Timeline for Binary operation on subsets of rings
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2015 at 2:58 | answer | added | Pace Nielsen | timeline score: 1 | |
| Dec 30, 2012 at 15:34 | comment | added | Todd Trimble | Oh, of course you're right, shatich: it's associative. I made a silly mistake; my apologies. | |
| Dec 30, 2012 at 7:15 | comment | added | user30230 | @David White: Thats what you think and you might be wrong. | |
| Dec 30, 2012 at 7:13 | comment | added | user30230 | @Todd Trimble: yes, $*$ is associative. | |
| Dec 30, 2012 at 3:10 | comment | added | Benjamin Steinberg | I think the answer to your question is that there is no "natural operation" a commutative semigroup of idempotents is automatically a meet-semilattice and inside of a ring forms a boolean algebra. You are trying to take the additive part of the corresponding boolean ring. | |
| Dec 30, 2012 at 3:08 | comment | added | Benjamin Steinberg | Don't you mean x+y-xy? You are supposed to think of + as union and product as intersection and you are trying to construct the symmetric difference. | |
| Dec 30, 2012 at 0:07 | comment | added | Gerhard Paseman | Note that for such an x, xx is an idempotent, so you already have such an operation on a subset of T. Gerhard "Ask Me About System Design" Paseman, 2012.12.29 | |
| Dec 29, 2012 at 23:41 | comment | added | Todd Trimble | Are you certain the operation that you named $\ast$ is associative? | |
| Dec 29, 2012 at 23:26 | comment | added | David White | I think you've asked too many questions too quickly. This is like 3 or 4 already today. I'd normally restrict myself to 1 per day, so I could actually respond to comments people leave and think through the answers they give. Also, if you post too many questions too quickly you'll have lower chances of all being answered. | |
| Dec 29, 2012 at 23:15 | history | asked | user30230 | CC BY-SA 3.0 |