Timeline for Jacobson radical = intersection of all maximal two-sided ideals
Current License: CC BY-SA 2.5
10 events
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| Dec 27, 2010 at 22:03 | comment | added | Gerald Edgar | Of course, you could cite this MO page. True, but what if this MO page is changed? | |
| Dec 27, 2010 at 18:34 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Dec 26, 2010 at 18:34 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Dec 25, 2010 at 22:57 | comment | added | darij grinberg | You mean $A\diagup \mathrm{rad}\left(A\right)$ rather than $A$, right? (Otherwise, nice comment; will have a closer look at it.) | |
| Dec 25, 2010 at 22:23 | comment | added | Kevin McGerty | I think you can avoid using the description of simple $k$-algebras: you should only really need semisimplicity of $A/\text{rad}(A)$ -- once you know that then $A$ will be a (finite) direct sum of its minimal two-sided ideals, so then the maximal two-sided ideals will be the direct sum of all but one of these, so you're done. | |
| Dec 25, 2010 at 22:03 | vote | accept | darij grinberg | ||
| Dec 25, 2010 at 22:03 | comment | added | darij grinberg | Thanks, Ben, this does it, although I hoped to avoid Artin-Wedderburn. Is it really a restatement, i. e., equivalent to Artin-Wedderburn? | |
| Dec 25, 2010 at 21:58 | comment | added | darij grinberg | For Artin rings, the other containment follows from the fact that the Jacobson radical is nilpotent (which is clear by the Jordan-Hölder decomposition - here the Artinianity(sp?) comes in). | |
| Dec 25, 2010 at 21:25 | comment | added | Kevin Ventullo | This shows that Jacobson contains Brown-McCoy. Is the other containment obvious? | |
| Dec 25, 2010 at 20:16 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |