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Suppose $(R,+,\cdot)$ be a ring and there is two elements, $a$ and $b$ are in the ring such that $ab$ is a zero divisor.

My attempt has two parts.
1 - if $a$ or $b$ equal to $0$ ($0$ is the zero element of the ring), then it is obvious that $a.b = 0$ and $b.a = 0$. then for every $x ∈ R, a.b.x = 0$ so $a.b$ is a right zero divisor. Then $xba = 0$ so $b.a$ is right zero divisor. Similarly we can see that if ab is left zero divisor then $ba$ is right zero divisor. So for this case , if $ab$ is zero divisor, then $ba$ is also a zero divisor.(I think ring R should have at least one non-zero element so that $x≠0$, otherwise zero divisor won't have meaning in such a ring ?)

2 - if $a$ and $b$ does not equal $0$.
so from the assumptions, we know that ab is zero divisor. It means that it is both right and left divisor. So there exists some $z ∈ R, z≠ 0, s.t, abz = 0$.then if we consider the element $bz$, it results that $ba.bz = b.(abz) = 0$. Similarly , there exist some $z' ∈ R, z' ≠ 0, s.t, z'ab = 0$. Then $z'a.ba = (z'ab).a = 0$.

But it now suffices to show that $z'a ≠ 0$ and $bz ≠ 0$ so that $ba$ can be a zero divisor. How can I show that? or there is some counterexamples that shows $ba$ is not necessarily zero divisor even hen ab is a zero divisor?

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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Feb 11, 2023 at 15:20
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    $\begingroup$ Your attempt at what? You say there is a ring $R$, and two elements with a property... and that's it. The subject line is not part of the post, and your post should be understandable without it. You don't start a letter on the envelope, you don't start a post on the subject line. $\endgroup$ Feb 12, 2023 at 1:02
  • $\begingroup$ I wanted to proof this statement and tried to do it, but I failed the last part and suspected that there should be some counterexamples and tried to make up a counterexample and couldn't do it $\endgroup$
    – M.Arya
    Feb 12, 2023 at 6:12
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    $\begingroup$ @M.Arya Having an example of a ring where $ba=1$ but $ab\neq 1$ suffices. EdwardH's comment below uses this but one might overlook it while reading the details of the construction. The thing is that even when $ab\neq 1$ for sure $(ab)^2=ab$, so that $(1-ab)ab=ab(1-ab)=0$, demonstrating $ab$ is a zero divisor, even if $ba$ is the identity (and hence clearly not a zero divisor.) $\endgroup$
    – rschwieb
    Feb 12, 2023 at 14:28
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    $\begingroup$ @AnneBauval Done... $\endgroup$
    – rschwieb
    Feb 12, 2023 at 16:32

2 Answers 2

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Hint: in the free algebra $A=\Bbb Z\langle X,Y,Z\rangle,$ let $I$ be the ideal generated by $YZ$ and $ZX.$ Prove that in the quotient $A/I,$ the image of $XY$ is a zero divisor but the image of $YX$ is not.

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    $\begingroup$ So you changed it to a "Hint". I think you should say more. Answers should be complete without handwaving, and if someone asks you to justify key claims we expect hinters to do so. Please do so. $\endgroup$ Feb 11, 2023 at 17:18
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    $\begingroup$ This is a perfectly fine answer. Everybody is perfectly able to enforce his own standards of what an answer should be when writing his own answers. And claiming this to be handwaving is really below any sensible standards, really... $\endgroup$ Feb 12, 2023 at 1:39
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    $\begingroup$ @Mariano This site is meant for answers - not guesses, half-baked derivations, "hints" etc. If a user cannot provide requested details for a "hint" then it is not an answer. I've lost count of prior "hints" that were incorrecr, misleading and/or led nowhere. One should not have to read minds to evaluate an answer. $\endgroup$ Feb 12, 2023 at 3:20
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    $\begingroup$ It is actually very reassuring that you've taken up the responsibility of patrolling around answers to make sure the posters can actually provide the requested (by you) details and sprinkle all apprarences of the word hint with scare quotes. I honestly cannot imagine anyone more apt for that position. Have fun $\endgroup$ Feb 12, 2023 at 3:48
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    $\begingroup$ For example, here you are, upholding the rule invented by yourself (and quite recently!) that Posters of Hints Shalt Immediately Respond With Full Detail When Prompted Lest You Think They Be Handwaving — much in the style of challenges of honor of yore — which was agreed in the celebrated MSE Concilium. And I for one am thankful for that expended effort. Half baked derivations are the worst, the worst! $\endgroup$ Feb 12, 2023 at 5:19
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Suppose you have any ring in which $ba=1$ but $ab\neq 1$. Then in particular $(ab)^2=ab$ and $(1-ab)ab=ab(1-ab)=0$ so that $ab$ is a zero divisor. But $ba=1$ is certainly not.

There are many ways to get such a ring, but the one I like the best is the ring of linear transformations of a countable dimensional vector space. This is closely related to the other one often given using sequence shifts.

Paraphrasing the link above:

For a fixed basis $\{b_0,b_1,\ldots\}$, the "right shift" $𝑎$ sending $b_i\mapsto b_{i+1}$ and the "left shift" $𝑏$ on the basis elements sending $b_i\mapsto b_{i-1}$ for $i>0$, and $b(b_0)=0$ satisfies $ba=1$ and $ab\neq 1$ since $ab(b_0)=0$.

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  • $\begingroup$ @OP The shift / representation theory method is a standard technique to obtain a ring that is not Dedekind finite (i.e. where $xy = 1\Rightarrow yx=1$ fails). You may find it instructive to peruse prior posts on Dedekind finite rings (that may even lead to a dupe - I have not checked). $\endgroup$ Feb 12, 2023 at 19:19
  • $\begingroup$ For another example of such representation-based ring constrcutions see van der Waerden's trick, e.g. as used to construct polynomial rings here and here. $\endgroup$ Feb 12, 2023 at 20:03
  • $\begingroup$ @rschwieb Thank you for having satisfied my wish. I find it unfair that it's my answer which is (at the moment...) accepted, instead of yours. But you are probably above these pettiness. $\endgroup$ Feb 15, 2023 at 11:35
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    $\begingroup$ @AnneBauval It doesn't matter to me in a case like this. A quotient algebra is certainly another natural way to go. $\endgroup$
    – rschwieb
    Feb 15, 2023 at 15:13

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