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. $\square$

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