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Let $A$ be a Noetherian local ring and $I$ be an ideal of $A$. An element $x\in I$ is said to be an $A$-superficial element if there exists a natural number $c$ such that $(I^{n+1}:x)\cap I^c=I^n$ for all $n\geq c$.

We know that if the residue field of a Noetherian local ring $R$ is infinite then all ideal $I$ of $R$ contain an $R$-superficial element. I want an example of a ring of finite residue field with no superficial element for an ideal $I$. This is very useful in commutative algebra. For example to reduce dimension and depth.

rschwieb
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Samarendra sahoo
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  • Could you include a definition of "superficial element"? Based on 5 minutes trying to find a concise definition and not finding one, I think people would find it useful. – rschwieb Nov 08 '23 at 23:05
  • Let A be a Noetherian local ring and I be an ideal of A. x\in I is said to be A-superficial element if there exist natural number c such that (I^{n+1}:x)\cap I^c=I^n for all n\geq c. It has been proved that if residue field is infinite then there always exist A superficial element for all non zero ideal. – Samarendra sahoo Nov 10 '23 at 04:52
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    Thanks for the additional information! You should edit it into the question, not the comments. I have done it for you this time so you can see how to format such things. It is as easy as wrapping math in dollar signs. Click the "edited" timestamp to see what was changed. – rschwieb Nov 10 '23 at 14:47
  • You can find such an example in Huneke and Swanson textbook on integral closure. – user26857 Nov 11 '23 at 21:01
  • Are there any simple example other than that? – Samarendra sahoo Nov 13 '23 at 16:13

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