The Stacks project

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 (

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 ( 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 ( Then $S' \amalg T' \to S$ does the job. $\square$

[1] Namely, any closed subscheme of $S'_1 \times _ S X$ flat and of relative dimension $\leq r$ over $S'_1$ may be viewed as a closed subscheme of $S' \times _ S X$ flat and of relative dimension $\leq r$ over $S'$.

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 0H59. Beware of the difference between the letter 'O' and the digit '0'.