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Let $R$ be a commutative Noetherian ring with unit and let $I$ be a fixed ideal. I am sorry if the following turns out to be a very silly question.

1) Suppose $\operatorname{Ass}(R/I)$ are all minimal, then what can we say about $R/I$?

One conclusion is that $I$ has a unique primary decomposition. If $R/I$ is CM, then $\operatorname{Ass}(R/I)$ are all minimal, but I don't think the converse is true. Can someone provide a counterexample?

2) Suppose $\operatorname{Ass}(R/I)$ are all minimal and have the same height. I still dont think we can conclude that $R/I$ is CM. Can someone provide a counterexample? Theorem 2.1.6 of Bruns and Herzog, CM Rings, gives a result in this direction but it needs additional hypothesis. It says that if every ideal $I$ is generated by height $I$ elements and $\operatorname{Ass}(R/I)$ are all minimal, then $R$ is CM.

It would be nice if someone can say something nice about the ring in the special situation described in 1) and 2).

Thanks.

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1 Answer 1

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The ideals defined by 1) are called unmixed, while the ideals defined by 2) are called height unmixed. (Note that having the same height implies minimal for the associated prime ideals, so 2) implies 1).)

Unmixedness of $I$ is not enough for $R/I$ to be CM: let $R=K[X,Y,Z]$ and $I=(X)\cap(Y,Z)$. This example shows also that 1) doesn't imply 2).

Height unmixedness of $I$ is not enough for $R/I$ to be CM: let $R=K[X_1,\dots,X_5]$ and $I=(X_1,X_4)\cap(X_2,X_5)$.

However, if $R$ is local and $\dim R/I=1$, then $R/I$ is CM iff $I$ is unmixed.

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  • $\begingroup$ I've preferred to give some nontrivial examples, although for $I$ a prime ideal in a polynomial ring $R$ over a field we can have $R/I$ not CM, and $I$ is trivialy (height) unmixed. $\endgroup$
    – user26857
    May 6, 2013 at 9:30
  • $\begingroup$ I know that the 2nd example is not CM because i remember doing that in the past. I am trying to check it, but no luck yet. I am trying to find a system of parameters which does not form a regular sequence. $\endgroup$
    – messi
    May 6, 2013 at 10:01
  • $\begingroup$ @messi The second example comes from simplicial complexes. It can be shown that is not CM by elementary methods, but another method is to compute its Hilbert series and observe that this has at the numerator a negative coefficient. $\endgroup$
    – user26857
    May 6, 2013 at 10:08
  • $\begingroup$ oh ok, but i think in the past i used the method i described above. I hope i can do it using my method in the past. Thanks $\endgroup$
    – messi
    May 6, 2013 at 10:10
  • $\begingroup$ I want to confirm something, in your 2nd example can we take $R=k[x_1,x_2,x_3,x_4]$ and $I=(x_1,x_2)\cap (x_3,x_4)$? $\endgroup$
    – messi
    May 6, 2013 at 10:47

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