TITLE: Suppose all the Associated Primes are Minimal.
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.