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Let $A$ be a ring and $A[[T]]$ the formal power series over $A$. Then, one can show that $\Omega^1_{A[[T]]/A}$ is not finitely generated over $A[[t]]$. Now, in $\Omega^1_{A[[T]]/A}$ I am trying to show that the intersection $\cap_{n \geq 1} T^i \Omega^1_{A[[T]]/A} $ is non-zero in general, and that it can even be non-zero if $A$ is a field of characteristic 0.

I can show that the intersection is non-zero if $A$ is not an integral domain, but if $A$ is a field of characteristic 0, then I don't see how to proceed. Any hints or solutions?

Edit:

Possible directions to consider might be to look at the relation $A[[t]]=A[t]+(t^n)A[[t]]$ for any $n >0$. But once again, I dont' neccesarily see how to construct an element in the intersection directly! Any thoughts are welcome.

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  • $\begingroup$ For any power series $f(T)=\sum_{n\ge 0} a_nT^n$, define $f'(T)=\sum_{n\ge 1} na_nT^{n-1}$ and consider the differential $f'(T)dT -df(T)$. $\endgroup$ – Cantlog Mar 23 '14 at 21:46
  • $\begingroup$ You still have to prove that $f'dT-df\ne 0$ in general or find such an $f$. $\endgroup$ – Cantlog Mar 23 '14 at 22:09
  • $\begingroup$ @Cantlog proving that it is not zero in general shouldn't be that hard, right? It is zero if we can write $df(T) =f'(T)dT$ and this can't be true for formal power series with infinitely many non-zero coefficients (we can apply the linearity rule as many times as we want, but still we wouldn't have equality). $\endgroup$ – user101036 Mar 23 '14 at 22:18
  • $\begingroup$ Maybe it is not that hard, but it really needs a proof. Intuition is not enough. For example, if $A$ has characteristic $p$, and $f$ is a $p$-th power in $A[[t]]$, then $f'dT=df=0$ even if $f$ is not polynomial. $\endgroup$ – Cantlog Mar 24 '14 at 12:01
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    $\begingroup$ If $f'dT-df=0$ for all $f$, then $\Omega^1_{A[[T]]/A}=A[[T]]dT$, which contradicts the fact that the latter is not finitely generated over $A[[T]]$. If you agree with this, please feel free to write down a complete solution. $\endgroup$ – Cantlog Mar 24 '14 at 20:06

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