Lemma 62.6.9. Let $f : X \to S$ be a finite type morphism of schemes with $S$ Noetherian. Let $r \geq 0$. Let $\alpha $ be a relative $r$-cycle on $X/S$. Then there is a proper, completely decomposed (More on Morphisms, Definition 37.78.1) morphism $g : S' \to S$ such that $g^*\alpha $ is in the image of (62.6.8.1).

**Proof.**
By Noetherian induction, we may assume the result holds for the pullback of $\alpha $ by any closed immersion $g : S' \to S$ which is not an isomorphism.

Let $S_1 \subset S$ be an irreducible component (viewed as an integral closed subscheme). Let $S_2 \subset S$ be the closure of the complement of $S'$ (viewed as a reduced closed subscheme). If $S_2 \not= \emptyset $, then the result holds for the pullback of $\alpha $ by $S_1 \to S$ and $S_2 \to S$. If $g_1 : S'_1 \to S_1$ and $g_2 : S'_2 \to S_2$ are the corresponding completely decomposed proper morphisms, then $S' = S'_1 \amalg S'_2 \to S$ is a completely decomposed proper morphism and we see the result holds for $S$^{1} . Thus we may assume $S' \to S$ is bijective and we reduce to the case described in the next paragraph.

Assume $S$ is integral. Let $\eta \in S$ be the generic point and let $K = \kappa (\eta )$ be the function field of $S$. Then $\alpha _\eta $ is an $r$-cycle on $X_ K$. Write $\alpha _\eta = \sum n_ i[Y_ i]$. Taking the closure of $Y_ i$ we obtain integral closed subschemes $Z_ i \subset X$ whose base change to $\eta $ is $Y_ i$. By generic flatness (for example Morphisms, Proposition 29.27.1), we see that $Z_ i$ is flat over a nonempty open $U$ of $S$ for each $i$. Applying More on Flatness, Lemma 38.31.1 we can find a $U$-admissible blowing up $g : S' \to S$ such that the strict transform $Z'_ i \subset X_{S'}$ of $Z_ i$ is flat over $S'$. Then $\beta = \sum n_ i[Z'_ i/X_{S'}/S']_ r$ is in the image of (62.6.8.1) and $\beta = g^*\alpha $ by Lemma 62.6.6.

However, this does not finish the proof as $S' \to S$ may not be completely decomposed. This is easily fixed: denoting $T \subset S$ the complement of $U$ (viewed as a closed subscheme), by Noetherian induction we can find a completely decomposed proper morphism $T' \to T$ such that $(T' \to S)^*\alpha $ is in the image of (62.6.8.1). Then $S' \amalg T' \to S$ does the job. $\square$

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