Timeline for Semigroup ideals of a ring or an algebra
Current License: CC BY-SA 4.0
11 events
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| Nov 13, 2021 at 11:11 | comment | added | Onur Oktay | Just to add the reference: "Let $R$ be a simple ring that is not a division ring. Then, $R$ is purely infinite if and only if $RxR=R$ for all nonzero $x\in R$ if and only if $R$ is $0$-simple as a semigroup." dx.doi.org/10.1023/A:1016358107918 | |
| Nov 12, 2021 at 16:43 | comment | added | Benjamin Steinberg | No I meant for a simple ring a being 0-simple as a semigruop is equivalent to being a division ring or purely infinite. | |
| Nov 12, 2021 at 16:31 | comment | added | Onur Oktay | @BenjaminSteinberg Thank you for your kind reply. Do I understand the converse correctly: "if every nonzero semigroup ideal spans the algebra R, then R is simple" ? I doubt it, since the sets of the form $RxR$, for some $x\in R$, are semigroup ideals of R that span principal ideals. Thus, R possess no nontrivial principal ideals, so it must be simple. On the other hand, clearly not all spanning sets can be semigroup ideals. Am I missing something crucial? | |
| Nov 12, 2021 at 13:15 | comment | added | Benjamin Steinberg | Let me point out for a simple ring every nonzero semigroup ideal spans the algebra but it is not usually the case that the converse holds. The converse holds for division rings and purely infinite simple rings. | |
| Nov 11, 2021 at 15:06 | comment | added | Onur Oktay | @SalvoTringali Thank you for your reply and for the book reference. | |
| Nov 11, 2021 at 14:20 | comment | added | rschwieb | Crossposted | |
| Nov 11, 2021 at 11:16 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Nov 11, 2021 at 7:23 | comment | added | Salvo Tringali | You may want to read about "multiplicative ideal theory". In particular, I would recommend to look at Halter-Koch's book on ideal systems. | |
| Nov 11, 2021 at 3:55 | history | edited | LSpice | CC BY-SA 4.0 |
R -> $R$
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| Nov 11, 2021 at 3:54 | comment | added | LSpice | It seems to me that this is just saying that $B$ is a sub-bi-$R$-module of $R$. EDIT: Oh, you're not assuming that $B$ is closed under addition. In that case, as you say, it seems that the additive structure of $R$ is playing no role, so this is really a question about semigroups. | |
| Nov 11, 2021 at 3:27 | history | asked | Onur Oktay | CC BY-SA 4.0 |