Timeline for Is there any non-commutative ring such that every element other than the identity is a zero divisor?
Current License: CC BY-SA 4.0
17 events
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| Sep 26, 2022 at 5:32 | comment | added | Salvo Tringali | @SidharthGhoshal The additive group of the ring need be abelian (that's part of the def of a ring): It's the multiplicative monoid that need be non-commutative. | |
| Sep 25, 2022 at 22:39 | comment | added | Sidharth Ghoshal | So can the underlying group operation be non commutative? or must that be commutative and only the ring operation gets to be non commutative? | |
| Jul 2, 2021 at 13:59 | vote | accept | Salvo Tringali | ||
| Jun 20, 2021 at 20:58 | history | became hot network question | |||
| Jun 20, 2021 at 20:34 | comment | added | Benjamin Steinberg | The problem seems to reduce to does there exist a primitive unital ring where every non-identity element is a two-sided zero divisor. First note that if every nonidentity element of $R$ is a zero divisor, then the same is true for $R/I$ for any ideal $I$. Since $J(R)=0$ (the radical), $R$ is a subdirect product of primitive rings and hence is commutative iff each of these primitive quotients are. I don’t see how to use Jacobson’s density theorem to get a contradiction to noncommutativity if the primitive ring is not artinian. | |
| Jun 20, 2021 at 20:28 | comment | added | JoshuaZ | @Wojowu Yeah, good point. Units really wreck this approach. | |
| Jun 20, 2021 at 19:24 | comment | added | Wojowu | I feel like units are going to very annoying to deal with here. A unit in a ring $R$ can't become a zero divisor in any overring $R'$, so an argument like that of JoshuaZ can't work. They also mess up any attempt to work with nilpotents - if $x$ is nilpotent, then $1+x$ is automatically a unit, so no attempt like that in the (now deleted) answer of Donu Arapura can work either. | |
| Jun 20, 2021 at 16:41 | comment | added | JoshuaZ | Does this work? Given a finite non-commutative ring $R$, we define $R_0=R$, and $R_i=R_{i-1}[a_1,a_2,a_3 \cdots...]$ where the $a_i$ are defined to be the be elements we adjoin to make every element in $R_{i-1}$ a zero divisor. (This requires some listing of $R_i$ which may be infinite, but since it is countable this isn't too bad in terms of the required set theory). We then set $R_\infty$ to be the union of the $R_i$. | |
| Jun 20, 2021 at 16:35 | answer | added | Salvo Tringali | timeline score: 12 | |
| Jun 20, 2021 at 13:24 | comment | added | Benjamin Steinberg | Sorry I missed that part of the OP. | |
| Jun 20, 2021 at 12:50 | comment | added | Salvo Tringali | @BenjaminSteinberg Sorry, I'm not sure to understand your comment. The OP cites a paper in the AMM where it's shown that any right artinian $\mathcal O$-ring is commutative. And yes, the Jacobson radical of any $\mathcal O$-ring is trivial. So what? I'm missing the point, I think. Clearly, you don't mean that an $\mathcal O$-ring is necessarily non-artinian. Do you mean that a non-commutative $\mathcal O$-ring is necessarily non-artinian and this can proved in a more direct way than done in the aforementioned AMM paper? | |
| Jun 20, 2021 at 12:42 | comment | added | Benjamin Steinberg | Such a thing would have to not be Artinian since it has a zero Jacobson radical | |
| Jun 20, 2021 at 10:45 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
clarified the "zero divisor" means "two-sided zero divisor" (upon request of YCor)
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| Jun 20, 2021 at 10:44 | comment | added | Salvo Tringali | Yes, let me make it clear in the OP. | |
| Jun 20, 2021 at 10:44 | comment | added | YCor | By "zero divisor" you mean "both left and right zero divisor"? | |
| Jun 20, 2021 at 10:38 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
added 1 character in body
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| Jun 20, 2021 at 10:33 | history | asked | Salvo Tringali | CC BY-SA 4.0 |