Timeline for Is simplicity preserved under completion of the base ring?

7 events

when toggle format	what		by	license	comment
May 10, 2011 at 15:50	comment	added	TonyS		Ah, okay. So in general $\widehat{R}$ is the completion of $R$ with respect to $rad(R)$? And in our our case this is isomorphic to $\widehat{A}\otimes_A R$. I didn't know that. You can post this as an answer, then i can click this to be the accepted answer :-).
May 10, 2011 at 14:54	comment	added	user91132		Since $R$ is module finite over $A$, $\widehat{R} \cong \widehat{A} \otimes_A R$. Hence $\widehat{A} \otimes_A M \cong \widehat{A} \otimes_A (R \otimes_R M) \cong (\widehat{A} \otimes_A R) \otimes_R M \cong \widehat{R} \otimes_R M$. So the natural map $M \to \widehat{R} \otimes_R M$ that sends $m \to 1 \otimes m$ is also an isomorphism, and it's $R$-linear because $r.(1 \otimes m) = r\otimes m = 1 \otimes rm$.
May 10, 2011 at 14:44	comment	added	TonyS		Yes, thanks. That is what i meant, i modified the question and added some of my thoughts on the problem.
May 10, 2011 at 14:41	history	edited	TonyS	CC BY-SA 3.0	added 869 characters in body; edited tags
May 9, 2011 at 21:31	comment	added	Pete L. Clark		@Tilman: the latter, I should think.
May 9, 2011 at 21:23	comment	added	Tilman		Are you asking if the cardinalities of the iso classes of simple left $R$-modules and simple left $\hat R$-modules are the same? Or are you asking if the map $M \mapsto \hat R \otimes_R M$ induces such a bijection?
May 9, 2011 at 15:33	history	asked	TonyS	CC BY-SA 3.0