Timeline for Is the normalisation of an integral noetherien dimension one ring a finite morphism?

8 events

when toggle format	what		by	license	comment
Dec 17, 2011 at 0:33	vote	accept	name
Dec 17, 2011 at 0:33	vote	accept	name
Dec 17, 2011 at 0:33
Dec 16, 2011 at 18:54	comment	added	Dustin Cartwright		@Parsa, the Krull-Akizuki theorem tells you that $B$ is Noetherian, which is weaker than being finite over $A$.
Dec 16, 2011 at 18:27	answer	added	Mahdi Majidi-Zolbanin		timeline score: 4
Dec 16, 2011 at 18:08	answer	added	Qing Liu		timeline score: 7
Dec 16, 2011 at 16:30	comment	added	Mahdi Majidi-Zolbanin		A note: when $A$ is local, if $A\rightarrow B$ is a finite morphism, where $B$ is the integral closure of $A$ in $K$, then $B$ is a semi-local Dedekind domain, and hence it's a PID.
Dec 16, 2011 at 16:24	comment	added	Parsa		Isn't this a consequence of the Krull-Akizuki theorem?
Dec 16, 2011 at 15:03	history	asked	name	CC BY-SA 3.0