The Stacks project

76.3 Relatively pure modules

This section is the analogue of More on Flatness, Section 38.16.

Definition 76.3.1. In Situation 76.2.1.

  1. We say $\mathcal{F}$ is pure above $y$ if none of the equivalent conditions of Lemma 76.2.5 hold.

  2. We say $\mathcal{F}$ is universally pure above $y$ if there does not exist any impurity of $\mathcal{F}$ above $y$.

  3. We say that $X$ is pure above $y$ if $\mathcal{O}_ X$ is pure above $y$.

  4. We say $\mathcal{F}$ is universally $Y$-pure, or universally pure relative to $Y$ if $\mathcal{F}$ is universally pure above $y$ for every $y \in |Y|$.

  5. We say $\mathcal{F}$ is $Y$-pure, or pure relative to $Y$ if $\mathcal{F}$ is pure above $y$ for every $y \in |Y|$.

  6. We say that $X$ is $Y$-pure or pure relative to $Y$ if $\mathcal{O}_ X$ is pure relative to $Y$.

The obligatory lemmas follow.

Lemma 76.3.2. In Situation 76.2.1.

  1. $\mathcal{F}$ is universally pure above $y$, and

  2. for every morphism $(Y', y') \to (Y, y)$ of pointed algebraic spaces the pullback $\mathcal{F}_{Y'}$ is pure above $y'$.

In particular, $\mathcal{F}$ is universally pure relative to $Y$ if and only if every base change $\mathcal{F}_{Y'}$ of $\mathcal{F}$ is pure relative to $Y'$.

Proof. This is formal. $\square$

Lemma 76.3.3. In Situation 76.2.1. Let $(Y', y') \to (Y, y)$ be a morphism of pointed algebraic spaces. If $Y' \to Y$ is quasi-finite at $y'$ and $\mathcal{F}$ is pure above $y$, then $\mathcal{F}_{Y'}$ is pure above $y'$.

Proof. It $(T \to Y', t' \leadsto t, \xi )$ is an impurity of $\mathcal{F}_{Y'}$ above $y'$ with $T \to Y'$ quasi-finite at $t$, then $(T \to Y, t' \to t, \xi )$ is an impurity of $\mathcal{F}$ above $y$ with $T \to Y$ quasi-finite at $t$, see Morphisms of Spaces, Lemma 66.27.3. Hence the lemma follows immediately from the definition of purity. $\square$

Purity satisfies flat descent.

Lemma 76.3.4. In Situation 76.2.1. Let $(Y_1, y_1) \to (Y, y)$ be a morphism of pointed algebraic spaces. Assume $Y_1 \to Y$ is flat at $y_1$.

  1. If $\mathcal{F}_{Y_1}$ is pure above $y_1$, then $\mathcal{F}$ is pure above $y$.

  2. If $\mathcal{F}_{Y_1}$ is universally pure above $y_1$, then $\mathcal{F}$ is universally pure above $y$.

Proof. This is true because impurities go up along a flat base change, see Lemma 76.2.4. For example part (1) follows because by any impurity $(T \to Y, t' \leadsto t, \xi )$ of $\mathcal{F}$ above $y$ with $T \to Y$ quasi-finite at $t$ by the lemma leads to an impurity $(T_1 \to Y_1, t_1' \leadsto t_1, \xi _1)$ of the pullback $\mathcal{F}_1$ of $\mathcal{F}$ to $X_1 = Y_1 \times _ Y X$ over $y_1$ such that $T_1$ is ├ętale over $Y_1 \times _ Y T$. Hence $T_1 \to Y_1$ is quasi-finite at $t_1$ because ├ętale morphisms are locally quasi-finite and compositions of locally quasi-finite morphisms are locally quasi-finite (Morphisms of Spaces, Lemmas 66.39.5 and 66.27.3). Similarly for part (2). $\square$

Lemma 76.3.5. In Situation 76.2.1. Let $i : Z \to X$ be a closed immersion and assume that $\mathcal{F} = i_*\mathcal{G}$ for some finite type, quasi-coherent sheaf $\mathcal{G}$ on $Z$. Then $\mathcal{G}$ is (universally) pure above $y$ if and only if $\mathcal{F}$ is (universally) pure above $y$.

Proof. This follows from Divisors on Spaces, Lemma 70.4.9. $\square$

Lemma 76.3.6. In Situation 76.2.1.

  1. If the support of $\mathcal{F}$ is proper over $Y$, then $\mathcal{F}$ is universally pure relative to $Y$.

  2. If $f$ is proper, then $\mathcal{F}$ is universally pure relative to $Y$.

  3. If $f$ is proper, then $X$ is universally pure relative to $Y$.

Proof. First we reduce (1) to (2). Namely, let $Z \subset X$ be the scheme theoretic support of $\mathcal{F}$ (Morphisms of Spaces, Definition 66.15.4). Let $i : Z \to X$ be the corresponding closed immersion and write $\mathcal{F} = i_*\mathcal{G}$ for some finite type quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$. In case (1) $Z \to Y$ is proper by assumption. Thus by Lemma 76.3.5 case (1) reduces to case (2).

Assume $f$ is proper. Let $(g : T \to Y, t' \leadsto t, \xi )$ be an impurity of $\mathcal{F}$ above $y$. Since $f$ is proper, it is universally closed. Hence $f_ T : X_ T \to T$ is closed. Since $f_ T(\xi ) = t'$ this implies that $t \in f(\overline{\{ \xi \} })$ which is a contradiction. $\square$


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