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Definition. Call two elements of a commutative ring associates iff each divides the other. Call them strong associates if there exists a unit that can be multiplied by the first to yield the second. (In an integral domain, these are equivalent.)

Every element of a commutative ring is the strong associate of at most one idempotent (exercise). Now consider the commutative ring $\mathbb{Z}/6\mathbb{Z}$. Its idempotents are $\{1,3,4\}.$ Hence not every element of this ring is idempotent. However, every element is the strong associate of some idempotent; $2$ is a strong associate of $4$ (since $2 \times 5 = 4$ and $5$ is a unit) and $5$ is a strong associate of $1$ (since $5$ is a unit.)

However, there exist rings in which some elements are not associates of an idempotent, not even weakly. An example is $\mathbb{Z}/n^2\mathbb{Z}$ whenever $n \geq 2.$ Observe that $n$ is not the associate of an idempotent in this ring, not even weakly. Because if it were, then $n^2$ would be an associate of this idempotent, hence $0$ would be an associate of this idempotent, hence this idempotent would be $0$, hence $n$ would be an associate of $0$, hence $n$ would be $0$.

Question. For which $n \in \mathbb{N}$ is it the case that every element of $\mathbb{Z}/n\mathbb{Z}$ is the strong associate of an idempotent? Precisely the square-free ones, perhaps?

(Side question: is there a name for those commutative rings in which every element is the strong associate of some idempotent? These generalize both fields and Boolean rings simultaneously.)

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By using CRT we find that the number of elements in $\mathbb Z/n\mathbb Z$ which are products of an idempotent by an invertible is $N=[1+p_1^{k_1-1}(p_1-1)]\cdots[1+p_r^{k_r-1}(p_r-1)]$, where $n=p_1^{k_1}\cdots p_r^{k_r}$. (Recall that the idempotents in $\mathbb Z/p^k\mathbb Z$ are trivial, and the number of invertibles is $p^k-p^{k-1}$.)

The question reduces to when $N=n$. This clearly happens iff $n$ is square-free.

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  • $\begingroup$ Any thoughts about the "side question"? $\endgroup$ – goblin Jan 10 '15 at 16:32
  • $\begingroup$ @goblin No idea. $\endgroup$ – user26857 Jan 10 '15 at 16:52

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