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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Sign up to join this communityA 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?
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.