Timeline for In what sense are fields an algebraic theory?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2015 at 2:30 | comment | added | goblin GONE | @DinakarMuthiah, you can try: "the object $F$ is a field iff every morphism out either has codomain equal to the terminal object, or else its a monomorphism." Note that in the category of rings, the "fields" in this sense are precisely the simple rings, so they're more general than division rings. | |
| Mar 7, 2010 at 22:26 | vote | accept | Omar Antolín-Camarena | ||
| Mar 7, 2010 at 22:26 | |||||
| Nov 23, 2009 at 18:46 | vote | accept | Omar Antolín-Camarena | ||
| Nov 23, 2009 at 18:46 | |||||
| Oct 29, 2009 at 8:02 | comment | added | Andrew Stacey | There's probably some way to make sense of that ("Anyone saying something is impossible is more than likely to be interrupted by some idiot doing it."), but I think it would be so contrived that none of the intuition from group objects or ring objects would carry over. The problem is that the inverse is defined on the complement of a subset and that's not obvious how to generalise to an arbitrary category. | |
| Oct 29, 2009 at 0:08 | comment | added | Dinakar Muthiah | Is there a way to talk about a "field object" in a category? | |
| Oct 28, 2009 at 16:00 | comment | added | Omar Antolín-Camarena | I didn't know about meadows, thanks for telling me about them! | |
| Oct 28, 2009 at 8:05 | history | answered | Andrew Stacey | CC BY-SA 2.5 |