## 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.

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