The Stacks project

Lemma 52.22.5. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume

  1. $A$ is a local ring which has a dualizing complex,

  2. all irreducible components of $X$ have the same dimension,

  3. the scheme $X \setminus Y$ is Cohen-Macaulay,

  4. $I$ is generated by $d$ elements,

  5. $\dim (X) - \dim (Z) > d + 2$, and

  6. for $y \in U \cap Y$ the module $\mathcal{F}_ y^\wedge $ is finite locally free outside $V(I\mathcal{O}_{X, y}^\wedge )$, for example if $\mathcal{F}_ n$ is a finite locally free $\mathcal{O}_ U/I^ n\mathcal{O}_ U$-module.

Then $(\mathcal{F}_ n)$ extends to $X$. In particular if $A$ is $I$-adically complete, then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module.

Proof. We will show that the hypotheses (1), (2), (3)(b) of Proposition 52.22.4 are satisfied. This is clear for (1) and (2).

Let $y \in U \cap Y$ and let $\mathfrak p$ be a prime $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ with $\mathfrak p \not\in V(I\mathcal{O}_{X, y}^\wedge )$. The last condition shows that $\text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) = \text{depth}((\mathcal{O}_{X, y}^\wedge )_\mathfrak p)$. Since $X \setminus Y$ is Cohen-Macaulay we see that $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p$ is Cohen-Macaulay. Thus we see that

\begin{align*} & \text{depth}((\mathcal{F}^\wedge _ y)_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) + \delta ^ Y_ Z(y) \\ & = \dim ((\mathcal{O}_{X, y}^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) + \delta ^ Y_ Z(y) \\ & = \dim (\mathcal{O}_{X, y}^\wedge ) + \delta ^ Y_ Z(y) \end{align*}

The final equality because $\mathcal{O}_{X, y}$ is equidimensional by the second condition. Let $\delta (y) = \dim (\overline{\{ y\} })$. This is a dimension function as $A$ is a catenary local ring. By Lemma 52.18.1 we have $\delta ^ Y_ Z(y) \geq \delta (y) - \dim (Z)$. Since $X$ is equidimensional we get

\[ \dim (\mathcal{O}_{X, y}^\wedge ) + \delta ^ Y_ Z(y) \geq \dim (\mathcal{O}_{X, y}^\wedge ) + \delta (y) - \dim (Z) = \dim (X) - \dim (Z) \]

Thus we get the desired inequality and we win. $\square$

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Your email address will not be published.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 0EJP. Beware of the difference between the letter 'O' and the digit '0'.