A commutative noetherian ring $R$ is Gorenstein if $R$ has finite injective dimension.
Obviously, if $R$ is Gorenstein, then $R$ localized at any prime ideal $P$ is also Gorenstein. But I don't know whether the converse holds?
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1$\begingroup$ I just rolled back an "edit" from @u1359111 which in my view defaced the original question. Why do people approve such edits? $\endgroup$– Yemon ChoiCommented Jul 26, 2014 at 13:25
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$\begingroup$ Yemon Choi: as you see in the answer, which is accepted, TmobiusX really want to ask: is there a commutative Noetherian ring R all of whose localizations have finite injective dimension but such that R itself does not? $\endgroup$– trxCommented Jul 26, 2014 at 13:28
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$\begingroup$ @u1359111 I feel we should respect people's original wordings. I know that Pete Clark answered your version of the question but to me this does not justify disinterring old questions and changing them $\endgroup$– Yemon ChoiCommented Jul 26, 2014 at 13:30
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$\begingroup$ @Yemon Choi" not for Pete Clark's answer but since it is accepted: "So I am guessing you really want to ask: is there a commutative Noetherian ring R all of whose localizations have finite injective dimension but such that R itself does not?" $\endgroup$– trxCommented Jul 26, 2014 at 13:36
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2$\begingroup$ People sometimes accept answers for reasons strictly other than that it answered the question as originally asked or intended. For example, it might mean something like, "thanks; this answer put me on the right track", or "I see, that's actually the better question". We don't know what OP was thinking when he/she accepted the answer, since OP's only reaction was "Well done! Thank you..." (invisible now to those below a certain level of rep). I'm with Yemon in that one should avoid stamping one's own interpretation on a question based on what the accepted answer says. $\endgroup$– Todd TrimbleCommented Jul 26, 2014 at 14:05
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In all of the standard references I know, Gorenstein is either only defined for local rings or a not necessarily local ring is defined to be Gorenstein if all of its localizations at maximal ideals are Gorenstein (which implies that its localizations at all prime ideals are Gorenstein). So I am guessing you really want to ask: is there a commutative Noetherian ring $R$ all of whose localizations have finite injective dimension but such that $R$ itself does not?
I believe that the answer is "yes" and that a counterexample is given by Nagata's (in)famous example of a Noetherian ring of infinite Krull dimension. See (5.96) in Lam's Lectures on modules and rings for an explanation why Nagata's example is regular, hence locally Gorenstein, i.e., locally of finite injective dimension.
Furthermore, the proof of the Theorem at the bottom of p. 7 of
http://www.math.hawaii.edu/~lee/homolog/Goren.pdf
states that the injective dimension of a ring is the supremum of the injective dimensions of its local rings $R_{\mathfrak{m}}$. In this case each $R_{\mathfrak{m}}$ is regular, hence Gorenstein, hence its injective dimension is simply equal to its Krull dimension, i.e., to the height of $\mathfrak{m}$. It follows that the injective dimension of $R$ itself is infinite.
Conversely, Lee's handout contains a proof that the answer is "no" if $R$ is Noetherian of finite Krull dimension.
answered May 2, 2010 at 8:29
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$\begingroup$ Did you mean to say that the answer is "yes" to the existence question you posed? $\endgroup$ Commented May 2, 2010 at 10:11
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