Lemma 47.23.2. Let $A$ be a Noetherian local ring. If $A$ has a dualizing complex, then the formal fibres of $A$ are Gorenstein.

Proof. Let $\mathfrak p$ be a prime of $A$. The formal fibre of $A$ at $\mathfrak p$ is isomorphic to the formal fibre of $A/\mathfrak p$ at $(0)$. The quotient $A/\mathfrak p$ has a dualizing complex (Lemma 47.15.9). Thus it suffices to check the statement when $A$ is a local domain and $\mathfrak p = (0)$. Let $\omega _ A^\bullet$ be a dualizing complex for $A$. Then $\omega _ A^\bullet \otimes _ A A^\wedge$ is a dualizing complex for the completion $A^\wedge$ (Lemma 47.22.1). Then $\omega _ A^\bullet \otimes _ A K$ is a dualizing complex for the fraction field $K$ of $A$ (Lemma 47.15.6). Hence $\omega _ A^\bullet \otimes _ A K$ is isomorphic ot $K[n]$ for some $n \in \mathbf{Z}$. Similarly, we conclude a dualizing complex for the formal fibre $A^\wedge \otimes _ A K$ is

$\omega _ A^\bullet \otimes _ A A^\wedge \otimes _{A^\wedge } (A^\wedge \otimes _ A K) = (\omega _ A^\bullet \otimes _ A K) \otimes _ K (A^\wedge \otimes _ A K) \cong (A^\wedge \otimes _ A K)[n]$

as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).