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?



Is there a non-Gorenstein ring but locally Gorenstein?
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.


