Lemma 47.21.6. Let $A$ be a Noetherian ring. The following are equivalent
$A$ has finite injective dimension as a module over itself,
$A$ is Gorenstein and finite dimensional.
In this case $A[0]$ is a dualizing complex.
Lemma 47.21.6. Let $A$ be a Noetherian ring. The following are equivalent
$A$ has finite injective dimension as a module over itself,
$A$ is Gorenstein and finite dimensional.
In this case $A[0]$ is a dualizing complex.
Proof. If $A$ has finite injective dimension as a module over $A$, then $A[0]$ is a dualizing complex over $A$ (see Definition 47.15.1) and we conclude $A$ has finite dimension by Lemma 47.17.4 and is Gorenstein by Lemma 47.21.4. Assume $A$ is Gorenstein and $\dim (A) \leq d < \infty $. Let $I \subset A$ be any ideal. We claim $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A/I, A) = 0$ for $i > d$; this proves that $A$ has finite injective dimension as a module over itself by More on Algebra, Lemma 15.71.2. Forming the $\mathop{\mathrm{Ext}}\nolimits $ groups commutes with localization (see for example More on Algebra, Lemma 15.101.2). Thus to prove vanishing we may assume $A$ is local of dimension $e < d$. As $A$ is Gorenstein we see that $\omega _ A^\bullet = A[e]$ is a normalized dualizing complex. Hence the claimed vanishing by Lemma 47.16.5 for example. $\square$
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