Lemma 42.21.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. Let $Z$ be a closed subscheme of $X \times \mathbf{P}^1$. Assume

$\dim _\delta (Z) \leq k + 1$,

$\dim _\delta (Z_0) \leq k$, $\dim _\delta (Z_\infty ) \leq k$, and

for any embedded point $\xi $ (Divisors, Definition 31.4.1) of $Z$ either $\xi \not\in Z_0 \cup Z_\infty $ or $\delta (\xi ) < k$.

Then $[Z_0]_ k \sim _{rat} [Z_\infty ]_ k$ as $k$-cycles on $X$.

**Proof.**
Let $\{ W_ i\} _{i \in I}$ be the collection of irreducible components of $Z$ which have $\delta $-dimension $k + 1$. Write

\[ [Z]_{k + 1} = \sum n_ i[W_ i] \]

with $n_ i > 0$ as per definition. Note that $\{ W_ i\} $ is a locally finite collection of closed subsets of $X \times _ S \mathbf{P}^1_ S$ by Divisors, Lemma 31.26.1. We claim that

\[ [Z_0]_ k = \sum n_ i[(W_ i)_0]_ k \]

and similarly for $[Z_\infty ]_ k$. If we prove this then the lemma follows from Lemma 42.21.1.

Let $Z' \subset X$ be an integral closed subscheme of $\delta $-dimension $k$. To prove the equality above it suffices to show that the coefficient $n$ of $[Z']$ in $[Z_0]_ k$ is the same as the coefficient $m$ of $[Z']$ in $\sum n_ i[(W_ i)_0]_ k$. Let $\xi ' \in Z'$ be the generic point. Set $\xi = (\xi ', 0) \in X \times _ S \mathbf{P}^1_ S$. Consider the local ring $A = \mathcal{O}_{X \times _ S \mathbf{P}^1_ S, \xi }$. Let $I \subset A$ be the ideal cutting out $Z$, in other words so that $A/I = \mathcal{O}_{Z, \xi }$. Let $t \in A$ be the element cutting out $X \times _ S D_0$ (i.e., the coordinate of $\mathbf{P}^1$ at zero pulled back). By our choice of $\xi ' \in Z'$ we have $\delta (\xi ) = k$ and hence $\dim (A/I) = 1$. Since $\xi $ is not an embedded point by assumption (3) we see that $A/I$ is Cohen-Macaulay. Since $\dim _\delta (Z_0) = k$ we see that $\dim (A/(t, I)) = 0$ which implies that $t$ is a nonzerodivisor on $A/I$. Finally, the irreducible closed subschemes $W_ i$ passing through $\xi $ correspond to the minimal primes $I \subset \mathfrak q_ i$ over $I$. The multiplicities $n_ i$ correspond to the lengths $\text{length}_{A_{\mathfrak q_ i}}(A/I)_{\mathfrak q_ i}$. Hence we see that

\[ n = \text{length}_ A(A/(t, I)) \]

and

\[ m = \sum \text{length}_ A(A/(t, \mathfrak q_ i)) \text{length}_{A_{\mathfrak q_ i}}(A/I)_{\mathfrak q_ i} \]

Thus the result follows from Lemma 42.3.2.
$\square$

## Comments (0)