The Stacks project

29.30 Syntomic morphisms

An algebra $A$ over a field $k$ is called a global complete intersection over $k$ if $A \cong k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $\dim (A) = n - c$. An algebra $A$ over a field $k$ is called a local complete intersection if $\mathop{\mathrm{Spec}}(A)$ can be covered by standard opens each of which are global complete intersections over $k$. See Algebra, Section 10.135. Recall that a ring map $R \to A$ is syntomic if it is of finite presentation, flat with local complete intersection rings as fibres, see Algebra, Definition 10.136.1.

Definition 29.30.1. Let $f : X \to S$ be a morphism of schemes.

  1. We say that $f$ is syntomic at $x \in X$ if there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset X$ of $x$ and affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ such that the induced ring map $R \to A$ is syntomic.

  2. We say that $f$ is syntomic if it is syntomic at every point of $X$.

  3. If $S = \mathop{\mathrm{Spec}}(k)$ and $f$ is syntomic, then we say that $X$ is a local complete intersection over $k$.

  4. A morphism of affine schemes $f : X \to S$ is called standard syntomic if there exists a global relative complete intersection $R \to R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ (see Algebra, Definition 10.136.5) such that $X \to S$ is isomorphic to

    \[ \mathop{\mathrm{Spec}}(R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)) \to \mathop{\mathrm{Spec}}(R). \]

In the literature a syntomic morphism is sometimes referred to as a flat local complete intersection morphism. It turns out this is a convenient class of morphisms. For example one can define a syntomic topology using these, which is finer than the smooth and étale topologies, but has many of the same formal properties.

A global relative complete intersection (which we used to define standard syntomic ring maps) is in particular flat. In More on Morphisms, Section 37.62 we will consider morphisms $X \to S$ which locally are of the form

\[ \mathop{\mathrm{Spec}}(R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)) \to \mathop{\mathrm{Spec}}(R). \]

for some Koszul-regular sequence $f_1, \ldots , f_ r$ in $R[x_1, \ldots , x_ n]$. Such a morphism will be called a local complete intersection morphism. Once we have this definition in place it will be the case that a morphism is syntomic if and only if it is a flat, local complete intersection morphism.

Note that there is no separation or quasi-compactness hypotheses in the definition of a syntomic morphism. Hence the question of being syntomic is local in nature on the source. Here is the precise result.

Lemma 29.30.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

  1. The morphism $f$ is syntomic.

  2. For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is syntomic.

  3. There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is syntomic.

  4. There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is syntomic, for all $j\in J, i\in I_ j$.

Moreover, if $f$ is syntomic then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is syntomic.

Proof. This follows from Lemma 29.14.3 if we show that the property “$R \to A$ is syntomic” is local. We check conditions (a), (b) and (c) of Definition 29.14.1. By Algebra, Lemma 10.136.3 being syntomic is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 10.136.17 being syntomic is stable under composition and trivially for any ring $R$ the ring map $R \to R_ f$ is syntomic. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 10.136.4. $\square$

Lemma 29.30.3. The composition of two morphisms which are syntomic is syntomic.

Proof. In the proof of Lemma 29.30.2 we saw that being syntomic is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 29.14.5 combined with the fact that being syntomic is a property of ring maps that is stable under composition, see Algebra, Lemma 10.136.17. $\square$

Lemma 29.30.4. The base change of a morphism which is syntomic is syntomic.

Proof. In the proof of Lemma 29.30.2 we saw that being syntomic is a local property of ring maps. Hence the lemma follows from Lemma 29.14.5 combined with the fact that being syntomic is a property of ring maps that is stable under base change, see Algebra, Lemma 10.136.3. $\square$

Proof. This is true because an open immersion is a local isomorphism. $\square$

Lemma 29.30.6. A syntomic morphism is locally of finite presentation.

Proof. True because a syntomic ring map is of finite presentation by definition. $\square$

Proof. True because a syntomic ring map is flat by definition. $\square$

Let $k$ be a field. Let $A$ be a local $k$-algebra essentially of finite type over $k$. Recall that $A$ is called a complete intersection over $k$ if we can write $A \cong R/(f_1, \ldots , f_ c)$ where $R$ is a regular local ring essentially of finite type over $k$, and $f_1, \ldots , f_ c$ is a regular sequence in $R$, see Algebra, Definition 10.135.5.

Lemma 29.30.9. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. The following are equivalent:

  1. $X$ is a local complete intersection over $k$,

  2. for every $x \in X$ there exists an affine open $U = \mathop{\mathrm{Spec}}(R) \subset X$ neighbourhood of $x$ such that $R \cong k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a global complete intersection over $k$, and

  3. for every $x \in X$ the local ring $\mathcal{O}_{X, x}$ is a complete intersection over $k$.

Proof. The corresponding algebra results can be found in Algebra, Lemmas 10.135.8 and 10.135.9. $\square$

The following lemma says locally any syntomic morphism is standard syntomic. Hence we can use standard syntomic morphisms as a local model for a syntomic morphism. Moreover, it says that a flat morphism of finite presentation is syntomic if and only if the fibres are local complete intersection schemes.

Lemma 29.30.10. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point with image $s = f(x)$. Let $V \subset S$ be an affine open neighbourhood of $s$. The following are equivalent

  1. The morphism $f$ is syntomic at $x$.

  2. There exist an affine open $U \subset X$ with $x \in U$ and $f(U) \subset V$ such that $f|_ U : U \to V$ is standard syntomic.

  3. The morphism $f$ is of finite presentation at $x$, the local ring map $\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$ is flat and $\mathcal{O}_{X, x}/\mathfrak m_ s \mathcal{O}_{X, x}$ is a complete intersection over $\kappa (s)$ (see Algebra, Definition 10.135.5).

Proof. Follows from the definitions and Algebra, Lemma 10.136.15. $\square$

Lemma 29.30.11. Let $f : X \to S$ be a morphism of schemes. If $f$ is flat, locally of finite presentation, and all fibres $X_ s$ are local complete intersections, then $f$ is syntomic.

Proof. Clear from Lemmas 29.30.9 and 29.30.10 and the isomorphisms of local rings $ \mathcal{O}_{X, x}/\mathfrak m_ s \mathcal{O}_{X, x} \cong \mathcal{O}_{X_ s, x} $. $\square$

Lemma 29.30.12. Let $f : X \to S$ be a morphism of schemes. Assume $f$ locally of finite type. Formation of the set

\[ T = \{ x \in X \mid \mathcal{O}_{X_{f(x)}, x} \text{ is a complete intersection over }\kappa (f(x))\} \]

commutes with arbitrary base change: For any morphism $g : S' \to S$, consider the base change $f' : X' \to S'$ of $f$ and the projection $g' : X' \to X$. Then the corresponding set $T'$ for the morphism $f'$ is equal to $T' = (g')^{-1}(T)$. In particular, if $f$ is assumed flat, and locally of finite presentation then the same holds for the open set of points where $f$ is syntomic.

Proof. Let $s' \in S'$ be a point, and let $s = g(s')$. Then we have

\[ X'_{s'} = \mathop{\mathrm{Spec}}(\kappa (s')) \times _{\mathop{\mathrm{Spec}}(\kappa (s))} X_ s \]

In other words the fibres of the base change are the base changes of the fibres. Hence the first part is equivalent to Algebra, Lemma 10.135.10. The second part follows from the first because in that case $T$ is the set of points where $f$ is syntomic according to Lemma 29.30.10. $\square$

Lemma 29.30.13. Let $R$ be a ring. Let $R \to A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ be a relative global complete intersection. Set $S = \mathop{\mathrm{Spec}}(R)$ and $X = \mathop{\mathrm{Spec}}(A)$. Consider the morphism $f : X \to S$ associated to the ring map $R \to A$. The function $x \mapsto \dim _ x(X_{f(x)})$ is constant with value $n - c$.

Proof. By Algebra, Definition 10.136.5 $R \to A$ being a relative global complete intersection means all nonzero fibre rings have dimension $n - c$. Thus for a prime $\mathfrak p$ of $R$ the fibre ring $\kappa (\mathfrak p)[x_1, \ldots , x_ n]/(\overline{f}_1, \ldots , \overline{f}_ c)$ is either zero or a global complete intersection ring of dimension $n - c$. By the discussion following Algebra, Definition 10.135.1 this implies it is equidimensional of dimension $n - c$. Whence the lemma. $\square$

Lemma 29.30.14. Let $f : X \to S$ be a syntomic morphism. The function $x \mapsto \dim _ x(X_{f(x)})$ is locally constant on $X$.

Proof. By Lemma 29.30.10 the morphism $f$ locally looks like a standard syntomic morphism of affines. Hence the result follows from Lemma 29.30.13. $\square$

Lemma 29.30.14 says that the following definition makes sense.

Definition 29.30.15. Let $d \geq 0$ be an integer. We say a morphism of schemes $f : X \to S$ is syntomic of relative dimension $d$ if $f$ is syntomic and the function $\dim _ x(X_{f(x)}) = d$ for all $x \in X$.

In other words, $f$ is syntomic and the nonempty fibres are equidimensional of dimension $d$.

Lemma 29.30.16. Let

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[dl]^ q \\ & S } \]

be a commutative diagram of morphisms of schemes. Assume that

  1. $f$ is surjective and syntomic,

  2. $p$ is syntomic, and

  3. $q$ is locally of finite presentation1.

Then $q$ is syntomic.

Proof. By Lemma 29.25.13 we see that $q$ is flat. Hence it suffices to show that the fibres of $Y \to S$ are local complete intersections, see Lemma 29.30.11. Let $s \in S$. Consider the morphism $X_ s \to Y_ s$. This is a base change of the morphism $X \to Y$ and hence surjective, and syntomic (Lemma 29.30.4). For the same reason $X_ s$ is syntomic over $\kappa (s)$. Moreover, $Y_ s$ is locally of finite type over $\kappa (s)$ (Lemma 29.15.4). In this way we reduce to the case where $S$ is the spectrum of a field.

Assume $S = \mathop{\mathrm{Spec}}(k)$. Let $y \in Y$. Choose an affine open $\mathop{\mathrm{Spec}}(A) \subset Y$ neighbourhood of $y$. Let $\mathop{\mathrm{Spec}}(B) \subset X$ be an affine open such that $f(\mathop{\mathrm{Spec}}(B)) \subset \mathop{\mathrm{Spec}}(A)$, containing a point $x \in X$ such that $f(x) = y$. Choose a surjection $k[x_1, \ldots , x_ n] \to A$ with kernel $I$. Choose a surjection $A[y_1, \ldots , y_ m] \to B$, which gives rise in turn to a surjection $k[x_ i, y_ j] \to B$ with kernel $J$. Let $\mathfrak q \subset k[x_ i, y_ j]$ be the prime corresponding to $y \in \mathop{\mathrm{Spec}}(B)$ and let $\mathfrak p \subset k[x_ i]$ the prime corresponding to $x \in \mathop{\mathrm{Spec}}(A)$. Since $x$ maps to $y$ we have $\mathfrak p = \mathfrak q \cap k[x_ i]$. Consider the following commutative diagram of local rings:

\[ \xymatrix{ \mathcal{O}_{X, x} \ar@{=}[r] & B_{\mathfrak q} & k[x_1, \ldots , x_ n, y_1, \ldots , y_ m]_{\mathfrak q} \ar[l] \\ \mathcal{O}_{Y, y} \ar@{=}[r] \ar[u] & A_{\mathfrak p} \ar[u] & k[x_1, \ldots , x_ n]_{\mathfrak p} \ar[l] \ar[u] } \]

We claim that the hypotheses of Algebra, Lemma 10.135.12 are satisfied. Conditions (1) and (2) are trivial. Condition (4) follows as $X \to Y$ is flat. Condition (3) follows as the rings $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X_ y, x} = \mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ are complete intersection rings by our assumptions that $f$ and $p$ are syntomic, see Lemma 29.30.10. The output of Algebra, Lemma 10.135.12 is exactly that $\mathcal{O}_{Y, y}$ is a complete intersection ring! Hence by Lemma 29.30.10 again we see that $Y$ is syntomic over $k$ at $y$ as desired. $\square$

[1] In fact, if $f$ is surjective, flat, and locally of finite presentation and $p$ is syntomic, then both $q$ and $f$ are syntomic, see Descent, Lemma 35.14.7.

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