Processing math: 100%

The Stacks project

Lemma 76.27.4. Let S be a scheme. Let f : X \to Y and g : Y \to Z be morphisms of algebraic spaces over S. Assume that the fibres of f, g, and g \circ f are locally Noetherian. Let x \in |X| with images y \in |Y| and z \in |Z|.

  1. If f is Gorenstein at x and g is Gorenstein at f(x), then g \circ f is Gorenstein at x.

  2. If f and g are Gorenstein, then g \circ f is Gorenstein.

  3. If g \circ f is Gorenstein at x and f is flat at x, then f is Gorenstein at x and g is Gorenstein at f(x).

  4. If f \circ g is Gorenstein and f is flat, then f is Gorenstein and g is Gorenstein at every point in the image of f.

Proof. Working étale locally this follows from the corresponding result for schemes, see Duality for Schemes, Lemma 48.25.6. Alternatively, we can use the equivalence of (a) and (b) in Lemma 76.27.3. Thus we consider the local homomorphism of Noetherian local rings

\mathcal{O}_{Y, \overline{y}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{Y, \overline{y}} \longrightarrow \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{X, \overline{x}}

whose fibre is

\mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}}

and we use Dualizing Complexes, Lemma 47.21.8. \square


Comments (0)


Your email address will not be published.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.