Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For any $A$-algebra $B$ ( commutative ring with 1 ), we have the existence of $\Omega_{B/A}$, the module of relative differentials of $B$ over $A$, which can be defined by an universal property. In the case $A = k$ being a field and $B = k[[X]]$ being the formal power series ring over $k$, $\Omega_{B/A}$ is not always a finite $B$-module. ( And I don't know if it is a free $B$-module. ) For example, if char$(k) = 0$, since $ k[[X]] $ has a infinite subset whose elements are algebraic independent over $k$, one can show that $\Omega_{B/A}$ is not a finite $B$-module. I have seen another notion in a book as following: the "universal finite differential module" $\Omega^f_{B/A}$ is a $B$-module with an $A$-derivation $d : B \rightarrow \Omega^f_{B/A}$ such that for any $A$-derivation $d^{'} : B \rightarrow M$ with $M$ being a finite generated $B$-module, there exists a $B$-module homomorphism $\phi : \Omega^f_{B/A} \rightarrow M$ such that $ d^{'} = \phi \circ d$. With this definition, one can show that, in the case of formal power series ring, $\Omega^f_{B/A}$ is a free $B$-module of rank $1$ with $dX$ as a basis, and for any $f \in k[[X]], df = f' dX$, here $f'$ is defined by the natural way, which is a result I can't deduce for $\Omega_{B/A}$. One need to use Krull Intersection Theorem or the structure theorem for finitely generated modules over a $PID$.

Now I try to compute $\Omega^f_{B/A}$ for $B = k ((X))$, the field of quotients of $k[[X]]$, but I can't get the result. The problem is that the $A$-derivation $d'$ is not necessary continuous with respect to the $(X)$-adic topology.

My question is that: do we have $\Omega^f_{B/A}$, for $B = k((X))$, is a free $k((X))$-module of rank $1$ with $dX$ as a basis?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

This is not a direct answer to your question, but a few comments about the topic.

First, in general it is not true that $\Omega^1_{B/A}$ is a free module. In fact, assuming $A \to B$ is flat and finitely generated, $\Omega^1_{B/A}$ is locally free if and only if $A \to B$ is a smooth ring map.

If in your situation (or any other adic formally smooth situation) you take the completion of $\Omega^1_{B/A}$ then you are guaranteed to get a projective module of the correct rank (and in your particular case you would indeed get the free module of rank 1 you expected). This is proved in EGA 0.IV, theorem 20.4.9. I think that the module of finite derivations should coincide with this completion. I recommend you check the book "Kahler Differentials" by Ernst Kunz which discusses the module of universal finite derivations.

Edit: remark 1.8 in http://arxiv.org/PS_cache/alg-geom/pdf/9510/9510007v4.pdf claims that indeed this completion coincides with the module of universal finite derivations, thus providing a positive answer to your question.

share|improve this answer
    
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 $\Omega^{f}_{B/A}$. ) –  user565739 Feb 18 '11 at 23:31
    
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} –  user565739 Feb 18 '11 at 23:41
3  
"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. –  Laurent Moret-Bailly Feb 19 '11 at 9:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.