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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?

rschwieb
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M.Arya
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    Here's a MathJax tutorial :) – Shaun Feb 11 '23 at 15:20
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    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. – Arturo Magidin Feb 12 '23 at 01:02
  • 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 – M.Arya Feb 12 '23 at 06:12
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    @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.) – rschwieb Feb 12 '23 at 14:28
  • @rschwieb Brilliant! Will you please edit it as an answer? (together with the usual ;-) example of such a ring and $a,b$). – Anne Bauval Feb 12 '23 at 14:33
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    @AnneBauval Done... – rschwieb Feb 12 '23 at 16:32
  • @M.Arya Please have a look at the tips for formatting posts in the FAQ. Using italics for math does not work well. It's as simple as wrapping most stuff in dollar signs. – rschwieb Feb 12 '23 at 16:36

2 Answers2

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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.

Anne Bauval
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    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. – Bill Dubuque Feb 11 '23 at 17:18
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    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... – Mariano Suárez-Álvarez Feb 12 '23 at 01:39
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    @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. – Bill Dubuque Feb 12 '23 at 03:20
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    Well, astound us with one of your technicolorly dazzling answers and show us how its done! Please! Enlighten us! – Mariano Suárez-Álvarez Feb 12 '23 at 03:44
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    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 – Mariano Suárez-Álvarez Feb 12 '23 at 03:48
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    @Mariano Sigh, back to your old ways of sarcastic jabs and personal attacks on those whose views differ from yours. In case you haven't noticed, there are many users who expend great effort to attempt to uphold site quality. Instead of harassing them as you do above, you should instead be thanking them for their painstaking effort. Oh for the day when that might occur... – Bill Dubuque Feb 12 '23 at 04:19
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    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! – Mariano Suárez-Álvarez Feb 12 '23 at 05:19
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    @Anne Bauval , Thanks for the answer, but I am an undergraduate student and this is my 2nd algebra course and still do not know what free algebra and Ideal are. We are just at basic stuffs like what are the rings/subrings and their properties. I was thinking about some simpler rings such as matrices and etc.. – M.Arya Feb 12 '23 at 06:15
  • @Edward H, thanks ! I finally could get it . It makes much more sense than before ! – M.Arya Feb 12 '23 at 13:26
  • @M.Arya See my comments on rschwieb's answer for more on the representation theoretic approach in the prior comment by EdwardH – Bill Dubuque Feb 12 '23 at 19:45
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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" $$ sending $b_i\mapsto b_{i+1}$ and the "left shift" $$ 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 $ab(b_0)=0$.

rschwieb
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  • @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). – Bill Dubuque Feb 12 '23 at 19:19
  • 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. – Bill Dubuque Feb 12 '23 at 20:03
  • @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. – Anne Bauval Feb 15 '23 at 11:35
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    @AnneBauval It doesn't matter to me in a case like this. A quotient algebra is certainly another natural way to go. – rschwieb Feb 15 '23 at 15:13