The Stacks project

48.21 Dimension functions

We need a bit more information about how the dimension functions change when passing to a scheme of finite type over another.

Lemma 48.21.1. Let $S$ be a Noetherian scheme and let $\omega _ S^\bullet $ be a dualizing complex. Let $X$ be a scheme of finite type over $S$ and let $\omega _ X^\bullet $ be the dualizing complex normalized relative to $\omega _ S^\bullet $. If $x \in X$ is a closed point lying over a closed point $s$ of $S$, then $\omega _{X, x}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{X, x}$ provided that $\omega _{S, s}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{S, s}$.

Proof. We may replace $X$ by an affine neighbourhood of $x$, hence we may and do assume that $f : X \to S$ is separated. Then $\omega _ X^\bullet = f^!\omega _ S^\bullet $. We have to show that $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\kappa (x), \omega _{X, x}^\bullet )$ is sitting in degree $0$. Let $i_ x : x \to X$ denote the inclusion morphism which is a closed immersion as $x$ is a closed point. Hence $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\kappa (x), \omega _{X, x}^\bullet )$ represents $i_ x^!\omega _ X^\bullet $ by Lemma 48.17.4. Consider the commutative diagram

\[ \xymatrix{ x \ar[r]_{i_ x} \ar[d]_\pi & X \ar[d]^ f \\ s \ar[r]^{i_ s} & S } \]

By Morphisms, Lemma 29.20.3 the extension $\kappa (x)/\kappa (s)$ is finite and hence $\pi $ is a finite morphism. We conclude that

\[ i_ x^!\omega _ X^\bullet = i_ x^! f^! \omega _ S^\bullet = \pi ^! i_ s^! \omega _ S^\bullet \]

Thus if $\omega _{S, s}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{S, s}$, then $i_ s^!\omega _ S^\bullet = \kappa (s)[0]$ by the same reasoning as above. We have

\[ R\pi _*(\pi ^!(\kappa (s)[0])) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ s}(R\pi _*(\kappa (x)[0]), \kappa (s)[0]) = \widetilde{\mathop{\mathrm{Hom}}\nolimits _{\kappa (s)}(\kappa (x), \kappa (s))} \]

The first equality by Example 48.3.9 applied with $L = \kappa (x)[0]$. The second equality holds because $\pi _*$ is exact. Thus $\pi ^!(\kappa (s)[0])$ is supported in degree $0$ and we win. $\square$

Lemma 48.21.2. Let $S$ be a Noetherian scheme and let $\omega _ S^\bullet $ be a dualizing complex. Let $f : X \to S$ be of finite type and let $\omega _ X^\bullet $ be the dualizing complex normalized relative to $\omega _ S^\bullet $. For all $x \in X$ we have

\[ \delta _ X(x) - \delta _ S(f(x)) = \text{trdeg}_{\kappa (f(x))}(\kappa (x)) \]

where $\delta _ S$, resp. $\delta _ X$ is the dimension function of $\omega _ S^\bullet $, resp. $\omega _ X^\bullet $, see Lemma 48.2.7.

Proof. We may replace $X$ by an affine neighbourhood of $x$. Hence we may and do assume there is a compactification $X \subset \overline{X}$ over $S$. Then we may replace $X$ by $\overline{X}$ and assume that $X$ is proper over $S$. We may also assume $X$ is connected by replacing $X$ by the connected component of $X$ containing $x$. Next, recall that both $\delta _ X$ and the function $x \mapsto \delta _ S(f(x)) + \text{trdeg}_{\kappa (f(x))}(\kappa (x))$ are dimension functions on $X$, see Morphisms, Lemma 29.52.3 (and the fact that $S$ is universally catenary by Lemma 48.2.7). By Topology, Lemma 5.20.3 we see that the difference is locally constant, hence constant as $X$ is connected. Thus it suffices to prove equality in any point of $X$. By Properties, Lemma 28.5.9 the scheme $X$ has a closed point $x$. Since $X \to S$ is proper the image $s$ of $x$ is closed in $S$. Thus we may apply Lemma 48.21.1 to conclude. $\square$

Lemma 48.21.3. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Let $x \in X$ with image $y \in Y$. Then

\[ H^ i(f^!\mathcal{O}_ Y)_ x \not= 0 \Rightarrow - \dim _ x(X_ y) \leq i. \]

Proof. Since the statement is local on $X$ we may assume $X$ and $Y$ are affine schemes. Write $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(R)$. Then $f^!\mathcal{O}_ Y$ corresponds to the relative dualizing complex $\omega _{A/R}^\bullet $ of Dualizing Complexes, Section 47.25 by Remark 48.17.5. Thus the lemma follows from Dualizing Complexes, Lemma 47.25.7. $\square$

Lemma 48.21.4. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Let $x \in X$ with image $y \in Y$. If $f$ is flat, then

\[ H^ i(f^!\mathcal{O}_ Y)_ x \not= 0 \Rightarrow - \dim _ x(X_ y) \leq i \leq 0. \]

In fact, if all fibres of $f$ have dimension $\leq d$, then $f^!\mathcal{O}_ Y$ has tor-amplitude in $[-d, 0]$ as an object of $D(X, f^{-1}\mathcal{O}_ Y)$.

Proof. Arguing exactly as in the proof of Lemma 48.21.3 this follows from Dualizing Complexes, Lemma 47.25.8. $\square$

Lemma 48.21.5. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Let $x \in X$ with image $y \in Y$. Assume

  1. $\mathcal{O}_{Y, y}$ is Cohen-Macaulay, and

  2. $\text{trdeg}_{\kappa (f(\xi ))}(\kappa (\xi )) \leq r$ for any generic point $\xi $ of an irreducible component of $X$ containing $x$.

Then

\[ H^ i(f^!\mathcal{O}_ Y)_ x \not= 0 \Rightarrow - r \leq i \]

and the stalk $H^{-r}(f^!\mathcal{O}_ Y)_ x$ is $(S_2)$ as an $\mathcal{O}_{X, x}$-module.

Proof. After replacing $X$ by an open neighbourhood of $x$, we may assume every irreducible component of $X$ passes through $x$. Then arguing exactly as in the proof of Lemma 48.21.3 this follows from Dualizing Complexes, Lemma 47.25.9. $\square$

Lemma 48.21.6. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. If $f$ is flat and quasi-finite, then

\[ f^!\mathcal{O}_ Y = \omega _{X/Y}[0] \]

for some coherent $\mathcal{O}_ X$-module $\omega _{X/Y}$ flat over $Y$.

Proof. Consequence of Lemma 48.21.4 and the fact that the cohomology sheaves of $f^!\mathcal{O}_ Y$ are coherent by Lemma 48.17.6. $\square$

Lemma 48.21.7. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. If $f$ is Cohen-Macaulay (More on Morphisms, Definition 37.20.1), then

\[ f^!\mathcal{O}_ Y = \omega _{X/Y}[d] \]

for some coherent $\mathcal{O}_ X$-module $\omega _{X/Y}$ flat over $Y$ where $d$ is the locally constant function on $X$ which gives the relative dimension of $X$ over $Y$.

Proof. The relative dimension $d$ is well defined and locally constant by Morphisms, Lemma 29.29.4. The cohomology sheaves of $f^!\mathcal{O}_ Y$ are coherent by Lemma 48.17.6. We will get flatness of $\omega _{X/Y}$ from Lemma 48.21.4 if we can show the other cohomology sheaves of $f^!\mathcal{O}_ Y$ are zero.

The question is local on $X$, hence we may assume $X$ and $Y$ are affine and the morphism has relative dimension $d$. If $d = 0$, then the result follows directly from Lemma 48.21.6. If $d > 0$, then we may assume there is a factorization

\[ X \xrightarrow {g} \mathbf{A}^ d_ Y \xrightarrow {p} Y \]

with $g$ quasi-finite and flat, see More on Morphisms, Lemma 37.20.8. Then $f^! = g^! \circ p^!$. By Lemma 48.17.3 we see that $p^!\mathcal{O}_ Y \cong \mathcal{O}_{\mathbf{A}^ d_ Y}[-d]$. We conclude by the case $d = 0$. $\square$

Remark 48.21.8. Let $S$ be a Noetherian scheme endowed with a dualizing complex $\omega _ S^\bullet $. In this case Lemmas 48.21.3, 48.21.4, 48.21.6, and 48.21.7 are true for any morphism $f : X \to Y$ of finite type schemes over $S$ but with $f^!$ replaced by $f_{new}^!$. This is clear because in each case the proof reduces immediately to the affine case and then $f^! = f_{new}^!$ by Lemma 48.20.9.


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 0BV4. Beware of the difference between the letter 'O' and the digit '0'.