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Let $R$ be a Noetherian ring and $I$ a non-zero ideal of $R$. Let $x\notin I$. Could someone provide me a counterexample to the following:

$$\operatorname{ht}(I)\leq \operatorname{ht}(I+(x))\leq \operatorname{ht}(I)+1?$$

Here $\operatorname{ht}(I)$ denotes height of $I$. I know that the above holds for Cohen-Macaulay rings.

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up vote 4 down vote accepted

It's easily seen that the first inequality holds in every commutative noetherian ring.

Instead the second inequality fails: take $R=K[X,Y,Z]/(XY,XZ)$ and $I=(y,z)$ (here small letters denote the residue classes of indeterminates). Now observe that $\operatorname{ht}(I)=0$ and $\operatorname{ht}(I+(x))=2$. (Obviously $R$ is not Cohen-Macaulay.)

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Fantastic, Thanks. – messi Apr 29 '13 at 14:56
Yes, I am messi, and I would like to take back my old account, but I don't know how to do that. – messi Apr 29 '13 at 15:34

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