The structure of the module of Kähler differentials of R[[x]] over R

Question

It seems that $\Omega_{R[[x]]/R}^1$ is rather big. For example, take $R$ to be the rational numbers.

We can see that $\Omega_{R((x))/R}^1$ is a $R((x))$-vector space of infinite rank. As by results from Matsumura's book, a basis of $\Omega_{R((x))/R}^1$ coincides with a transcedence basis of $R((x))$ over $R$----remember that we take $R$ to be the rationals which is of char $0$. Yet this has nothing to do with the torsion in $\Omega_{R[[x]]/R}^1$. And what can we say for structures of $\Omega_{R[[x]]/R}^1$ for general $R$'s?

Answer 1

• If $R$ is a field of characteristic $0$, then the differentials of a transcendence basis of $R((x))$ over $R$ constitute a basis of $\Omega^1_{R((x))/R}$ over $R((x))$.

In this sense this module is big: If $d$ denotes the transcendence degree, then we have $|R|^{\aleph_0} = max(|R|,\aleph_0,d)$. Thus if $R$ is countable, we have $d = |R|^{\aleph_0}$. But I don't know $d$ if $|R| = |R|^{\aleph_0}$. Since $\Omega^1$ commutes with localization, this also shows that $\Omega^1_{R[[x]]/R}$ is very big.

• There is an exact sequence of $R[[x]]$-modules $0 \to \cap_ {n \geq 1} x^n \Omega^1_{R[[x]]/R} \to \Omega^1_{R[[x]]/R} \to R[[x]] \to 0$. Here the map on the right is induced by the formal derivative $R[[x]] \to R[[x]]$.

For a proof (thanks owk!), we use the exact sequence of $A/I$-modules $I/I^2 \to \Omega^1_{A/R} \otimes A/I \to \Omega^1_{(A/I)/R} \to 0$, where $A$ is a $R$-algebra and $I \subseteq A$ an ideal.

If we put $A=R[[x]]$ and $I=(x^n)$, it follows that $0 \to x^n \Omega^1_{R[[x]]/R} \to \Omega^1_{R[[x]]/R} \to \Omega^1_{R[[x]]/x^n / R} \to 0 (*)$ is exact. If we put $A = R[x], I = (x^n)$, we get a exact sequence $0 \to x^n R[x] \to R[x] \to \Omega^1_{R[[x]]/x^n / R} \to 0$. Using this, take the inverse limit of $(*)$ and get the desired sequence.

• If $M$ is a $R[[x]]$-module which is complete and separated with respect to the $x$-adic topology, then the map $Hom(\Omega^1_{R[[x]]/R},M)=Der_R(R[[x]],M) \to M, \delta \mapsto \delta(x)$ is bijective. For a proof, remark that the leibniz equality ensures that $\delta$ is continuous.

I think that $\Omega^1_{R[[x]]/R}$ is a rather pathological object. But the last remark shows that it is simple free of rank $1$ (as in the polynomial case) if we restrict to complete and separated modules, which also fits to $R[[x]]$ as $R$-module.

Answer 2

In the adic case, when things are rarely finitely generated, what people are usually working with is the - universally finite module of differentials. A derivation $d:R\rightarrow M$ is called finite if $RdR$ is a finitely generated $R$-module. Universally finite derivation is a derivation $d$ for which any other finite derivation could be obtained by composing $d$ with a linear map. Then one defines $\Omega^1_{R/k}$ to be $RdR$ where $d$ is a Universally finite derivation. For example, if $char k = 0$, and $R = k[[x_1,\dots,x_n]]$, then $\Omega^1_{R/k} = \oplus_{i=1}^n Rd x_i$.

More details on universally finite derivations can be found at the book "Kahler Differentials" by Ernst Kunz.