The Stacks project

Proof. This proof is very similar to the proof of Discriminants, Proposition 49.13.2 and we suggest the reader look at that proof first. We fix $p \geq 0$ throughout the proof.

Let us reformulate the statement. Consider the category $\mathcal{C}$ whose objects, denoted $Y/X$, are locally quasi-finite syntomic morphism $f : Y \to X$ of schemes and whose morphisms $b/a : Y'/X' \to Y/X$ are commutative diagrams

which induce an isomorphism of $Y'$ with an open subscheme of $X' \times _ X Y$. The lemma means that for every object $Y/X$ of $\mathcal{C}$ we have maps $c^ p_{Y/X}$ with property (1) and for every morphism $b/a : Y'/X' \to Y/X$ of $\mathcal{C}$ we have $b^*c^ p_{Y/X} = c^ p_{Y'/X'}$ via the identifications $b^*\det (\mathop{N\! L}\nolimits _{Y/X}) = \det (\mathop{N\! L}\nolimits _{Y'/X'})$ (Discriminants, Section 49.13) and $b^*\Omega ^ p_{Y/X} = \Omega ^ p_{Y'/X'}$ (Lemma 50.2.1).

Given $Y/X$ in $\mathcal{C}$ and $y \in Y$ we can find an affine open $V \subset Y$ and $U \subset X$ with $f(V) \subset U$ such that there exists some maps

with property (1). This follows from picking affine opens as in Discriminants, Lemma 49.10.1 part (5) and Remark 50.18.2. If $\Omega ^ p_{X/\mathbf{Z}}$ is finite locally free and annihilator of the section $\delta (\mathop{N\! L}\nolimits _{Y/X})$ is zero, then these local maps are unique and automatically glue!

Let $\mathcal{C}_{nice} \subset \mathcal{C}$ denote the full subcategory of $Y/X$ such that

1. $X$ is of finite type over $\mathbf{Z}$,

2. $\Omega _{X/\mathbf{Z}}$ is locally free, and

3. the annihilator of $\delta (\mathop{N\! L}\nolimits _{Y/X})$ is zero.

By the remarks in the previous paragraph, we see that for any object $Y/X$ of $\mathcal{C}_{nice}$ we have a unique map $c^ p_{Y/X}$ satisfying condition (1). If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}_{nice}$, then $b^*c^ p_{Y/X}$ is equal to $c^ p_{Y'/X'}$ because $b^*\delta (\mathop{N\! L}\nolimits _{Y/X}) = \delta (\mathop{N\! L}\nolimits _{Y'/X'})$ (see Discriminants, Section 49.13). In other words, we have solved the problem on the full subcategory $\mathcal{C}_{nice}$. For $Y/X$ in $\mathcal{C}_{nice}$ we continue to denote $c^ p_{Y/X}$ the solution we've just found.

Consider morphisms

in $\mathcal{C}$ such that $Y_1/X_1$ and $Y_2/X_2$ are objects of $\mathcal{C}_{nice}$. Claim. $b_1^*c^ p_{Y_1/X_1} = b_2^*c^ p_{Y_2/X_2}$. We will first show that the claim implies the lemma and then we will prove the claim.

Let $d, n \geq 1$ and consider the locally quasi-finite syntomic morphism $Y_{n, d} \to X_{n, d}$ constructed in Discriminants, Example 49.10.5. Then $Y_{n, d}$ and $Y_{n, d}$ are irreducible schemes of finite type and smooth over $\mathbf{Z}$. Namely, $X_{n, d}$ is a spectrum of a polynomial ring over $\mathbf{Z}$ and $Y_{n, d}$ is an open subscheme of such. The morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite syntomic and étale over a dense open, see Discriminants, Lemma 49.10.6. Thus $\delta (\mathop{N\! L}\nolimits _{Y_{n, d}/X_{n, d}})$ is nonzero: for example we have the local description of $\delta (\mathop{N\! L}\nolimits _{Y/X})$ in Discriminants, Remark 49.13.1 and we have the local description of étale morphisms in Morphisms, Lemma 29.36.15 part (8). Now a nonzero section of an invertible module over an irreducible regular scheme has vanishing annihilator. Thus $Y_{n, d}/X_{n, d}$ is an object of $\mathcal{C}_{nice}$.

Let $Y/X$ be an arbitrary object of $\mathcal{C}$. Let $y \in Y$. By Discriminants, Lemma 49.10.7 we can find $n, d \geq 1$ and morphisms

of $\mathcal{C}$ such that $V \subset Y$ and $U \subset X$ are open. Thus we can pullback the canonical morphism $c^ p_{Y_{n, d}/X_{n, d}}$ constructed above by $b$ to $V$. The claim guarantees these local isomorphisms glue! Thus we get a well defined global maps $c^ p_{Y/X}$ with property (1). If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}$, then the claim also implies that the similarly constructed map $c^ p_{Y'/X'}$ is the pullback by $b$ of the locally constructed map $c^ p_{Y/X}$. Thus it remains to prove the claim.

In the rest of the proof we prove the claim. We may pick a point $y \in Y$ and prove the maps agree in an open neighbourhood of $y$. Thus we may replace $Y_1$, $Y_2$ by open neighbourhoods of the image of $y$ in $Y_1$ and $Y_2$. Thus we may assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine. We may write $X = \mathop{\mathrm{lim}}\nolimits X_\lambda $ as a cofiltered limit of affine schemes of finite type over $X_1 \times X_2$. For each $\lambda $ we get

If we take limits we obtain

By Limits, Lemma 32.4.11 we can find a $\lambda $ and opens $V_{1, \lambda } \subset Y_1 \times _{X_1} X_\lambda $ and $V_{2, \lambda } \subset X_\lambda \times _{X_2} Y_2$ whose base change to $X$ recovers $Y$ (on both sides). After increasing $\lambda $ we may assume there is an isomorphism $V_{1, \lambda } \to V_{2, \lambda }$ whose base change to $X$ is the identity on $Y$, see Limits, Lemma 32.10.1. Then we have the commutative diagram

Thus it suffices to prove the claim for the lower row of the diagram and we reduce to the case discussed in the next paragraph.

Assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine of finite type over $\mathbf{Z}$. Write $X = \mathop{\mathrm{Spec}}(A)$, $X_ i = \mathop{\mathrm{Spec}}(A_ i)$. The ring map $A_1 \to A$ corresponding to $X \to X_1$ is of finite type and hence we may choose a surjection $A_1[x_1, \ldots , x_ n] \to A$. Similarly, we may choose a surjection $A_2[y_1, \ldots , y_ m] \to A$. Set $X'_1 = \mathop{\mathrm{Spec}}(A_1[x_1, \ldots , x_ n])$ and $X'_2 = \mathop{\mathrm{Spec}}(A_2[y_1, \ldots , y_ m])$. Observe that $\Omega _{X'_1/\mathbf{Z}}$ is the direct sum of the pullback of $\Omega _{X_1/\mathbf{Z}}$ and a finite free module. Similarly for $X'_2$. Set $Y'_1 = Y_1 \times _{X_1} X'_1$ and $Y'_2 = Y_2 \times _{X_2} X'_2$. We get the following diagram

Since $X'_1 \to X_1$ and $X'_2 \to X_2$ are flat, the same is true for $Y'_1 \to Y_1$ and $Y'_2 \to Y_2$. It follows easily that the annihilators of $\delta (\mathop{N\! L}\nolimits _{Y'_1/X'_1})$ and $\delta (\mathop{N\! L}\nolimits _{Y'_2/X'_2})$ are zero. Hence $Y'_1/X'_1$ and $Y'_2/X'_2$ are in $\mathcal{C}_{nice}$. Thus the outer morphisms in the displayed diagram are morphisms of $\mathcal{C}_{nice}$ for which we know the desired compatibilities. Thus it suffices to prove the claim for $Y'_1/X'_1 \leftarrow Y/X \rightarrow Y'_2/X'_2$. This reduces us to the case discussed in the next paragraph.

Assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine of finite type over $\mathbf{Z}$ and $X \to X_1$ and $X \to X_2$ are closed immersions. Consider the open embeddings $Y_1 \times _{X_1} X \supset Y \subset X \times _{X_2} Y_2$. There is an open neighbourhood $V \subset Y$ of $y$ which is a standard open of both $Y_1 \times _{X_1} X$ and $X \times _{X_2} Y_2$. This follows from Schemes, Lemma 26.11.5 applied to the scheme obtained by glueing $Y_1 \times _{X_1} X$ and $X \times _{X_2} Y_2$ along $Y$; details omitted. Since $X \times _{X_2} Y_2$ is a closed subscheme of $Y_2$ we can find a standard open $V_2 \subset Y_2$ such that $V_2 \times _{X_2} X = V$. Similarly, we can find a standard open $V_1 \subset Y_1$ such that $V_1 \times _{X_1} X = V$. After replacing $Y, Y_1, Y_2$ by $V, V_1, V_2$ we reduce to the case discussed in the next paragraph.

Assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine of finite type over $\mathbf{Z}$ and $X \to X_1$ and $X \to X_2$ are closed immersions and $Y_1 \times _{X_1} X = Y = X \times _{X_2} Y_2$. Write $X = \mathop{\mathrm{Spec}}(A)$, $X_ i = \mathop{\mathrm{Spec}}(A_ i)$, $Y = \mathop{\mathrm{Spec}}(B)$, $Y_ i = \mathop{\mathrm{Spec}}(B_ i)$. Then we can consider the affine schemes

Observe that $X' = X_1 \amalg _ X X_2$ and $Y' = Y_1 \amalg _ Y Y_2$, see More on Morphisms, Lemma 37.14.1. By More on Algebra, Lemma 15.5.1 the rings $A'$ and $B'$ are of finite type over $\mathbf{Z}$. By More on Algebra, Lemma 15.6.4 we have $B' \otimes _ A A_1 = B_1$ and $B' \times _ A A_2 = B_2$. In particular a fibre of $Y' \to X'$ over a point of $X' = X_1 \amalg _ X X_2$ is always equal to either a fibre of $Y_1 \to X_1$ or a fibre of $Y_2 \to X_2$. By More on Algebra, Lemma 15.6.8 the ring map $A' \to B'$ is flat. Thus by Discriminants, Lemma 49.10.1 part (3) we conclude that $Y'/X'$ is an object of $\mathcal{C}$. Consider now the commutative diagram

Now we would be done if $Y'/X'$ is an object of $\mathcal{C}_{nice}$, but this is almost never the case. Namely, then pulling back $c^ p_{Y'/X'}$ around the two sides of the square, we would obtain the desired conclusion. To get around the problem that $Y'/X'$ is not in $\mathcal{C}_{nice}$ we note the arguments above show that, after possibly shrinking all of the schemes $X, Y, X_1, Y_1, X_2, Y_2, X', Y'$ we can find some $n, d \geq 1$, and extend the diagram like so:

and then we can use the already given argument by pulling back from $c^ p_{Y_{n, d}/X_{n, d}}$. This finishes the proof. $\square$