Z}_n^{\times}$ is the set which you've written? – ZFR May 20 '18 at 19:39
• $\Bbb{Z}_n^\times$ is everything without $p$ or $q$ in its factorization. Letting in $p$ means you have to be able to multiply by any power of $p$, so now you have everything without $q$ in its factorization. – C Monsour May 20 '18 at 19:44