# Does the existence of zero divisors imply an infinite factorization for some nonzero element?

We know that zero factors infinitely into nonzero nonunits in a ring with zero divisors: $$0=ab=a^2b=a^3b=\ldots$$ and nontrivial idempotents also factor infinitely into nonzero nonunits: $$e=e^2=e^3=e^4=\ldots$$ Idempotents are special cases of zero divisors, so I'm looking to see if the presence of zero divisors in a commutative ring always implies that some nonzero element in the ring factors infinitely into nonzero nonunits.

Before I spend time cracking away at this, is the above statement true in general? Or is there a counterexample for what I'm trying to prove?

My work so far looks like:

Suppose $$0=ab$$, so that $$a,b$$ are nonzero zero divisors. Then: \begin{aligned} a&=a\\ a&=a+0\\ a&=a+kab&\text{ where }k\text{ is any integer}\\ a&=a(1+kb)\\ \end{aligned} shows that $$a$$ factors into itself and $$(1+kb)$$ for all integers $$k$$. Thus we can write \begin{aligned} a&=a(1+kb)\\ a&=a(1+kb)^2\\ a&=a(1+kb)^3\\ \end{aligned} and so on. Note that since $$kb$$ is a zero divisor, $$kb$$ cannot be $$-1$$ (which is always a unit) and thus every $$1+kb$$ is nonzero. So the problem is reduced to checking if $$1+kb$$ is a nonunit for an arbitrary zero divisor $$b$$, for some integer $$k$$. Alternatively, in the case that every $$1+kb$$ is a unit, I could try to prove the existence of a nontrivial idempotent in the ring.

The result you are trying to prove is not correct: take $$R=K[\varepsilon]$$, the ring of dual numbers over some field $$K$$. Elements of $$R$$ are of the form $$a+\varepsilon b$$ with $$a,b\in K$$, and we impose that $$\varepsilon^2=0$$.
Then $$\varepsilon$$ is nilpotent, therefore a zero divisor. The units of $$R$$ are exactly the elements of the form $$a+\varepsilon b$$ with $$a\neq 0$$, so any product of non-units has the form $$\varepsilon b\cdot \varepsilon b'=0$$, and you cannot have "infinite factorizations" as you wish.
Actually, already your first sentence is incorrect, since in your sequence of equalities $$0=ab=a^2b=a^3b=\dots$$, you can in general have $$a$$ nilpotent, in which case $$a^k=0$$ when $$k$$ is high enough. (Actually this is how I constructed my counter-example: I chose a ring in which all non-units are nilpotent.)