Timeline for rings in which every element is a sum of two commuting idempotents
Current License: CC BY-SA 3.0
19 events
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| Sep 18, 2013 at 21:15 | comment | added | Benjamin Steinberg | @user40021, look at Will's argument. It is not hard to see that in a subdirect product of copies $Z/3$ each prime ideal $P$ is maximal and $R/P\cong Z/3$. So we can identify $R$ with continuous functions $R\to Z/3$ via the Gelfand transformation. I.e., we can think of Spec(R) as Hom(R,Z/3) with the topology of pointwise convergence and the each element of $r$ gives a continuous map by evaluation. | |
| Sep 18, 2013 at 21:09 | comment | added | user40021 | @EricWofsey Would you explain how do you consider elements as continuous maps? That is an interesting point of view but I cannot see how you are doing that. | |
| Sep 18, 2013 at 21:09 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Sep 18, 2013 at 21:06 | comment | added | Benjamin Steinberg | My so-called ``trivial proof'' is the only one that does not use the axiom of choice (or at least the ultrafilter theorem). To get a subdirect decomposition or to prove that reduced is equivalent to the intersection of all prime ideals requires some form of choice (namely the ultrafilter lemma). | |
| Sep 18, 2013 at 21:04 | comment | added | Benjamin Steinberg | If the characteristic of $R$ is 2 or 3, we are done. If the characteristic is $6$, then $1=3+4$ and $3,4$ are central idempotents of $R$. Thus $R\cong 3R\times 4R$. The ring $3R$ has characteristic $2$ (note the identity of $3R$ is 3) and $4R$ has characteristic $3$ (it has identity 4). | |
| Sep 18, 2013 at 20:57 | comment | added | user40021 | How do you conclude that "Thus we have $R$ is a product $R_1\times R_2$ of a ring $R_1$ of characteristic $2$ and $R_2$ of characteristic $3$" ? | |
| Sep 18, 2013 at 19:03 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Sep 18, 2013 at 19:02 | comment | added | Benjamin Steinberg | I don't think my argument uses commutativity anymore. The direct product decomposition comes because 1 is a sum of orthogonal idempotents in Z_6 and z commutes with its square. | |
| Sep 18, 2013 at 18:53 | comment | added | Benjamin Steinberg | @WillSawin, Thanks. My answer basically came from interacting with you and Clement and should be considered all our answers. | |
| Sep 18, 2013 at 18:52 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Sep 18, 2013 at 18:51 | comment | added | Will Sawin | The proof can be simplified in one direction. My equation $z^3-3z^2+2z=0$ is just $(z-1)^3=(z-1)$. | |
| Sep 18, 2013 at 18:48 | comment | added | Benjamin Steinberg | @EricWofsey, this is Will's version of the argument. My first version was to reduce to the finitely generated case, where the ring is finite and no Zariski topology is needed. Now I have the elementary proof. | |
| Sep 18, 2013 at 18:46 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Sep 18, 2013 at 18:33 | comment | added | Eric Wofsey | Any subdirect product of $\mathbb{Z}/3$ has this property. Any element $z\in R$ can be considered as a continuous function $z:\operatorname{Spec} R\to \mathbb{Z}/3$. For $i\in\mathbb{Z}/3$, let $A_i=z^{-1}(\{i\})$. Then these sets are clopen, and $z=1_{A_1\cup A_2}+1_{A_2}$. | |
| Sep 18, 2013 at 18:10 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Sep 18, 2013 at 17:57 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Sep 18, 2013 at 17:55 | comment | added | Benjamin Steinberg | ok, I was hasty. I was thinking of finite subdirect products. Let me fix. | |
| Sep 18, 2013 at 17:53 | comment | added | Todd Trimble | I am confused. Not every Boolean algebra is a direct product of copies of $\mathbb{Z}/2$. (Or maybe I'm not sure what "direct" in direct product means.) | |
| Sep 18, 2013 at 17:50 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |