The Stacks project

76.26 Cohen-Macaulay morphisms

This is the analogue of More on Morphisms, Section 37.22.

Lemma 76.26.1. The property of morphisms of germs of schemes

\begin{align*} & \mathcal{P}((X, x) \to (S, s)) = \\ & \text{the local ring } \mathcal{O}_{X_ s, x} \text{ of the fibre is Noetherian and Cohen-Macaulay} \end{align*}

is étale local on the source-and-target (Descent, Definition 35.33.1).

Proof. Given a diagram as in Descent, Definition 35.33.1 we obtain an étale morphism of fibres $U'_{v'} \to U_ v$ mapping $u'$ to $u$, see Descent, Lemma 35.33.5. Thus the strict henselizations of the local rings $\mathcal{O}_{U'_{v'}, u'}$ and $\mathcal{O}_{U_ v, u}$ are the same. We conclude by More on Algebra, Lemma 15.45.9. $\square$

Definition 76.26.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian (Divisors on Spaces, Definition 71.4.2).

  1. Let $x \in |X|$, and $y = f(x)$. We say that $f$ is Cohen-Macaulay at $x$ if $f$ is flat at $x$ and the equivalent conditions of Morphisms of Spaces, Lemma 67.22.5 hold for the property $\mathcal{P}$ described in Lemma 76.26.1.

  2. We say $f$ is a Cohen-Macaulay morphism if $f$ is Cohen-Macaulay at every point of $X$.

Here is a translation.

Lemma 76.26.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian. The following are equivalent

  1. $f$ is Cohen-Macaulay,

  2. $f$ is flat and for some surjective étale morphism $V \to Y$ where $V$ is a scheme, the fibres of $X_ V \to V$ are Cohen-Macaulay algebraic spaces, and

  3. $f$ is flat and for any étale morphism $V \to Y$ where $V$ is a scheme, the fibres of $X_ V \to V$ are Cohen-Macaulay algebraic spaces.

Given $x \in |X|$ with image $y \in |Y|$ the following are equivalent

  1. $f$ is Cohen-Macaulay at $x$, and

  2. $\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is flat and $\mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}}$ is Cohen-Macaulay.

Proof. Given an étale morphism $V \to Y$ where $V$ is a scheme choose a scheme $U$ and a surjective étale morphism $U \to X \times _ Y V$. Consider the commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

Let $u \in U$ with images $x \in |X|$, $y \in |Y|$, and $v \in V$. Then $f$ is Cohen-Macaulay at $x$ if and only if $U \to V$ is Cohen-Macaulay at $u$ (by definition). Moreover the morphism $U_ v \to X_ v = (X_ V)_ v$ is surjective étale. Hence the scheme $U_ v$ is Cohen-Macaulay if and only if the algebraic space $X_ v$ is Cohen-Macaulay. Thus the equivalence of (1), (2), and (3) follows from the corresponding equivalence for morphisms of schemes, see More on Morphisms, Lemma 37.22.2 by a formal argument.

Proof of equivalence of (a) and (b). The corresponding equivalence for flatness is Morphisms of Spaces, Lemma 67.30.8. Thus we may assume $f$ is flat at $x$ when proving the equivalence. Consider a diagram and $x, y, u, v$ as above. Then $\mathcal{O}_{Y, \overline{y}} \to \mathcal{O}_{X, \overline{x}}$ is equal to the map $\mathcal{O}_{V, v}^{sh} \to \mathcal{O}_{U, u}^{sh}$ on strict henselizations of local rings, see Properties of Spaces, Lemma 66.22.1. Thus we have

\[ \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}} = (\mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u})^{sh} \]

by Algebra, Lemma 10.156.4. Thus we have to show that the Noetherian local ring $\mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u}$ is Cohen-Macaulay if and only if its strict henselization is. This is More on Algebra, Lemma 15.45.9. $\square$

Lemma 76.26.4. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Assume that the fibres of $f$, $g$, and $g \circ f$ are locally Noetherian. Let $x \in |X|$ with images $y \in |Y|$ and $z \in |Z|$.

  1. If $f$ is Cohen-Macaulay at $x$ and $g$ is Cohen-Macaulay at $f(x)$, then $g \circ f$ is Cohen-Macaulay at $x$.

  2. If $f$ and $g$ are Cohen-Macaulay, then $g \circ f$ is Cohen-Macaulay.

  3. If $g \circ f$ is Cohen-Macaulay at $x$ and $f$ is flat at $x$, then $f$ is Cohen-Macaulay at $x$ and $g$ is Cohen-Macaulay at $f(x)$.

  4. If $f \circ g$ is Cohen-Macaulay and $f$ is flat, then $f$ is Cohen-Macaulay and $g$ is Cohen-Macaulay at every point in the image of $f$.

Proof. Working étale locally this follows from the corresponding result for schemes, see More on Morphisms, Lemma 37.22.4. Alternatively, we can use the equivalence of (a) and (b) in Lemma 76.26.3. Thus we consider the local homomorphism of Noetherian local rings

\[ \mathcal{O}_{Y, \overline{y}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{Y, \overline{y}} \longrightarrow \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{z}}\mathcal{O}_{X, \overline{x}} \]

whose fibre is

\[ \mathcal{O}_{X, \overline{x}}/ \mathfrak m_{\overline{y}}\mathcal{O}_{X, \overline{x}} \]

and we use Algebra, Lemma 10.163.3. $\square$

Lemma 76.26.5. Let $S$ be a scheme. Let $f : X \to Y$ be a flat morphism of locally Noetherian algebraic spaces over $S$. If $X$ is Cohen-Macaulay, then $f$ is Cohen-Macaulay and $\mathcal{O}_{Y, f(\overline{x})}$ is Cohen-Macaulay for all $x \in |X|$.

Proof. After translating into algebra using Lemma 76.26.3 (compare with the proof of Lemma 76.26.4) this follows from Algebra, Lemma 10.163.3. $\square$

Lemma 76.26.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume the fibres of $f$ are locally Noetherian. Let $Y' \to Y$ be locally of finite type. Let $f' : X' \to Y'$ be the base change of $f$. Let $x' \in |X'|$ be a point with image $x \in |X|$.

  1. If $f$ is Cohen-Macaulay at $x$, then $f' : X' \to Y'$ is Cohen-Macaulay at $x'$.

  2. If $f$ is flat at $x$ and $f'$ is Cohen-Macaulay at $x'$, then $f$ is Cohen-Macaulay at $x$.

  3. If $Y' \to Y$ is flat at $f'(x')$ and $f'$ is Cohen-Macaulay at $x'$, then $f$ is Cohen-Macaulay at $x$.

Proof. Denote $y \in |Y|$ and $y' \in |Y'|$ the image of $x'$. Choose a surjective étale morphism $V \to Y$ where $V$ is a scheme. Choose a surjective étale morphism $U \to X \times _ Y V$ where $U$ is a scheme. Choose a surjectiev étale morphism $V' \to Y' \times _ Y V$ where $V'$ is a scheme. Then $U' = U \times _ V V'$ is a scheme which comes equipped with a surjective étale morphism $U' \to X'$. Choose $u' \in U'$ mapping to $x'$. Denote $u \in U$ the image of $u'$. Then the lemma follows from the lemma for $U \to V$ and its base change $U' \to V'$ and the points $u'$ and $u$ (this follows from the definitions). Thus the lemma follows from the case of schemes, see More on Morphisms, Lemma 37.22.6. $\square$

Lemma 76.26.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is flat and locally of finite presentation. Let

\[ W = \{ x \in |X| : f\text{ is Cohen-Macaulay at }x\} \]

Then $W$ is open in $|X|$ and the formation of $W$ commutes with arbitrary base change of $f$: For any morphism $g : Y' \to Y$, consider the base change $f' : X' \to Y'$ of $f$ and the projection $g' : X' \to X$. Then the corresponding set $W'$ for the morphism $f'$ is equal to $W' = (g')^{-1}(W)$.

Proof. Choose a commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

with étale vertical arrows and $U$ and $V$ schemes. Let $u \in U$ with image $x \in |X|$. Then $f$ is Cohen-Macaulay at $x$ if and only if $U \to V$ is Cohen-Macaulay at $u$ (by definition). Thus we reduce to the case of the morphism $U \to V$. See More on Morphisms, Lemma 37.22.7. $\square$

Lemma 76.26.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is locally of finite presentation and Cohen-Macaulay. Then there exist open and closed subschemes $X_ d \subset X$ such that $X = \coprod _{d \geq 0} X_ d$ and $f|_{X_ d} : X_ d \to Y$ has relative dimension $d$.

Proof. Choose a commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

with étale vertical arrows and $U$ and $V$ schemes. Then $U \to V$ is locally of finite presentation and Cohen-Macaulay (immediate from our definitions). Thus we have a decomposition $U = \coprod _{d \geq 0} U_ d$ into open and closed subschemes with $f|_{U_ d} : U_ d \to V$ of relative dimension $d$, see Morphisms, Lemma 29.29.4. Let $u \in U$ with image $x \in |X|$. Then $f$ has relative dimension $d$ at $x$ if and only if $U \to V$ has relative dimension $d$ at $u$ (this follows from our definitions). In this way we see that $U_ d$ is the inverse image of a subset $X_ d \subset |X|$ which is necessarily open and closed. Denoting $X_ d$ the corresponding open and closed algebraic subspace of $X$ we see that the lemma is true. $\square$

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