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Do Gorenstein rings necessarily have finite projective dimensions?

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    $\begingroup$ What? Every ring $R$ has finite projective dimension as an $R$-module. Is that what you want to ask? If not, then what exactly? $\endgroup$
    – Robin Chapman
    Commented Aug 6, 2010 at 15:16
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    $\begingroup$ @robin projective dim of a ring $R$ is (sometimes) defined to be the supremum of the projective dim of $R$-modules. $\endgroup$
    – Manish Kumar
    Commented Aug 6, 2010 at 15:25
  • $\begingroup$ Of course. Sorry, I was being dumb (it happens quite often). $\endgroup$
    – ashpool
    Commented Aug 6, 2010 at 15:25
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    $\begingroup$ It is better to also edit the actual text of the question, not only the title. In any case, you surely did not want to ask «Do Gorenstein rings necessarily have a finite projective dimension (as a module over itself)?» because then the answer is trivially yes: every ring is projective over itself! $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 6, 2010 at 15:59
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    $\begingroup$ I have voted to close, this question is trivial as stated. $\endgroup$
    – Hailong Dao
    Commented Aug 6, 2010 at 17:13

2 Answers 2

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Take $A=k[x]/(x^2)$ for a field $k$. This is a self-injective $k$-algebra (that is, it is an injective module over itself), so it is Gorenstein. Yet the residue field is of infinite projective dimension.

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  • $\begingroup$ Thanks. How do I see that the residue field of A has infinite projective dimension? $\endgroup$
    – ashpool
    Commented Aug 6, 2010 at 15:59
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    $\begingroup$ @kwan: compute a projective resolution: it will be clear quite fast that the dimension is infinite. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 6, 2010 at 16:00
  • $\begingroup$ I see. I get an infinite minimal free resolution. Thanks! $\endgroup$
    – ashpool
    Commented Aug 6, 2010 at 16:35
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No. There is a result of Serre/Auslander-Buchsbaum which says a noethrian local ring is regular if and only if the residue field has finite free resolution. So just take any Gorenstein singularity as an example.

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    $\begingroup$ What is Gorenstein singularity? $\endgroup$
    – ashpool
    Commented Aug 6, 2010 at 15:58
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    $\begingroup$ A local Gorenstein ring that's not regular. $\endgroup$
    – Robin Chapman
    Commented Aug 6, 2010 at 16:04

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