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user26857
  • Member for 12 years, 7 months
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zero homology of augmented Koszul complex implies the sequence is regular?
Actually Bourbaki proved that, in general, from $H_1(\underline{\bf x},M)=0$ one can obtain that the sequence $\underline{\bf x}$ is $M$-quasi-regular. Now, if we want to be $M$-regular need to impose extra-conditions, the most common being $R$ noetherian, $M$ finite and the sequence contained in the Jacobson radical.
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zero homology of augmented Koszul complex implies the sequence is regular?
@Manos Sorry, I've thought that $I=(x(y-1),z(y-1))$. Please read my comment using this notation.
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zero homology of augmented Koszul complex implies the sequence is regular?
+1. Your example was the first that crossed my mind too (as a counterexample to Manos' question) since it is known as having a bad behaviour with respect to grade, that is, grade of $I$ is 1 while the grade of $I+yR$ is 3.
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zero homology of augmented Koszul complex implies the sequence is regular?
Matsumura's Corollary is wrong and the example given by @Youngsu shows this pretty clear.
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Reject
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Why Cohen-Macaulay rings have become important in commutative algebra?
It would be nice to know who introduced this terminology and when (Rees, Serre, Auslander-Buchsbaum, or earlier)?
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Depth of ideals in a commutative ring
I've make it clear the enounce.
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Approve
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Depth of intersection
Edited by texifing.
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Approve
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If the quotient of a local ring is regular, does that imply that the original ring must be regular?
@QiL'8 Maybe I'm missing something: in your case, if, for instance, $r=1$ and $R/(x)$ is regular, then the embedding dimension of $R/(x)$ equals $\dim R/(x)=\dim R-1$. I think now it follows that $x\notin\mathfrak m^2$. (What I want to say is that in the end the elements $x_i$ turn out to be outside of $\mathfrak m^2$, although you are not assuming this.)
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Local ring $(R,\mathfrak m)$ such that $\mathfrak m^2$ is the unique minimal ideal
Change the title to reflect better the question and some other small improvements
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