Timeline for An example of a commutative ring which is not SIP
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15 at 13:56 | comment | added | Ali Taghavi | Do I understand the definition of SIP correctly?: If $N_1$ and $N_2$ are complemented submodules of M then their intersection is a complemented submodule again | |
| Aug 14 at 14:52 | comment | added | Hussein Eid | Thanks. This is a good answer. But excuse me, I want to ask is every $R$-submodule of a commutative ring $R$ is an SIP-module? (i.e., is every ideal of $R$ is also an SIP-module?) | |
| Aug 14 at 14:47 | vote | accept | Hussein Eid | ||
| Aug 14 at 14:41 | comment | added | rschwieb | One can prove quite elementarily that for central idempotents $e, f$, we have $eR\cap fR=efR$. Since summands are precisely of that form, that settles things. | |
| Aug 14 at 13:33 | answer | added | Dave Benson | timeline score: 4 | |
| Aug 14 at 12:49 | history | asked | Hussein Eid | CC BY-SA 4.0 |