Timeline for Skew fraction fields of *-algebras
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2012 at 20:25 | comment | added | Tobias Fritz | Abstract nonsense also gives you a canonical isomorphism between $R^{op}S^{-1}$ and $(RS^{-1})^{op}$. I have expanded my answer to include this, and also switched my notation to yours. | |
| Nov 3, 2012 at 20:21 | history | edited | Tobias Fritz | CC BY-SA 3.0 |
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| Nov 3, 2012 at 10:37 | comment | added | Joakim Arnlind | Yes, I do know about the universal property, and that an $S$-inverting homomorphism between to rings give rise to a (unique) homomorphism between the corresponding fraction fields. But, I can't straighten out the details of your last comment; i.e. why $R^{op}S^{-1}$ is isomorphic to $(RS^{-1})^{op}$. I guess that one argues that $(RS^{-1})^{op}$ is a fraction field of $R^{op}$ and then uses universality to equate it to $R^{op}S^{-1}$. But why is $(RS^{-1})^{op}$ is a fraction field of $R^{op}$? Explicit check? | |
| Nov 2, 2012 at 19:49 | comment | added | Tobias Fritz | Do you know the universal property of localization? It states that if $T$ is any ring and $f:R\to T$ any ring homomorphism which sends every element of $S$ to an invertible element in $T$, then $f$ induces a unique homomorphism $R[S^{-1}]\to T$ which reproduces $f$ after composing with $R\to R[S^{-1}]$. In the case at hand, I have applied this to $R^{op}$ instead of $R$ itself, and also identified $R^{op}[S^{-1}]$ with $R[S^{-1}]^{op}$. (That these two are canonically isomorphic is again a consequence of the universal property.) | |
| Nov 2, 2012 at 11:29 | comment | added | Joakim Arnlind | Thanks! How is it that *:R^{op}->RS^{-1} induces *:(RS^{-1})^{op} -> RS^{-1} in your argument? | |
| Nov 2, 2012 at 9:21 | vote | accept | Joakim Arnlind | ||
| Nov 2, 2012 at 9:21 | |||||
| Nov 1, 2012 at 21:28 | history | answered | Tobias Fritz | CC BY-SA 3.0 |