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Definition: A nonempty subset S o f a ring R is multiplicative provided that $$a,b\in S \implies ab\in S$$

Examples. The set S of all elements in a nonzero ring with identity that are not zero divisors is multiplicative.

My attempt: Let $S=\{x\in R\mid x \text{ is not zero divisor}\}$. Suppose $a,b\in S$. Assume towards contradiction, $ab\notin S$. That is, $ab$ is left and right zero divisior. So $\exists c,d\in R\setminus \{0\}$ such that $c(ab)=(ab)d=0$. If $ca=0$, then $a$ is right zero divisor. Since $a$ is not zero divisor, $a$ is not left zero divisor and $(ab)d=a(bd)=0$ implies $bd=0$. So $b$ is left zero divisor. Similarly, if $ca\neq 0$, then $b$ is right zero divisor and $a$ is left zero divisor. How to progress from here?

Where to use $R$ has a multiplicative identity in the proof? If $R$ were a commutative ring, then proof is relatively easy. But we are not given $R$ is commutative.

user264745
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    I think that if $ab$ is not a zero divisor, then either exists $c\ne 0$, $cab=0$, or exists $d\ne 0$, $abd=0$. In the first case, "$b$ is not a zero divisor" implies $ca=0$, and "$a$ is not a zero-divisor" implies $c=0$, a contradiction. $abd=0$ can be handled in a totally similar way. – Asigan Dec 14 '23 at 12:24
  • @Asigan ab is not a zero divisor is what we are trying to prove. Author defined zero divisor as both left and right zero divisor. – user264745 Dec 14 '23 at 12:50
  • Strange. I'd view zero divisor in the non commutative setting as an element which is either (or possibly both, but who cares?) a left or right zero divisor. I'm not sure this theorem is correct if you require both to hold – FShrike Dec 14 '23 at 12:55
  • @FShrike I also think this example is not true. Wikipedia definition of zero divisor is left or right zero divisor. I again checked from book, it is written both left and right zero divisor. – user264745 Dec 14 '23 at 13:01
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    With your definitions, the theorem is false in general, and false in particular for any ring in which there exists left-not right zero divisors and right-not left zero divisors, such as the ring of endomorphisms of (say, integer) sequences. – SomeCallMeTim Dec 14 '23 at 13:04
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    You can read here for the concrete example (note, the notation is different than yours) https://en.wikipedia.org/wiki/Zero_divisor#One-sided_zero-divisor – SomeCallMeTim Dec 14 '23 at 13:05
  • @SomeCallMeTim Thank you for the link. – user264745 Dec 14 '23 at 13:16
  • @user264745 Sorry I made a fatal typo. The "not" should be deleted in "if $ab$ is $\underline{\text{not}}$ a zero divisor,". And as others have commented, many textbooks accept the convention that a "zero-divisor" is an either side one. – Asigan Dec 14 '23 at 15:01

2 Answers2

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The statement is intended for the set of elements which are "not a zero divisor on either side" (also sometimes called (left and right) "regular" elements).

If $a,b$ are both regular, we want to show $ab$ is regular too. Check both sides: let $c\in R$ and check that $cab=0$ or $abc=0$ implies $c=0$. Since $a$ and $b$ are regular on the left, $cab=0$ implies $ca=0$ implies $c=0$. Since $a$ and $b$ are regular on the right, $abc=0$ implies $bc=0$ implies $c=0$. Therefore $ab$ is regular on both sides, and is in your multiplicative set.

The problem in what you've written might be here:

Assume towards contradiction, $∉$. That is, $$ is left and right zero divisor.

From this statement one would have to infer that members of $S$ are either not a left zero divisor or not a right zero divisor. Or said another way, that $S$ is the union of right regular elements with left regular elements. But that set is not always multiplicatively closed. The link gives a standard example of a ring with elements $a,b$ such that $ab=1$ (so that both would be in the union) but $ba$ is an idempotent other than $1$ (so that it is a zero divisor on both sides using $1-ba$.)

rschwieb
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Following is definition of zero divisor given in book:

A nonzero element $a$ in a ring $R$ is said to be a left [resp. right] zero divisor if there exists a nonzero $b\in R$ such that $ab = 0$ [resp. $ba = 0$]. A zero divisor is an element of $R$ which is both a left and a right zero divisor.

It is easy to verify that a ring $R$ has no zero divisors if and only if the right and left cancellation laws hold in $R$; that is, for all $a,b,c\in R$ with $a\neq 0$, $$ab=ac \text{ or } ba=ca \implies b=c$$

If we assume zero divisor as both left and right zero divisor, then $\Rightarrow$ part of verification doesn’t hold. So I guess definition of zero divisor contain typo.

user264745
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  • Hungerford only uses multiplicative subsets in the context of commutative rings (despite the more general definition), so he was just being careless in the example (not in the definition of zero divisors). – Arturo Magidin Dec 14 '23 at 22:10