The Stacks project

Lemma 47.23.1. Properties (A), (B), (C), (D), and (E) of More on Algebra, Section 15.51 hold for $P(k \to R) =$“$R$ is a Gorenstein ring”.

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 $K/k$ 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 10.158.11. 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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BJN. Beware of the difference between the letter 'O' and the digit '0'.