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Let $R$ be a ring, $S$ is a multiplicative subset of $R$. $a$ is an arbitrary element of $S$. Should there be 2 element $b,c \in S, b, c \neq 1$ such that $a=b.c$?

If not please give a counter example. Thank you very much!

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Are you assuming that $S$ contains $1$ in your definition? – azarel Apr 14 '12 at 5:01
@azarel : Thanks. I have just edited. – Arsenaler Apr 14 '12 at 5:07
up vote 2 down vote accepted

No, there is no such requirement. Consider $R=\mathbb{Z}$ and let $S$ be the set of all positive integers not divisible by $2$. This is a multiplicative set, and contains $3$. However, any expression of $3$ as a product $3=bc$ with $b$ and $c$ positive integers will have $b=1$ or $c=1$.

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Consider $R=\Bbb Z$ and $S=a\Bbb Z$ with $a\not\in\{-1,0,1\}$, especially in light of prime factorization.

More specifically, say we take $R$ to be the integers and the multiples of some nonzero nonunit $a$ to be our multiplicative subset $S$. For a quick example, let $a=2$ and so $S$ is the set of even numbers closed under multiplication. Can $2$ be written as the product of two even numbers?

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Sorry, can you explain more clearly ? – Arsenaler Apr 14 '12 at 5:08

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