Timeline for rings in which every element is a sum of two commuting idempotents
Current License: CC BY-SA 3.0
23 events
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| Nov 26, 2018 at 0:22 | comment | added | darij grinberg | @WillSawin: Ah! I didn't realize you could do the invertibility check already in $\mathbb{F}_3\left[a\right] / \left(a^3-a\right)$. Nice trick. | |
| Nov 25, 2018 at 23:46 | comment | added | Will Sawin | @darijgrinberg If we know $a^3=a$ then to check a polynomial in $a$ is a unit, it suffices to check it is invertible for $a=0, 1, -1$ as then it is invertible in the formal ring $\mathbb F_3[a]/(a^3-a)$. One can easily check that for these polynomials. It should also be possible to check directly that they are their own inverses by squaring and dividing by $a^3-a$. | |
| Nov 25, 2018 at 20:36 | comment | added | darij grinberg | I am slow today. Why is every element a sum of two units in the characteristic-$3$ case? | |
| Nov 25, 2018 at 20:34 | comment | added | darij grinberg | Oh, I see why the ring decomposes. Namely, the Chinese Remainder Theorem factors the ring $\mathbb{Z} / \left(n+1\right)! \mathbb{Z}$ into a product of $\mathbb{Z} / p^k \mathbb{Z}$'s; if we take the idempotents in this ring corresponding to this factorization and send them canonically into our ring $R$, then the resulting idempotents in $R$ split $R$ into a corresponding product of $\mathbb{Z} / p^k \mathbb{Z}$-algebras. | |
| Nov 25, 2018 at 20:31 | comment | added | darij grinberg | Hi Will, I've just realized that the question I've posted at math.stackexchange.com/questions/3013336 is rather close to the question you've drive-by-answered in your EDIT. But I don't understand how you answered it. Why does the ring decompose as a product of $p$-power parts? (As a $\mathbb{Z}$-module, it does decompose this way, because the Chinese Remainder Theorem factors the ring $\mathbb{Z} / \left(n+1\right)! \mathbb{Z}$; but as a ring?) And how do you get $z^p - z = 0$ ? | |
| Sep 19, 2013 at 6:50 | vote | accept | CommunityBot | ||
| Sep 18, 2013 at 22:57 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Sep 18, 2013 at 22:02 | comment | added | user40021 | The first two lines are very very nice. I wonder what happens if every element is a sum of three mutually commuting idempotents. Do you have any idea? | |
| Sep 18, 2013 at 21:32 | comment | added | Harry Altman | Also, it's probably worth noting at the end that the converse holds: Given such a ring as $R\times S$, where $R$ is characteristic 2 and $S$ is characteristic 3, we can decompose $(x,y)$ as $(x,y^2)+(0,y-y^2)$, and both summands here are idempotent. | |
| Sep 18, 2013 at 21:30 | comment | added | Harry Altman | A question: Why is $1-a^2$ invertible? I see how to invert $a^2+a-1$, since $(a^2+a-1)(1-a-a^2)=1$, but not $1-a^2$. (Of course ultimately this is irrelevant as we can also show commutativity by other means.) | |
| Sep 18, 2013 at 21:07 | comment | added | Benjamin Steinberg | @user40021, I give the proof in the comments below my answer. | |
| Sep 18, 2013 at 20:55 | comment | added | user40021 | How do you conclude that "the ring is a product of a ring of characteristic 3 and a ring of characteristic 2" ? | |
| Sep 18, 2013 at 18:36 | comment | added | Benjamin Steinberg | Ok, I found the trivial proof. | |
| Sep 18, 2013 at 18:30 | comment | added | Benjamin Steinberg | @ClémentdeSeguinsPazzis, True. Now we need the one line proof that avoids going through reduced rings etc. | |
| Sep 18, 2013 at 18:23 | comment | added | Clément de Seguins Pazzis | It is noteworthy that the property is actually equivalent to having $\forall z \in R, \; z^3=z$ ! | |
| Sep 18, 2013 at 18:10 | comment | added | Benjamin Steinberg | @WillSawin, I just gave what I assume is an equivalent answer at the same time. | |
| Sep 18, 2013 at 18:07 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Sep 18, 2013 at 18:01 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Sep 18, 2013 at 17:54 | comment | added | Benjamin Steinberg | Ok, I had falsely assumed every subdirect product is direct for these rings. (I fixed my comment). | |
| Sep 18, 2013 at 17:50 | comment | added | Clément de Seguins Pazzis | Benjamin, there are subdirect products of division rings that are not direct products of division rings. Consider the subring of $(\mathbb{Z}/3)^\mathbb{N}$ consisting of the ultimately constant sequences. It is not a direct product of copies of $\mathbb{Z}/3$. | |
| Sep 18, 2013 at 17:46 | comment | added | Benjamin Steinberg | In fact, rings satisfying z3=z are I believe subdirect products of division rings. Since the only characteristic 3 division ring where each element is a sum of 2 commuting idempotents is Z/3, we get that the answer is products of copies of Z/2 and Z/3. | |
| Sep 18, 2013 at 17:29 | comment | added | Clément de Seguins Pazzis | If the ring has characteristic 3, then your equation yields $z^3=z$ for all $z$ in the ring, which classically yields that the ring is commutative. | |
| Sep 18, 2013 at 16:31 | history | answered | Will Sawin | CC BY-SA 3.0 |