Timeline for module of differentials of formal power series ring and of its field of quotiens

6 events

when toggle format	what		by	license	comment
Feb 23, 2011 at 15:39	vote	accept	user565739
Feb 19, 2011 at 9:06	comment	added	Laurent Moret-Bailly		"assuming $A\to B$ is flat and finitely generated,$\Omega_{B/A}$ is locally free if and only if $A\to B$ is a smooth ring map": of course "if" is correct, but "only if" is not: if $A$ has characteristic $p>0$, take $B=A[x]/(x^p-a)$ where $a$ is any element of $A$. This is never smooth but $\Omega_{B/A}$ is free of rank one. You have to add the condition that $\Omega_{B/A}$ has the right rank, namely the relative dimension.
Feb 18, 2011 at 23:41	comment	added	user565739		Sorry, in the above comment, I want to type: \Omega^{~}_{B/A} exists if and only if \Omega_{B/A} is finite and in this case, they are equal. So for $A=k$ and $B=k((X))$, \Omega^{~}_{B/A} doesn't exist. I assume that \Omega^{~}_{B/A} equals to \Omega^{f}_{B/A}
Feb 18, 2011 at 23:31	comment	added	user565739		Thanks, Liran. But the result is true for $k[[X]]$, but not for $k((X))$ over $k$. In fact, I took a look on Kunz's book, and in Example 11.2 in p. 172, he mentioned that if $B$ is a field, then $\Omega^{~}_{B/A}$ exists if and only if $\Omega_{B/A}$ is finite and in this case, $\Omega^{~}_{B/A} = \Omega_{B/A}$. So for $A=k$ and $B=k((X))$, $\Omega^{~}_{B/A}$ doesn't exists. ( I assume that $\Omega^{~}_{B/A}$ equals to $\Omega^{f}_{B/A}$. )
Feb 18, 2011 at 22:46	history	edited	the L	CC BY-SA 2.5	added 229 characters in body
Feb 18, 2011 at 19:32	history	answered	the L	CC BY-SA 2.5