Proof. Since we already know the result holds for Cohen-Macaulay instead of Gorenstein, we may in each step assume the ring we have is Cohen-Macaulay. This is not particularly helpful for the proof, but psychologically may be useful.
Part (A). Let $k \subset K$ be a finitely generated field extension. Let $R$ be a Gorenstein $k$-algebra. We can find a global complete intersection $A = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ over $k$ such that $K$ is isomorphic to the fraction field of $A$, see Algebra, Lemma 10.156.11. Then $R \to R \otimes _ k A$ is a relative global complete intersection. Hence $R \otimes _ k A$ is Gorenstein by Lemma 47.21.7. Thus $R \otimes _ k K$ is too as a localization.
Proof of (B). This is clear because a ring is Gorenstein if and only if all of its local rings are Gorenstein.
Part (C). Let $A \to B \to C$ be flat maps of Noetherian rings. Assume the fibres of $A \to B$ are Gorenstein and $B \to C$ is regular. We have to show the fibres of $A \to C$ are Gorenstein. Clearly, we may assume $A = k$ is a field. Then we may assume that $B \to C$ is a regular local homomorphism of Noetherian local rings. Then $B$ is Gorenstein and $C/\mathfrak m_ B C$ is regular, in particular Gorenstein (Lemma 47.21.3). Then $C$ is Gorenstein by Lemma 47.21.8.
Part (D). This follows from Lemma 47.21.8. Part (E) is immediate as the condition does not refer to the ground field. $\square$
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