3
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

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.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .