Lemma 37.54.4. Let $S$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes over $S$ with both $X$ and $Y$ of finite presentation over $S$. Then there exists a $t \geq 0$ and closed subschemes
\[ S \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset \]
with the following properties:
$S_ i \to S$ is defined by a finite type ideal sheaf,
$S_0 \subset S$ is a thickening, and
with $T_ i = S_ i \setminus S_{i + 1}$ and $f_ i$ the base change of $f$ to $T_ i$ we have: formation of the scheme theoretic image of $f_ i/T_ i$ commutes with arbitrary base change (see discussion above the lemma).
Proof.
We can find a commutative diagram
\[ \xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \ar[r] & S \ar[d] \\ U \ar[r] & V \ar[r] & W } \]
with cartesian squares such that $U$, $V$, $W$ are of finite type over $\mathbf{Z}$. Namely, first write $S$ as a cofiltered limit of finite type schemes over $\mathbf{Z}$ with affine transition morphisms using Limits, Proposition 32.5.4 and then descend the morphism $X \to Y$ using Limits, Lemma 32.10.1. This reduces us to the case discussed in the next paragraph.
Assume $S$ is Noetherian. In this case every quasi-coherent ideal is of finite type, hence we do not have to check the condition that $S_ i$ is cut out by a finite type ideal. Set $S_0 = S_{red}$ equal to the reduction of $S$. Let $\eta \in S_0$ be a generic point of an irreducible component of $S_0$. By Noetherian induction on the underlying topological space of $S_0$, we may assume the result holds for any closed subscheme of $S_0$ not containing $\eta $. Thus it suffices to show that there exists an open neighbourhood $U_0 \subset S_0$ such that the base change $f_0$ of $f$ to $U_0$ has property (3).
Let $R$ be a Noetherian domain. Let $f : X \to Y$ be a morphism of finite type schemes over $R$. By the discussion in the previous paragraph it suffices to show that after replacing $R$ by $R_ g$ for some $g \in R$ nonzero and $X$, $Y$ by their base changes to $R_ g$, formation of the scheme theoretic image of $f/R$ commutes with arbitrary base change.
Let $Y = V_1 \cup \ldots V_ n$ be an affine open covering. Let $U_ i = f^{-1}(V_ i)$. If the statement is true for each of the morphisms $U_ i \to V_ i$ over $R$, then it holds for $f$. Namely, the scheme theoretic image of $U_ i \to V_ i$ is the intersection of $V_ i$ with the scheme theoretic image of $f : X \to Y$ by Morphisms, Lemma 29.6.3. Thus we may assume $Y$ is affine.
Let $X = U_1 \cup \ldots U_ n$ be an affine open covering. Then the scheme theoretic image of $X \to Y$ is the same as the scheme theoretic imge of $\coprod U_ i \to Y$. Thus we may assume $X$ is affine.
Say $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ and $f$ corresponds to the $R$-algebra map $\varphi : A \to B$. Then the scheme theoretic image of $f$ is $\mathop{\mathrm{Spec}}(A/\mathop{\mathrm{Ker}}(\varphi ))$ and similarly after base change (by an affine morphism, but it is enough to check for those). Thus formation of the scheme theoretic image commutes with base change if $\mathop{\mathrm{Ker}}(\varphi \otimes _ R R') = \mathop{\mathrm{Ker}}(\varphi ) \otimes _ R R'$ for all ring maps $R \to R'$.
After replacing $R$, $A$, $B$ by $R_ g$, $A_ g$, $B_ g$ for a suitable nonzero $g$ in $R$, we may assume $A$ and $B$ are flat over $R$. By Lemma 37.54.3 we may also assume $B/A$ is a flat $R$-module. Then $0 \to \mathop{\mathrm{Ker}}(\varphi ) \to A \to B \to B/A \to 0$ is an exact sequence of flat $R$-modules, which implies the desired base change statement.
$\square$
Comments (0)