The Stacks project

49.11 Finite syntomic morphisms

This section is the analogue of Section 49.10 for finite syntomic morphisms.

Lemma 49.11.1. Let $f : Y \to X$ be a morphism of schemes. The following are equivalent

  1. $f$ is finite and syntomic,

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

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

  4. $f$ is finite and for every $x \in X$ there is an affine open $x \in U = \mathop{\mathrm{Spec}}(A) \subset X$ an integer $n$ and $f_1, \ldots , f_ n \in A[x_1, \ldots , x_ n]$ such that $f^{-1}(U)$ is isomorphic to the spectrum of $A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$,

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

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

Proof. The equivalence of (1), (2), (3), (5), and (6) and the implication (4) $\Rightarrow $ (1) follow immediately from Lemma 49.10.1. Assume the equivalent conditions (1), (2), (3), (5), (6) hold. Choose a point $x \in X$ and an affine open $U = \mathop{\mathrm{Spec}}(A)$ of $x$ in $X$ and say $x$ corresponds to the prime ideal $\mathfrak p \subset A$. Write $f^{-1}(U) = \mathop{\mathrm{Spec}}(B)$. Write $B = A[x_1, \ldots , x_ n]/I$. Since $\mathop{N\! L}\nolimits _{B/A}$ is perfect of tor-amplitude in $[-1, 0]$ by (6) we see that $I/I^2$ is a finite locally free $B$-module of rank $n$. Since $B_\mathfrak p$ is semi-local we see that $(I/I^2)_\mathfrak p$ is free of rank $n$, see Algebra, Lemma 10.78.7. Thus after replacing $A$ by a principal localization at an element not in $\mathfrak p$ we may assume $I/I^2$ is a free $B$-module of rank $n$. Thus by Algebra, Lemma 10.136.6 we can find a presentation of $B$ over $A$ with the same number of variables as equations. In other words, we may assume $B = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$. This proves (4). $\square$

Example 49.11.2. Let $d \geq 1$ be an integer. Consider variables $a_{ij}^ l$ for $1 \leq i, j, l \leq d$ and denote

\[ A_ d = \mathbf{Z}[a_{ij}^ k]/J \]

where $J$ is the ideal generated by the elements

\[ \left\{ \begin{matrix} \sum _ l a_{ij}^ la_{lk}^ m - \sum _ l a_{il}^ ma_{jk}^ l & \forall i, j, k, m \\ a_{ij}^ k - a_{ji}^ k & \forall i, j, k \\ a_{i1}^ j - \delta _{ij} & \forall i, j \end{matrix} \right. \]

where $\delta _{ij}$ indices the Kronecker delta function. We define an $A_ d$-algebra $B_ d$ as follows: as an $A_ d$-module we set

\[ B_ d = A_ d e_1 \oplus \ldots \oplus A_ d e_ d \]

The algebra structure is given by $A_ d \to B_ d$ mapping $1$ to $e_1$. The multiplication on $B_ d$ is the $A_ d$-bilinar map

\[ m : B_ d \times B_ d \longrightarrow B_ d, \quad m(e_ i, e_ j) = \sum a_{ij}^ k e_ k \]

It is straightforward to check that the relations given above exactly force this to be an $A_ d$-algebra structure. The morphism

\[ \pi _ d : Y_ d = \mathop{\mathrm{Spec}}(B_ d) \longrightarrow \mathop{\mathrm{Spec}}(A_ d) = X_ d \]

is the “universal” finite free morphism of rank $d$.

Lemma 49.11.3. With notation as in Example 49.11.2 there is an open subscheme $U_ d \subset X_ d$ with the following property: a morphism of schemes $X \to X_ d$ factors through $U_ d$ if and only if $Y_ d \times _{X_ d} X \to X$ is syntomic.

Proof. Recall that being syntomic is the same thing as being flat and a local complete intersection morphism, see More on Morphisms, Lemma 37.59.8. The set $W_ d \subset Y_ d$ of points where $\pi _ d$ is Koszul is open in $Y_ d$ and its formation commutes with arbitrary base change, see More on Morphisms, Lemma 37.59.21. Since $\pi _ d$ is finite and hence closed, we see that $Z = \pi _ d(Y_ d \setminus W_ d)$ is closed. Since clearly $U_ d = X_ d \setminus Z$ and since its formation commutes with base change we find that the lemma is true. $\square$

Lemma 49.11.4. With notation as in Example 49.11.2 and $U_ d$ as in Lemma 49.11.3 then $U_ d$ is smooth over $\mathop{\mathrm{Spec}}(\mathbf{Z})$.

Proof. Let us use More on Morphisms, Lemma 37.12.1 to show that $U_ d \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is smooth. Namely, suppose that $\mathop{\mathrm{Spec}}(A) \to U_ d$ is a morphism and $A' \to A$ is a small extension. Then $B = A \otimes _{A_ d} B_ d$ is a finite free $A$-algebra which is syntomic over $A$ (by construction of $U_ d$). By Smoothing Ring Maps, Proposition 16.3.2 there exists a syntomic ring map $A' \to B'$ such that $B \cong B' \otimes _{A'} A$. Set $e'_1 = 1 \in B'$. For $1 < i \leq d$ choose lifts $e'_ i \in B'$ of the elements $1 \otimes e_ i \in A \otimes _{A_ d} B_ d = B$. Then $e'_1, \ldots , e'_ d$ is a basis for $B'$ over $A'$ (for example see Algebra, Lemma 10.101.1). Thus we can write $e'_ i e'_ j = \sum \alpha _{ij}^ l e'_ l$ for unique elements $\alpha _{ij}^ l \in A'$ which satisfy the relations $\sum _ l \alpha _{ij}^ l \alpha _{lk}^ m = \sum _ l \alpha _{il}^ m \alpha _{jk}^ l$ and $\alpha _{ij}^ k = \alpha _{ji}^ k$ and $\alpha _{i1}^ j - \delta _{ij}$ in $A'$. This determines a morphism $\mathop{\mathrm{Spec}}(A') \to X_ d$ by sending $a_{ij}^ l \in A_ d$ to $\alpha _{ij}^ l \in A'$. This morphism agrees with the given morphism $\mathop{\mathrm{Spec}}(A) \to U_ d$. Since $\mathop{\mathrm{Spec}}(A')$ and $\mathop{\mathrm{Spec}}(A)$ have the same underlying topological space, we see that we obtain the desired lift $\mathop{\mathrm{Spec}}(A') \to U_ d$ and we conclude that $U_ d$ is smooth over $\mathbf{Z}$. $\square$

Lemma 49.11.5. With notation as in Example 49.11.2 consider the open subscheme $U'_ d \subset X_ d$ over which $\pi _ d$ is étale. Then $U'_ d$ is a dense subset of the open $U_ d$ of Lemma 49.11.3

Proof. By exactly the same reasoning as in the proof of Lemma 49.11.3, using Morphisms, Lemma 29.36.17, there is a maximal open $U'_ d \subset X_ d$ over which $\pi _ d$ is étale. Moreover, since an étale morphism is syntomic, we see that $U'_ d \subset U_ d$. To finish the proof we have to show that $U'_ d \subset U_ d$ is dense. Let $u : \mathop{\mathrm{Spec}}(k) \to U_ d$ be a morphism where $k$ is a field. Let $B = k \otimes _{A_ d} B_ d$ as in the proof of Lemma 49.11.4. We will show there is a local domain $A'$ with residue field $k$ and a finite syntomic $A'$ algebra $B'$ with $B = k \otimes _{A'} B'$ whose generic fibre is étale. Exactly as in the previous paragraph this will determine a morphism $\mathop{\mathrm{Spec}}(A') \to U_ d$ which will map the generic point into $U'_ d$ and the closed point to $u$, thereby finishing the proof.

By Lemma 49.11.1 part (4) we can choose a presentation $B = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$. Let $d'$ be the maximum total degree of the polynomials $f_1, \ldots , f_ n$. Let $Y_{n, d'} \to X_{n, d'}$ be as in Example 49.10.5. By construction there is a morphism $u' : \mathop{\mathrm{Spec}}(k) \to X_{n, d'}$ such that

\[ \mathop{\mathrm{Spec}}(B) \cong Y_{n, d'} \times _{X_{n, d'}, u'} \mathop{\mathrm{Spec}}(k) \]

Denote $A = \mathcal{O}_{X_{n, d'}, u'}^ h$ the henselization of the local ring of $X_{n, d'}$ at the image of $u'$. Then we can write

\[ Y_{n, d'} \times _{X_{n, d'}} \mathop{\mathrm{Spec}}(A) = Z \amalg W \]

with $Z \to \mathop{\mathrm{Spec}}(A)$ finite and $W \to \mathop{\mathrm{Spec}}(A)$ having empty closed fibre, see Algebra, Lemma 10.153.3 part (13) or the discussion in More on Morphisms, Section 37.40. By Lemma 49.10.6 the local ring $A$ is regular (here we also use More on Algebra, Lemma 15.45.10) and the morphism $Z \to \mathop{\mathrm{Spec}}(A)$ is étale over the generic point of $\mathop{\mathrm{Spec}}(A)$ (because it is mapped to the generic point of $X_{d, n'}$). By construction $Z \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(k) \cong \mathop{\mathrm{Spec}}(B)$. This proves what we want except that the map from residue field of $A$ to $k$ may not be an isomorphism. By Algebra, Lemma 10.159.1 there exists a flat local ring map $A \to A'$ such that the residue field of $A'$ is $k$. If $A'$ isn't a domain, then we choose a minimal prime $\mathfrak p \subset A'$ (which lies over the unique minimal prime of $A$ by flatness) and we replace $A'$ by $A'/\mathfrak p$. Set $B'$ equal to the unique $A'$-algebra such that $Z \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A') = \mathop{\mathrm{Spec}}(B')$. This finishes the proof. $\square$

Remark 49.11.6. Let $\pi _ d : Y_ d \to X_ d$ be as in Example 49.11.2. Let $U_ d \subset X_ d$ be the maximal open over which $V_ d = \pi _ d^{-1}(U_ d)$ is finite syntomic as in Lemma 49.11.3. Then it is also true that $V_ d$ is smooth over $\mathbf{Z}$. (Of course the morphism $V_ d \to U_ d$ is not smooth when $d \geq 2$.) Arguing as in the proof of Lemma 49.11.4 this corresponds to the following deformation problem: given a small extension $C' \to C$ and a finite syntomic $C$-algebra $B$ with a section $B \to C$, find a finite syntomic $C'$-algebra $B'$ and a section $B' \to C'$ whose tensor product with $C$ recovers $B \to C$. By Lemma 49.11.1 we may write $B = C[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ as a relative global complete intersection. After a change of coordinates with may assume $x_1, \ldots , x_ n$ are in the kernel of $B \to C$. Then the polynomials $f_ i$ have vanishing constant terms. Choose any lifts $f'_ i \in C'[x_1, \ldots , x_ n]$ of $f_ i$ with vanishing constant terms. Then $B' = C'[x_1, \ldots , x_ n]/(f'_1, \ldots , f'_ n)$ with section $B' \to C'$ sending $x_ i$ to zero works.

Lemma 49.11.7. Let $f : Y \to X$ be a morphism of schemes. If $f$ satisfies the equivalent conditions of Lemma 49.11.1 then for every $x \in X$ there exist a $d$ and a commutative diagram

\[ \xymatrix{ Y \ar[d] & V \ar[d] \ar[l] \ar[r] & V_ d \ar[d] \ar[r] & Y_ d \ar[d]^{\pi _ d}\\ X & U \ar[l] \ar[r] & U_ d \ar[r] & X_ d } \]

with the following properties

  1. $U \subset X$ is open and $V = f^{-1}(U)$,

  2. $\pi _ d : Y_ d \to X_ d$ is as in Example 49.11.2,

  3. $U_ d \subset X_ d$ is as in Lemma 49.11.3 and $V_ d = \pi _ d^{-1}(U_ d) \subset Y_ d$,

  4. where the middle square is cartesian.

Proof. Choose an affine open neighbourhood $U = \mathop{\mathrm{Spec}}(A) \subset X$ of $x$. Write $V = f^{-1}(U) = \mathop{\mathrm{Spec}}(B)$. Then $B$ is a finite locally free $A$-module and the inclusion $A \subset B$ is a locally direct summand. Thus after shrinking $U$ we can choose a basis $1 = e_1, e_2, \ldots , e_ d$ of $B$ as an $A$-module. Write $e_ i e_ j = \sum \alpha _{ij}^ l e_ l$ for unique elements $\alpha _{ij}^ l \in A$ which satisfy the relations $\sum _ l \alpha _{ij}^ l \alpha _{lk}^ m = \sum _ l \alpha _{il}^ m \alpha _{jk}^ l$ and $\alpha _{ij}^ k = \alpha _{ji}^ k$ and $\alpha _{i1}^ j - \delta _{ij}$ in $A$. This determines a morphism $\mathop{\mathrm{Spec}}(A) \to X_ d$ by sending $a_{ij}^ l \in A_ d$ to $\alpha _{ij}^ l \in A$. By construction $V \cong \mathop{\mathrm{Spec}}(A) \times _{X_ d} Y_ d$. By the definition of $U_ d$ we see that $\mathop{\mathrm{Spec}}(A) \to X_ d$ factors through $U_ d$. This finishes the proof. $\square$


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