In this section we produce an isomorphism between the determinant of the relative cotangent complex and the relative dualizing module for a locally quasi-finite syntomic morphism of locally Noetherian schemes. Following [1.4.4, Garel] we dub the isomorphism the Tate map. Our approach is to avoid doing local calculations as much as is possible.

Let $Y \to X$ be a locally quasi-finite syntomic morphism of schemes. We will use all the equivalent conditions for this notion given in Lemma 49.10.1 without further mention in this section. In particular, we see that $\mathop{N\! L}\nolimits _{Y/X}$ is a perfect object of $D(\mathcal{O}_ Y)$ with tor-amplitude in $[-1, 0]$. Thus we have a canonical invertible module $\det (\mathop{N\! L}\nolimits _{Y/X})$ on $Y$ and a global section

\[ \delta (\mathop{N\! L}\nolimits _{Y/X}) \in \Gamma (Y, \det (\mathop{N\! L}\nolimits _{Y/X})) \]

whose vertical arrows are locally quasi-finite syntomic and which induces an isomorphism of $Y'$ with an open of $X' \times _ X Y$. Then the canonical map

\[ Lb^*\mathop{N\! L}\nolimits _{Y/X} \longrightarrow \mathop{N\! L}\nolimits _{Y'/X'} \]

Let $Y \to X$ be a locally quasi-finite syntomic morphism of locally Noetherian schemes. By Remarks 49.2.11 and 49.4.7 we have a coherent $\mathcal{O}_ Y$-module $\omega _{Y/X}$ and a canonical global section

whose vertical arrows are locally quasi-finite syntomic and which induces an isomorphism of $Y'$ with an open of $X' \times _ X Y$. Then there is a canonical base change map

which is an isomorphism mapping $\tau _{Y/X}$ to $\tau _{Y'/X'}$. Namely, the base change map in the affine setting is (49.2.3.1), it is an isomorphism by Lemma 49.2.10, and it maps $\tau _{Y/X}$ to $\tau _{Y'/X'}$ by Lemma 49.4.4 part (1).

**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

\[ \xymatrix{ Y' \ar[d] \ar[r]_ b & Y \ar[d] \\ X' \ar[r]^ a & X } \]

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 open $V \subset Y$ and $U \subset X$ with $f(V) \subset U$ such that there exists some isomorphism

\[ \det (\mathop{N\! L}\nolimits _{Y/X})|_ V \longrightarrow \omega _{Y/X}|_ V \]

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

\[ Y_1/X_1 \xleftarrow {b_1/a_1} Y/X \xrightarrow {b_2/a_2} Y_2/X_2 \]

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

\[ Y/X \leftarrow V/U \xrightarrow {b/a} Y_{n, d}/X_{n, d} \]

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 $Y_2$. Thus we may assume there are morphisms

\[ Y_{n_1, d_1}/X_{n_1, d_1} \leftarrow Y_1/X_1 \quad \text{and}\quad Y_2/X_2 \rightarrow Y_{n_2, d_2}/X_{n_2, d_2} \]

These are morphisms of $\mathcal{C}_{nice}$ for which we know the desired compatibilities. Thus we may replace $Y_1/X_1$ by $Y_{n_1, d_1}/X_{n_1, d_1}$ and $Y_2/X_2$ by $Y_{n_2, d_2}/X_{n_2, d_2}$. This reduces us to the case that $Y_1, X_1, Y_2, X_2$ are of finite type over $\mathbf{Z}$. (The astute reader will realize that this step wouldn't have been necessary if we'd defined $\mathcal{C}_{nice}$ to consist only of those objects $Y/X$ with $Y$ and $X$ of finite type over $\mathbf{Z}$.)

Assume $Y_1, X_1, Y_2, X_2$ are of finite type over $\mathbf{Z}$. After replacing $Y, X, Y_1, X_1, Y_2, X_2$ by suitable open neighbourhoods of the image of $y$ 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

\[ Y_1 \times _{X_1} X_\lambda \quad \text{and}\quad X_\lambda \times _{X_2} Y_2 \]

If we take limits we obtain

\[ \mathop{\mathrm{lim}}\nolimits Y_1 \times _{X_1} X_\lambda = Y_1 \times _{X_1} X \supset Y \subset X \times _{X_2} Y_2 = \mathop{\mathrm{lim}}\nolimits X_\lambda \times _{X_2} Y_2 \]

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

\[ \xymatrix{ & Y/X \ar[d] \ar[ld]_{b_1/a_1} \ar[rd]^{b_2/a_2} \\ Y_1/X_1 & V_{1, \lambda }/X_\lambda \ar[l] \ar[r] & Y_2/X_2 } \]

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 $X'_2 = \mathop{\mathrm{Spec}}(A_2[y_1, \ldots , y_ m])$. 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

\[ Y_1/X_1 \leftarrow Y'_1/X'_1 \leftarrow Y/X \rightarrow Y'_2/X'_2 \rightarrow Y_2/X_2 \]

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 $\tau _{Y'_1/X'_1}$ and $\tau _{Y'_2/X'_2}$ 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

\[ X' = \mathop{\mathrm{Spec}}(A_1 \times _ A A_2) = \mathop{\mathrm{Spec}}(A') \quad \text{and}\quad Y' = \mathop{\mathrm{Spec}}(B_1 \times _ B B_2) = \mathop{\mathrm{Spec}}(B') \]

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 Lemma 49.10.1 part (3) we conclude that $Y'/X'$ is an object of $\mathcal{C}$. Consider now the commutative diagram

\[ \xymatrix{ & Y/X \ar[ld]_{b_1/a_1} \ar[rd]^{b_2/a_2} \\ Y_1/X_1 \ar[rd] & & Y_2/X_2 \ar[ld] \\ & Y'/X' } \]

Now we would be done if $Y'/X'$ is an object of $\mathcal{C}_{nice}$. Namely, then pulling back $c_{Y'/X'}$ around the two sides of the square, we would obtain the desired conclusion. Now, in fact, it is true that $Y'/X'$ is an object of $\mathcal{C}_{nice}$^{1}. But it is amusing to note that we don't even need this. Namely, 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:

\[ \xymatrix{ & Y/X \ar[ld]_{b_1/a_1} \ar[rd]^{b_2/a_2} \\ Y_1/X_1 \ar[rd] & & Y_2/X_2 \ar[ld] \\ & Y'/X' \ar[d] \\ & Y_{n, d}/X_{n, d} } \]

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

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