## 49.10 Quasi-finite syntomic morphisms

This section discusses the fact that a quasi-finite syntomic morphism has an invertible relative dualizing module.

Lemma 49.10.1. Let $f : Y \to X$ be a morphism of schemes. The following are equivalent

1. $f$ is locally quasi-finite and syntomic,

2. $f$ is locally quasi-finite, flat, and a local complete intersection morphism,

3. $f$ is locally quasi-finite, flat, locally of finite presentation, and the fibres of $f$ are local complete intersections,

4. $f$ is locally quasi-finite and for every $y \in Y$ there are affine opens $y \in V = \mathop{\mathrm{Spec}}(B) \subset Y$, $U = \mathop{\mathrm{Spec}}(A) \subset X$ with $f(V) \subset U$ an integer $n$ and $h, f_1, \ldots , f_ n \in A[x_1, \ldots , x_ n]$ such that $B = A[x_1, \ldots , x_ n, 1/h]/(f_1, \ldots , f_ n)$,

5. for every $y \in Y$ there are affine opens $y \in V = \mathop{\mathrm{Spec}}(B) \subset Y$, $U = \mathop{\mathrm{Spec}}(A) \subset X$ with $f(V) \subset U$ such that $A \to B$ is a relative global complete intersection of the form $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$,

6. $f$ is locally quasi-finite, flat, locally of finite presentation, and $\mathop{N\! L}\nolimits _{X/Y}$ has tor-amplitude in $[-1, 0]$, and

7. $f$ is flat, locally of finite presentation, $\mathop{N\! L}\nolimits _{X/Y}$ is perfect of rank $0$ with tor-amplitude in $[-1, 0]$,

Proof. The equivalence of (1) and (2) is More on Morphisms, Lemma 37.60.8. The equivalence of (1) and (3) is Morphisms, Lemma 29.30.11.

If $A \to B$ is as in (4), then $B = A[x, x_1, \ldots , x_ n]/(xh - 1, f_1, \ldots , f_ n]$ is a relative global complete intersection by see Algebra, Definition 10.136.5. Thus (4) implies (5). It is clear that (5) implies (4).

Condition (5) implies (1): by Algebra, Lemma 10.136.14 a relative global complete intersection is syntomic and the definition of a relative global complete intersection guarantees that a relative global complete intersection on $n$ variables with $n$ equations is quasi-finite, see Algebra, Definition 10.136.5 and Lemma 10.122.2.

Either Algebra, Lemma 10.136.15 or Morphisms, Lemma 29.30.10 shows that (1) implies (5).

More on Morphisms, Lemma 37.60.17 shows that (6) is equivalent to (1). If the equivalent conditions (1) – (6) hold, then we see that affine locally $Y \to X$ is given by a relative global complete intersection $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ with the same number of variables as the number of equations. Using this presentation we see that

$\mathop{N\! L}\nolimits _{B/A} =\left( (f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2 \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} B \text{d} x_ i\right)$

By Algebra, Lemma 10.136.13 the module $(f_1, \ldots , f_ n)/(f_1, \ldots , f_ n)^2$ is free with generators the congruence classes of the elements $f_1, \ldots , f_ n$. Thus $\mathop{N\! L}\nolimits _{B/A}$ has rank $0$ and so does $\mathop{N\! L}\nolimits _{Y/X}$. In this way we see that (1) – (6) imply (7).

Finally, assume (7). By More on Morphisms, Lemma 37.60.17 we see that $f$ is syntomic. Thus on suitable affine opens $f$ is given by a relative global complete intersection $A \to B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$, see Morphisms, Lemma 29.30.10. Exactly as above we see that $\mathop{N\! L}\nolimits _{B/A}$ is a perfect complex of rank $n - m$. Thus $n = m$ and we see that (5) holds. This finishes the proof. $\square$

Lemma 49.10.2. Invertibility of the relative dualizing module.

1. If $A \to B$ is a quasi-finite flat homomorphism of Noetherian rings, then $\omega _{B/A}$ is an invertible $B$-module if and only if $\omega _{B \otimes _ A \kappa (\mathfrak p)/\kappa (\mathfrak p)}$ is an invertible $B \otimes _ A \kappa (\mathfrak p)$-module for all primes $\mathfrak p \subset A$.

2. If $Y \to X$ is a quasi-finite flat morphism of Noetherian schemes, then $\omega _{Y/X}$ is invertible if and only if $\omega _{Y_ x/x}$ is invertible for all $x \in X$.

Proof. Proof of (1). As $A \to B$ is flat, the module $\omega _{B/A}$ is $A$-flat, see Lemma 49.2.9. Thus $\omega _{B/A}$ is an invertible $B$-module if and only if $\omega _{B/A} \otimes _ A \kappa (\mathfrak p)$ is an invertible $B \otimes _ A \kappa (\mathfrak p)$-module for every prime $\mathfrak p \subset A$, see More on Morphisms, Lemma 37.16.7. Still using that $A \to B$ is flat, we have that formation of $\omega _{B/A}$ commutes with base change, see Lemma 49.2.10. Thus we see that invertibility of the relative dualizing module, in the presence of flatness, is equivalent to invertibility of the relative dualizing module for the maps $\kappa (\mathfrak p) \to B \otimes _ A \kappa (\mathfrak p)$.

Part (2) follows from (1) and the fact that affine locally the dualizing modules are given by their algebraic counterparts, see Remark 49.2.11. $\square$

Lemma 49.10.3. Let $k$ be a field. Let $B = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ be a global complete intersection over $k$ of dimension $0$. Then $\omega _{B/k}$ is invertible.

Proof. By Noether normalization, see Algebra, Lemma 10.115.4 we see that there exists a finite injection $k \to B$, i.e., $\dim _ k(B) < \infty$. Hence $\omega _{B/k} = \mathop{\mathrm{Hom}}\nolimits _ k(B, k)$ as a $B$-module. By Dualizing Complexes, Lemma 47.15.8 we see that $R\mathop{\mathrm{Hom}}\nolimits (B, k)$ is a dualizing complex for $B$ and by Dualizing Complexes, Lemma 47.13.3 we see that $R\mathop{\mathrm{Hom}}\nolimits (B, k)$ is equal to $\omega _{B/k}$ placed in degree $0$. Thus it suffices to show that $B$ is Gorenstein (Dualizing Complexes, Lemma 47.21.4). This is true by Dualizing Complexes, Lemma 47.21.7. $\square$

Lemma 49.10.4. Let $f : Y \to X$ be a morphism of locally Noetherian schemes. If $f$ satisfies the equivalent conditions of Lemma 49.10.1 then $\omega _{Y/X}$ is an invertible $\mathcal{O}_ Y$-module.

Proof. We may assume $A \to B$ is a relative global complete intersection of the form $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ and we have to show $\omega _{B/A}$ is invertible. This follows in combining Lemmas 49.10.2 and 49.10.3. $\square$

Example 49.10.5. Let $n \geq 1$ and $d \geq 1$ be integers. Let $T$ be the set of multi-indices $E = (e_1, \ldots , e_ n)$ with $e_ i \geq 0$ and $\sum e_ i \leq d$. Consider the ring

$A = \mathbf{Z}[a_{i, E} ; 1 \leq i \leq n, E \in T]$

In $A[x_1, \ldots , x_ n]$ consider the elements $f_ i = \sum _{E \in T} a_{i, E} x^ E$ where $x^ E = x_1^{e_1} \ldots x_ n^{e_ n}$ as is customary. Consider the $A$-algebra

$B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$

Denote $X_{n, d} = \mathop{\mathrm{Spec}}(A)$ and let $Y_{n, d} \subset \mathop{\mathrm{Spec}}(B)$ be the maximal open subscheme such that the restriction of the morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A) = X_{n, d}$ is quasi-finite, see Algebra, Lemma 10.123.13.

Lemma 49.10.6. With notation as in Example 49.10.5 the schemes $X_{n, d}$ and $Y_{n, d}$ are regular and irreducible, the morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite and syntomic, and there is a dense open subscheme $V \subset Y_{n, d}$ such that $Y_{n, d} \to X_{n, d}$ restricts to an étale morphism $V \to X_{n, d}$.

Proof. The scheme $X_{n, d}$ is the spectrum of the polynomial ring $A$. Hence $X_{n, d}$ is regular and irreducible. Since we can write

$f_ i = a_{i, (0, \ldots , 0)} + \sum \nolimits _{E \in T, E \not= (0, \ldots , 0)} a_{i, E} x^ E$

we see that the ring $B$ is isomorphic to the polynomial ring on $x_1, \ldots , x_ n$ and the elements $a_{i, E}$ with $E \not= (0, \ldots , 0)$. Hence $\mathop{\mathrm{Spec}}(B)$ is an irreducible and regular scheme and so is the open $Y_{n, d}$. The morphism $Y_{n, d} \to X_{n, d}$ is locally quasi-finite and syntomic by Lemma 49.10.1. To find $V$ it suffices to find a single point where $Y_{n, d} \to X_{n, d}$ is étale (the locus of points where a morphism is étale is open by definition). Thus it suffices to find a point of $X_{n, d}$ where the fibre of $Y_{n, d} \to X_{n, d}$ is nonempty and étale, see Morphisms, Lemma 29.36.15. We choose the point corresponding to the ring map $\chi : A \to \mathbf{Q}$ sending $f_ i$ to $1 + x_ i^ d$. Then

$B \otimes _{A, \chi } \mathbf{Q} = \mathbf{Q}[x_1, \ldots , x_ n]/(x_1^ d - 1, \ldots , x_ n^ d - 1)$

which is a nonzero étale algebra over $\mathbf{Q}$. $\square$

Lemma 49.10.7. Let $f : Y \to X$ be a morphism of schemes. If $f$ satisfies the equivalent conditions of Lemma 49.10.1 then for every $y \in Y$ there exist $n, d$ and a commutative diagram

$\xymatrix{ Y \ar[d] & V \ar[d] \ar[l] \ar[r] & Y_{n, d} \ar[d] \\ X & U \ar[l] \ar[r] & X_{n, d} }$

where $U \subset X$ and $V \subset Y$ are open, where $Y_{n, d} \to X_{n, d}$ is as in Example 49.10.5, and where the square on the right hand side is cartesian.

Proof. By Lemma 49.10.1 we can choose $U$ and $V$ affine so that $U = \mathop{\mathrm{Spec}}(R)$ and $V = \mathop{\mathrm{Spec}}(S)$ with $S = R[y_1, \ldots , y_ n]/(g_1, \ldots , g_ n)$. With notation as in Example 49.10.5 if we pick $d$ large enough, then we can write each $g_ i$ as $g_ i = \sum _{E \in T} g_{i, E}y^ E$ with $g_{i, E} \in R$. Then the map $A \to R$ sending $a_{i, E}$ to $g_{i, E}$ and the map $B \to S$ sending $x_ i \to y_ i$ give a cocartesian diagram of rings

$\xymatrix{ S & B \ar[l] \\ R \ar[u] & A \ar[l] \ar[u] }$

which proves the lemma. $\square$

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