Timeline for Derivations annihilated by powers of the augmentation ideal

7 events

when toggle format	what		by	license	comment
Nov 9, 2015 at 10:33	vote	accept	grok
Nov 8, 2015 at 9:01	answer	added	darij grinberg		timeline score: 1
Nov 4, 2015 at 22:41	comment	added	grok		Thanks a lot! Would you mind typing your answer as an answer, so you would get appropriate credit?
Nov 4, 2015 at 21:30	comment	added	darij grinberg		Actually there is an even simpler argument: First prove $\delta\left(I^{m+1}\right) \subseteq I^m$ (for instance, by induction over $m$, or by the generalized Leibniz identity as above); thus, $\delta\left(I^{m+1}\right) f = 0$. Then, argue using $\delta\left(if\right) = \delta\left(i\right) f + i \delta\left(f\right)$ for $i \in I^{m+1}$.
Nov 4, 2015 at 21:28	comment	added	darij grinberg		(Notice that $R$ needs not be commutative here, and $\varpi$ can be any two-sided ideal, not necessarily the augmentation ideal.)
Nov 4, 2015 at 21:27	comment	added	darij grinberg		To the non-bonus question: The Leibniz identity (generalized to multiple factors) yields $\delta\left(i_1 i_2 \cdots i_{m+1} f\right) = \sum_{k=1}^{m+1} i_1 i_2 \cdots i_{k-1} \delta\left(i_k\right) i_{k+1} i_{k+2} \cdots i_{m+1} f + i_1 i_2 \cdots i_{m+1} \delta\left(f\right)$ for all $i_1, i_2, \ldots, i_{m+1} \in \varpi$. Argue that both the left hand side and each addend of the sum on the right hand side are zero. Conclude that the second addend of the right hand side is zero, and thus $\varpi^{m+1} \delta\left(f\right) = 0$.
Nov 4, 2015 at 21:21	history	asked	grok	CC BY-SA 3.0