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Harry Gindi
Harry Gindi
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  • 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.

  • 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.

  • 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.

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Martin Brandenburg
Martin Brandenburg
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  • 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.