The Stacks project

Proof. Let us reformulate the statement of the proposition. Consider the category $\mathcal{C}$ whose objects, denoted $Y/X$, are locally quasi-finite syntomic morphism $Y \to X$ of locally Noetherian 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 proposition means that for every object $Y/X$ of $\mathcal{C}$ we have an isomorphism $c_{Y/X} : \det (\mathop{N\! L}\nolimits _{Y/X}) \to \omega _{Y/X}$ with $c_{Y/X}(\delta (\mathop{N\! L}\nolimits _{Y/X})) = \tau _{Y/X}$ and for every morphism $b/a : Y'/X' \to Y/X$ of $\mathcal{C}$ we have $b^*c_{Y/X} = c_{Y'/X'}$ via the identifications $b^*\det (\mathop{N\! L}\nolimits _{Y/X}) = \det (\mathop{N\! L}\nolimits _{Y'/X'})$ and $b^*\omega _{Y/X} = \omega _{Y'/X'}$ described above.

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

mapping $\delta (\mathop{N\! L}\nolimits _{Y/X})|_ V$ to $\tau _{Y/X}|_ V$. This follows from picking affine opens as in Lemma 49.10.1 part (5), the affine local description of $\delta (\mathop{N\! L}\nolimits _{Y/X})$ in Remark 49.13.1, and Lemma 49.12.4. If the annihilator of the section $\tau _{Y/X}$ is zero, then these local maps are unique and automatically glue. Hence if the annihilator of $\tau _{Y/X}$ is zero, then there is a unique isomorphism $c_{Y/X} : \det (\mathop{N\! L}\nolimits _{Y/X}) \to \omega _{Y/X}$ with $c_{Y/X}(\delta (\mathop{N\! L}\nolimits _{Y/X})) = \tau _{Y/X}$. If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}$ and the annihilator of $\tau _{Y'/X'}$ is zero as well, then $b^*c_{Y/X}$ is the unique isomorphism $c_{Y'/X'} : \det (\mathop{N\! L}\nolimits _{Y'/X'}) \to \omega _{Y'/X'}$ with $c_{Y'/X'}(\delta (\mathop{N\! L}\nolimits _{Y'/X'})) = \tau _{Y'/X'}$. This follows formally from the fact that $b^*\delta (\mathop{N\! L}\nolimits _{Y/X}) = \delta (\mathop{N\! L}\nolimits _{Y'/X'})$ and $b^*\tau _{Y/X} = \tau _{Y'/X'}$.

We can summarize the results of the previous paragraph as follows. Let $\mathcal{C}_{nice} \subset \mathcal{C}$ denote the full subcategory of $Y/X$ such that the annihilator of $\tau _{Y/X}$ is zero. Then we have solved the problem on $\mathcal{C}_{nice}$. For $Y/X$ in $\mathcal{C}_{nice}$ we continue to denote $c_{Y/X}$ 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. $b_1^*c_{Y_1/X_1} = b_2^*c_{Y_2/X_2}$. We will first show that the claim implies the proposition 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 Example 49.10.5. Then $Y_{n, d}$ is an irreducible regular scheme and the morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite syntomic and étale over a dense open, see Lemma 49.10.6. Thus $\tau _{Y_{n, d}/X_{n, d}}$ is nonzero for example by Lemma 49.9.6. 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 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 $c_{Y_{n, d}/X_{n, d}}$ constructed above by $b$ to $V$. The claim guarantees these local isomorphisms glue! Thus we get a well defined global isomorphism $c_{Y/X} : \det (\mathop{N\! L}\nolimits _{Y/X}) \to \omega _{Y/X}$ with $c_{Y/X}(\delta (\mathop{N\! L}\nolimits _{Y/X})) = \tau _{Y/X}$. If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}$, then the claim also implies that the similarly constructed map $c_{Y'/X'}$ is the pullback by $b$ of the locally constructed map $c_{Y/X}$. 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 $Y_2$. Therefore we may assume $Y, X, Y_1, X_1, Y_2, X_2$ are affine, say they are the spectra of rings $B, A, B_1, A_1, B_2, A_2$. Picture

By assumption the spectrum of $B$ is an affine open of both the spectrum of $A \otimes _{A_1} B_1$ and $A \otimes _{A_2} B_2$. Shrinking more we may assume there exist elements $g_ i \in A \otimes _{A_ i} B_ i$ such that our maps give isomorphisms $(A \otimes _{A_ i} B_ i)_{g_ i} = B$, see Properties, Lemma 28.29.9. Let $x_\alpha $, $y_\beta $ be a sufficiently large collection of variables such that we may choose surjections

of $A_1$ and $A_2$-algebras. Then we can choose lifts $h_ i \in A'_ i \otimes _{A_ i} B_ i$ of $g_ i \in A \otimes _{A_ i} B_ i$ and we consider the diagram

By construction the two squares are cocartesian. Next, we consider the ring map

By More on Algebra, Lemma 15.6.4 we have $A'_1 \otimes _{A'} B' = (A'_1 \otimes _{A_1} B_1)_{h_1}$ and $A'_2 \otimes _{A'} B' = (A'_2 \otimes _{A_2} B_2)_{h_2}$. In particular the fibres of the morphism $Y' = \mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A') = X'$ are open subschemes of base changes of the fibres of the maps $Y_ i \to X_ i$, hence finite and local complete intersections (see Lemma 49.10.1). By More on Algebra, Lemma 15.7.7 the ring map $A' \to B'$ is flat and of finite presentation. Thus by Lemma 49.10.1 part (3) we conclude that $Y' \to X'$ is locally quasi-finite and syntomic. Thus we obtain the commutative diagram

where $Y'_ i/X'_ i$ corresponds to $A'_ i \to (A'_ i \otimes _{A_ i} B_ i)_{h_ i}$.

In this paragraph, we fix the issue that $A'$ and $A'_ i$ may not be Noetherian by a limit argument; we suggest the reader skip this. We can find a finitely generated $\mathbf{Z}$-subalgebra $A'' \subset A'$ and a quasi-finite and syntomic ring map $A'' \to B''$ such that $B' = A' \otimes _{A''} B''$. This follows from Limits, Lemmas 32.10.1, 32.8.16, and 32.18.2. Then the ring maps $A'' \to A'_1 = A_1[x_\alpha ]$ will factor through a finite polynomial subalgebra $A_ i[x_1, \ldots , x_ n$ and the ring map $A'' \to A'_2 = A_2[y_\beta ]$ will factor through $A_2[y_1, \ldots , y_ m]$. After increasing the sets of variables, we may also assume that $h_1 \in A_1[x_1, \ldots , x_ n] \otimes _{A_1} B_1$ and $h_2 \in A_2[y_1, \ldots , y_ m] \otimes _{A_2} B_2$. Increasing the sets of variables one more time, we can assume that the maps $B'' \to (A'_ i \otimes _{A_ i} B_ i)_{h_ i}$ factor through $(A_1[x_1, \ldots , x_ n] \otimes _{A_1} B_1)_{h_1}$ and $(A_2[y_1, \ldots , y_ m] \otimes _{A_2} B_2)_{h_2}$. Replacing $Y'$, $X'$, $Y'_1$, $X'_1$, $Y'_2$, $X'_2$ by the spectra of $B''$, $A''$, $(A_1[x_1, \ldots , x_ n] \otimes _{A_1} B_1)_{h_1}$, $A_1[x_1, \ldots , x_ n]$, $(A_2[y_1, \ldots , y_ m] \otimes _{A_2} B_2)_{h_2}$, $A_2[y_1, \ldots , y_ m]$, we see that we get a diagram as above where all schemes are Noetherian.

Note that $Y'_ i/X'_ i$ is an object of $\mathcal{C}_{nice}$ as it obtained by taking an open subscheme of the product of an affine space with $Y_ i/X_ i$. In particular, the pullback of $c_{Y_ i/X_ i}$ via $(b_ i/a_ i)$ is the same as the pullback of $c_{Y'_ i/X'_ i}$ via $Y/X \to Y'_ i/X'_ i$. Now we would be done if $Y'/X'$ is an object of $\mathcal{C}_{nice}$ (but this is likely not the case). Namely, then pulling back $c_{Y'/X'}$ 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 $c_{Y_{n, d}/X_{n, d}}$. This finishes the proof. $\square$