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?


1 Answer 1


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.

  • $\begingroup$ 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 $\Omega^{f}_{B/A}$. ) $\endgroup$
    – user565739
    Feb 18, 2011 at 23:31
  • $\begingroup$ 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} $\endgroup$
    – user565739
    Feb 18, 2011 at 23:41
  • 4
    $\begingroup$ "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. $\endgroup$ Feb 19, 2011 at 9:06

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