Timeline for Characterizing nilpotents in a ring by a universal property
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2010 at 20:30 | comment | added | Pace Nielsen | Let $R$ be the subring of $S=\mathbb{Q}$ with odd denominators. Then $R$ embeds in $S$, but the Jacobson radical maps to units in $S$. | |
| Feb 3, 2010 at 19:26 | comment | added | darij grinberg | And I still believe that the original question is correct for commutative rings. | |
| Feb 3, 2010 at 19:25 | comment | added | darij grinberg | "Since ring homomorphisms preserve the Jacobson radical". Wikipedia says that if $f:R\to S$ is a surjective ring homomorphism, then $f(J(R))\subseteq J(S)$, which is somewhat weaker. Are you sure that your answer still works then? | |
| Feb 3, 2010 at 19:19 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
added 60 characters in body; added 66 characters in body
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| Feb 3, 2010 at 19:14 | comment | added | darij grinberg | Somehow I doubt your solution. Why is "image under homomorphism never unit" equivalent to "lies in every maximal ideal"? Doesn't the homomorphism $A\to A_{\left\{1,a,a^2,...\right\}}$ (where $A_{\left\{1,a,a^2,...\right\}}$ means localization at the multiplicative subset \left\{1,a,a^2,...\right\}$) contradict this? | |
| Feb 3, 2010 at 19:06 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |