## 42.14 Flat pullback

In the following we use $f^{-1}(Z)$ to denote the scheme theoretic inverse image of a closed subscheme $Z \subset Y$ for a morphism of schemes $f : X \to Y$. We recall that the scheme theoretic inverse image is the fibre product

$\xymatrix{ f^{-1}(Z) \ar[r] \ar[d] & X \ar[d] \\ Z \ar[r] & Y }$

and it is also the closed subscheme of $X$ cut out by the quasi-coherent sheaf of ideals $f^{-1}(\mathcal{I})\mathcal{O}_ X$, if $\mathcal{I} \subset \mathcal{O}_ Y$ is the quasi-coherent sheaf of ideals corresponding to $Z$ in $Y$. (This is discussed in Schemes, Section 26.4 and Lemma 26.17.6 and Definition 26.17.7.)

Definition 42.14.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a morphism. Assume $f$ is flat of relative dimension $r$.

1. Let $Z \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$. We define $f^*[Z]$ to be the $(k+r)$-cycle on $X$ to the scheme theoretic inverse image

$f^*[Z] = [f^{-1}(Z)]_{k+r}.$

This makes sense since $\dim _\delta (f^{-1}(Z)) = k + r$ by Lemma 42.13.1.

2. Let $\alpha = \sum n_ i [Z_ i]$ be a $k$-cycle on $Y$. The flat pullback of $\alpha$ by $f$ is the sum

$f^* \alpha = \sum n_ i f^*[Z_ i]$

where each $f^*[Z_ i]$ is defined as above. The sum is locally finite by Lemma 42.13.2.

3. We denote $f^* : Z_ k(Y) \to Z_{k + r}(X)$ the map of abelian groups so obtained.

An open immersion is flat. This is an important though trivial special case of a flat morphism. If $U \subset X$ is open then sometimes the pullback by $j : U \to X$ of a cycle is called the restriction of the cycle to $U$. Note that in this case the maps

$j^* : Z_ k(X) \longrightarrow Z_ k(U)$

are all surjective. The reason is that given any integral closed subscheme $Z' \subset U$, we can take the closure of $Z$ of $Z'$ in $X$ and think of it as a reduced closed subscheme of $X$ (see Schemes, Lemma 26.12.4). And clearly $Z \cap U = Z'$, in other words $j^*[Z] = [Z']$ whence the surjectivity. In fact a little bit more is true.

Lemma 42.14.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $U \subset X$ be an open subscheme, and denote $i : Y = X \setminus U \to X$ as a reduced closed subscheme of $X$. For every $k \in \mathbf{Z}$ the sequence

$\xymatrix{ Z_ k(Y) \ar[r]^{i_*} & Z_ k(X) \ar[r]^{j^*} & Z_ k(U) \ar[r] & 0 }$

is an exact complex of abelian groups.

Proof. First assume $X$ is quasi-compact. Then $Z_ k(X)$ is a free $\mathbf{Z}$-module with basis given by the elements $[Z]$ where $Z \subset X$ is integral closed of $\delta$-dimension $k$. Such a basis element maps either to the basis element $[Z \cap U]$ or to zero if $Z \subset Y$. Hence the lemma is clear in this case. The general case is similar and the proof is omitted. $\square$

Lemma 42.14.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X, Y, Z$ be locally of finite type over $S$. Let $f : X \to Y$ and $g : Y \to Z$ be flat morphisms of relative dimensions $r$ and $s$. Then $g \circ f$ is flat of relative dimension $r + s$ and

$f^* \circ g^* = (g \circ f)^*$

as maps $Z_ k(Z) \to Z_{k + r + s}(X)$.

Proof. The composition is flat of relative dimension $r + s$ by Morphisms, Lemma 29.29.3. Suppose that

1. $W \subset Z$ is a closed integral subscheme of $\delta$-dimension $k$,

2. $W' \subset Y$ is a closed integral subscheme of $\delta$-dimension $k + s$ with $W' \subset g^{-1}(W)$, and

3. $W'' \subset Y$ is a closed integral subscheme of $\delta$-dimension $k + s + r$ with $W'' \subset f^{-1}(W')$.

We have to show that the coefficient $n$ of $[W'']$ in $(g \circ f)^*[W]$ agrees with the coefficient $m$ of $[W'']$ in $f^*(g^*[W])$. That it suffices to check the lemma in these cases follows from Lemma 42.13.1. Let $\xi '' \in W''$, $\xi ' \in W'$ and $\xi \in W$ be the generic points. Consider the local rings $A = \mathcal{O}_{Z, \xi }$, $B = \mathcal{O}_{Y, \xi '}$ and $C = \mathcal{O}_{X, \xi ''}$. Then we have local flat ring maps $A \to B$, $B \to C$ and moreover

$n = \text{length}_ C(C/\mathfrak m_ AC), \quad \text{and} \quad m = \text{length}_ C(C/\mathfrak m_ BC) \text{length}_ B(B/\mathfrak m_ AB)$

Hence the equality follows from Algebra, Lemma 10.52.14. $\square$

Lemma 42.14.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X, Y$ be locally of finite type over $S$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$.

1. Let $Z \subset Y$ be a closed subscheme with $\dim _\delta (Z) \leq k$. Then we have $\dim _\delta (f^{-1}(Z)) \leq k + r$ and $[f^{-1}(Z)]_{k + r} = f^*[Z]_ k$ in $Z_{k + r}(X)$.

2. Let $\mathcal{F}$ be a coherent sheaf on $Y$ with $\dim _\delta (\text{Supp}(\mathcal{F})) \leq k$. Then we have $\dim _\delta (\text{Supp}(f^*\mathcal{F})) \leq k + r$ and

$f^*[{\mathcal F}]_ k = [f^*{\mathcal F}]_{k+r}$

in $Z_{k + r}(X)$.

Proof. The statements on dimensions follow immediately from Lemma 42.13.1. Part (1) follows from part (2) by Lemma 42.10.3 and the fact that $f^*\mathcal{O}_ Z = \mathcal{O}_{f^{-1}(Z)}$.

Proof of (2). As $X$, $Y$ are locally Noetherian we may apply Cohomology of Schemes, Lemma 30.9.1 to see that $\mathcal{F}$ is of finite type, hence $f^*\mathcal{F}$ is of finite type (Modules, Lemma 17.9.2), hence $f^*\mathcal{F}$ is coherent (Cohomology of Schemes, Lemma 30.9.1 again). Thus the lemma makes sense. Let $W \subset Y$ be an integral closed subscheme of $\delta$-dimension $k$, and let $W' \subset X$ be an integral closed subscheme of dimension $k + r$ mapping into $W$ under $f$. We have to show that the coefficient $n$ of $[W']$ in $f^*[{\mathcal F}]_ k$ agrees with the coefficient $m$ of $[W']$ in $[f^*{\mathcal F}]_{k+r}$. Let $\xi \in W$ and $\xi ' \in W'$ be the generic points. Let $A = \mathcal{O}_{Y, \xi }$, $B = \mathcal{O}_{X, \xi '}$ and set $M = \mathcal{F}_\xi$ as an $A$-module. (Note that $M$ has finite length by our dimension assumptions, but we actually do not need to verify this. See Lemma 42.10.1.) We have $f^*\mathcal{F}_{\xi '} = B \otimes _ A M$. Thus we see that

$n = \text{length}_ B(B \otimes _ A M) \quad \text{and} \quad m = \text{length}_ A(M) \text{length}_ B(B/\mathfrak m_ AB)$

Thus the equality follows from Algebra, Lemma 10.52.13. $\square$

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