The Stacks project

Proof. Say $M^0$ and $M^{-1}$ are free of rank $n$. For every $p \geq 0$ there is a unique surjection

with the following two properties; (i) we have $ \pi _ p(k_1 \wedge \ldots \wedge k_ p \wedge l_1 \wedge \ldots \wedge l_ n) = k_1 \wedge \ldots \wedge k_ p \otimes m_1 \wedge \ldots \wedge m_ n $ if $k_1, \ldots , k_ p \in K^0$ and $l_1, \ldots , l_ n \in L^0$ map to $m_1, \ldots , m_ n \in M^0$ and (ii) the kernel of $\pi _ p$ is the submodule generated by wedges $l_1 \wedge \ldots \wedge l_{p + n}$ such that $> p$ of the $l_ j$ are in $K^0$. Pick a basis $e_1, \ldots , e_ n$ for $L^{-1} = M^{-1}$. Then we define our map by the rule

where $\eta \in \wedge ^ p(H^0(L^\bullet ))$ and $\tilde\eta \in \wedge ^ p(L^0)$ is a lift of $\eta $. The expression on the right hand side makes sense because $\det (M^\bullet ) = \wedge ^ n(M^0) \otimes \wedge ^ n(M^{-1})^{\otimes -1}$. The expression on the right is independent of the choice of $\tilde\eta $ since $\partial (\xi ) \wedge \partial (e_1) \wedge \ldots \wedge \partial (e_ n) = 0$ for any $\xi \in L^{-1}$. A straightforward calculation shows that the map is independent of the choice of the basis of $L^{-1} = M^{-1}$. It is immediate from the definition of $\delta (M^\bullet )$ in More on Algebra, Section 15.123 that $c^0(1) = \delta (M^\bullet )$. Finally, the rule in statement (2) can be directly verified from the definition. $\square$

Proof. This follows from the explicit description of the map given in the proof of Lemma 50.18.1. Note that we need the assumption that $M_1^ j \otimes _{R_1} R_2 \to M_2^ j$ are isomorphisms to see that the basis we use for $M_1^{-1}$ maps to a basis for $M_2^{-1}$ (and hence in checking the compatibility in both constructions we can use the “same” basis). $\square$

Proof. The map

is surjective because $Y'$ is an open subscheme of the fibre product; the corresponding algebra statement is that $\Omega _{B \otimes _ A C/\mathbf{Z}}$ is generated as a module by the images of $\Omega _{B/\mathbf{Z}}$ and $\Omega _{C/\mathbf{Z}}$. Thus for all $p$ the map

is surjective. Conditions (1) and (2) combined with the commutativity of the diagram in the lemma, implies that the map $c^ p_{Y'/X'}$ is determined (but existence is not immediate). $\square$

Proof. It is clear that we may replace $Y'$ by $X' \times _ X Y$ in order to prove this. Pick affine opens $V \subset Y$ and $U \subset X$ with $V$ containing the image of $y'$ as in Discriminants, Lemma 49.10.1 part (5). Pick an affine open $U' \subset X'$ containing the image of $y'$ and mapping into $U$. After replacing $X, Y, X', Y'$ by $U, V, U', U' \times _ U V$ we may assume we are in the situation described in the next paragraph.

Assume that $X' = \mathop{\mathrm{Spec}}(A')$, $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(B)$, and $Y' = \mathop{\mathrm{Spec}}(A' \otimes _ A B)$ with $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ a relative global complete intersection over $A$. Then also $B' = A'[x_1, \ldots , x_ n]/(f'_1, \ldots , f'_ n)$ is a relative global complete intersection over $A'$ where $f'_ i$ is the image of $f_ i$ in $A'[x_1, \ldots , x_ n]$, see Algebra, Lemma 10.136.9. The construction in Remark 50.18.3 provides us with the maps $c^ p_{V/U}$ and $c^ p_{V'/U'}$. These maps are compatible by Lemma 50.18.2 and the compatibility of the presentations of $B$ and $B'$ over $A$ and $A'$. Some details omitted. $\square$

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.

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 $c^ p_{Y/X}$, $p \geq 0$ with properties (1) and (2), and for every morphism $b/a : Y'/X' \to Y/X$ of $\mathcal{C}$ we have that $c^ p_{Y/X}$ and $c^ p_{Y'/X'}$ are compatible as in Lemma 50.18.4.

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 properties (1) and (2). This follows from picking affine opens as in Discriminants, Lemma 49.10.1 part (5) and then using the construction in Remark 50.18.3.

Note that the étale locus of $f$ is exactly the set of points where $\delta (\mathop{N\! L}\nolimits _{Y/X})$ does not vanish, see discussion in Discriminants, Section 49.13. In particular $f^*\Omega ^ p_{X/\mathbf{Z}} \to \Omega ^ p_{Y/\mathbf{Z}}$ becomes an isomorphism after inverting $\delta (\mathop{N\! L}\nolimits _{Y/X})$. We conclude that if $\Omega ^ p_{X/\mathbf{Z}}$ is finite locally free and the annihilator of $\delta (\mathop{N\! L}\nolimits _{Y/X})$ in $\mathcal{O}_ Y$ is zero, then the local maps constructed in the previous paragraph are unique and automatically glue!

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

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

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

By the remarks in the previous paragraph, we see that for any object $Y/X$ of $\mathcal{C}_{nice}$ we have unique maps $c^ p_{Y/X}$, $p \geq 0$ satisfying conditions (1) and (2). If $b/a : Y'/X' \to Y/X$ is a morphism of $\mathcal{C}_{nice}$, then the maps $c^ p_{Y/X}$ and $c^ p_{Y'/X'}$ are compatible as in Lemma 50.18.4: namely, locally there do exist compatible maps by Lemma 50.18.5 and the uniqueness just mentioned shows these maps agree with $c^ p_{Y/X}$ and $c^ p_{Y'/X'}$. 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.

In fact, more generally, suppose we have a morphism $b/a : Y'/X' \to Y/X$ of $\mathcal{C}$ such that $Y/X$ is an object of $\mathcal{C}_{nice}$. The same argument, but this time using the uniqueness in Lemma 50.18.4, tells us there are maps $c^ p_{Y'/X'}$, $p \geq 0$ satisfying conditions (1) and (2) compatible with the already constructed maps $c^ p_{Y/X}$. Let us call this the base change $(b/a)^*c^ p_{Y/X}$.

Consider morphisms

in $\mathcal{C}$ such that $Y_1/X_1$ and $Y_2/X_2$ are objects of $\mathcal{C}_{nice}$. Claim. The two base changes $(b_ i/a_ i)^*c^ p_{Y_ i/X_ i}$, $i = 1, 2$ are equal. 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. Then we have the base change $c^ p_{V/U} = (b/a)^*c^ p_{Y_{n, d}/X_{n, d}}$. The claim guarantees these locally constructed maps $c^ p_{V/U}$ glue! Thus we get a well defined global maps $c^ p_{Y/X}$ with properties (1) and (2). Finally, let $b/a : Y'/X' \to Y/X$ be an arbitray morphism of $\mathcal{C}$. We have the maps $c^ p_{Y'/X'}$, $p \geq 0$ and $c^ p_{Y/X}$, $p \geq 0$ constructed in this paragraph. To check they are compatible as in Lemma 50.18.4 we may work locally on $Y'/X'$ and $Y/X$. Thus we may assume that there exists a morphism $Y/X \to Y_{n, d}/X_{n, d}$. By construction, both the family $c^ p_{Y'/X'}$ and the family $c^ p_{X/Y}$ are compatible with the map towards $Y_{n, d}/X_{n, d}$. It follows easily from this (and the uniqueness in Lemma 50.18.4) that $c^ p_{Y'/X'}$ is compatible with $c^ p_{Y/X}$. 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 $Y_2$. 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 Discriminants, 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 Discriminants, Lemma 49.10.1 part (3) we conclude that $Y'/X'$ is an object of $\mathcal{C}$. Consider now the commutative diagram

where $Y'_ i/X'_ i$ corresponds to $A'_ i \to (A'_ i \otimes _{A_ i} B_ i)_{h_ i}$. 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 infinite dimensional affine space with $Y_ i/X_ i$; small detail omitted. In particular, the pullback of $c^ p_{Y_ i/X_ i}$ via $(b_ i/a_ i)$ is the same as the pullback of $c^ p_{Y'_ i/X'_ i}$ 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 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. $\square$