Timeline for The structure of the module of Kähler differentials of R[[x]] over R
Current License: CC BY-SA 2.5
6 events
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| Apr 14, 2010 at 15:20 | comment | added | BCnrd | Martin, then to paraphrase Clinton, it all depends on what the meaning of "it" is (for last sentence in your answer): $\Omega^1_{ R[[x]]/R}$ is what it is, regardless of what modules we map it into. Like considering a non-complete local noetherian ring $A$ and its local maps into complete local noetherian rings. That functor on the category of complete guys is represented by completion of $A$. I wouldn't say this "shows that it is complete", since $A$ isn't complete. So best to use a different notation, such as $\widehat{\Omega}$, when considering module of cont. Kahler difftls. | |
| Apr 13, 2010 at 22:36 | comment | added | Martin Brandenburg | This is exactly what I meant with "if we restrict to ... modules". By that I mean that we change the category in which the universal property holds. | |
| Apr 13, 2010 at 13:55 | comment | added | BCnrd | The end of this answer is not quite right: the module of "continuous" Kahler differentials is free of rank 1, and the given argument shows (for $R$ a $\mathbf{Q}$-algebra) that this is the $x$-adic completion of the module of "algebraic" differentials (for $R[[x]]$ as an $R$-algebra viewed discretely). The "algebraic" module is indeed pathological, but since it is not in the category of complete and separated modules we cannot expect to "know" it from knowing how it maps into objects of that type (i.e., the Yoneda nonsense doesn't adapt without replacing module with $x$-adic completion). | |
| Apr 13, 2010 at 12:38 | comment | added | Harry Gindi | Fixed LaTeX. Also, this is a great answer! | |
| Apr 13, 2010 at 12:38 | history | edited | Harry Gindi | CC BY-SA 2.5 |
Fixed LaTeX
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| Apr 13, 2010 at 12:02 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |