Timeline for Rings in which every non-unit is a zero divisor
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 22, 2018 at 19:34 | comment | added | Pierre-Yves Gaillard | See Atiyah & MacDonald, Exercise 3.9, parts (ii) and (iii), p. 44. | |
| May 28, 2014 at 0:30 | comment | added | CPM | I am interested in a property for a ring $R$ called strongly associate. $R$ is strongly associate if $(a)=(b)$ implies $a= \lambda b$ for some unit $\lambda$. This is a properly weaker property than domainlike, presimplifiable, quasi-local properties. Artinian rings and principal ideal rings both satisfy this property. Hence, $\mathbb{Z}/n\mathbb{Z}$ as well. I was wondering if the property here that $R= Z(R) \cup U(R)$ is enough to conclude that $R$ is strongly associate. Seems unlikely, but I was unable to come up with an example. | |
| May 18, 2012 at 10:18 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Oct 20, 2010 at 21:30 | comment | added | Martin Brandenburg | The characterization with total quotient rings is a) trivial, b) useless when we actually test the property. I wonder about the upvotes ;). @Erman: I also don't know if your ring $R$ is a iterated (product or directed union) of artinian rings. Perhaps as a first step we have to show that every subring of $R$ is noetherian? | |
| Oct 18, 2010 at 22:24 | comment | added | Pete L. Clark | +1: I think this is indeed the best possible answer. | |
| Oct 18, 2010 at 21:01 | vote | accept | lhf | ||
| Oct 18, 2010 at 21:00 | comment | added | lhf | I've found "full quotient ring" in MathSciNet which it's probably the same as "total rings of fractions" since it is mentioned in en.wikipedia.org/wiki/Total_quotient_ring. | |
| Oct 18, 2010 at 18:57 | comment | added | Daniel Erman | Nice answer. Regarding the question at the end of your response, here's an example which might be relevant: Let $R=k[x,y]_{(x,y)}/(x^2,xy)$. Then $R$ is a local ring with maximal ideal $(x,y)$. It is not Artinian, since it has dimension $1$. Since $R$ is a local ring, the non-units in $R$ are precisely the elements in $(x,y)$. And since $(x,y)$ is an associated prime of $R$, it follows that every element of $(x,y)$ is a zero-divisor. But it's not clear to me how to check if $R$ is/is not the directed union of Artinian rings. | |
| Oct 18, 2010 at 15:40 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Oct 18, 2010 at 15:30 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |