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Is there any criterion of regularity for rings in terms of jets?

More precisely: It is known that a local ring $B$ (with some hypothesis) is regular if and only if the module of differentials $\Omega_{B/k}=I/I^2$ is free of rank $\dim B$.

If we now consider $I/I^{n+1}$ with $n>1$, the regularity of $B$ still implies that $I/I^{n+1}$ is free (although its rank is greater than $\dim B$). For this it suffices to consider the exact sequence

$$0\rightarrow I^n/I^{n+1}\rightarrow I/I^{n+1}\rightarrow I/I^n\rightarrow 0$$

Using the previous result and by induction we are done.

The question is: Is there any chance for the other implication to be also true, that is, $I/I^{n+1}$ free implies $B$ regular?

I would appreciate also if somebody could give me some references on this subject!

Thanks in advance.

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    $\begingroup$ $I/I^{n+1}$ isn't annihilated by $I$ for $n \geq 1$, so isn't a $B$-module. In particular, your short exact sequence is not a s.e.s. of $B$-modules. $\endgroup$
    – Graham Leuschke
    Commented Mar 6, 2012 at 12:25
  • $\begingroup$ $I=Ker\phi$ where $\phi:B\otimes B\rightarrow B$, $\phi(b\otimes b')=bb'$. We can make $B\otimes B$ an $B$-module via multiplication on the first entry, so that $I$ and therefore $I/I^{n+1}$ is also a $B$-module. $\endgroup$
    – user20544
    Commented Mar 6, 2012 at 15:38
  • $\begingroup$ Why not the second entry? The two will not coincide. $\endgroup$
    – Graham Leuschke
    Commented Mar 6, 2012 at 18:52
  • $\begingroup$ Indeed, they will not coincide, but, why is this a problem? $\endgroup$
    – user20544
    Commented Apr 17, 2012 at 14:43

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