## 98.2 Conventions and abuse of language

We choose a big fppf site $\mathit{Sch}_{fppf}$. All schemes are contained in $\mathit{Sch}_{fppf}$. And all rings $A$ considered have the property that $\mathop{\mathrm{Spec}}(A)$ is (isomorphic) to an object of this big site.

We also fix a base scheme $S$, by the conventions above an element of $\mathit{Sch}_{fppf}$. The reader who is only interested in the absolute case can take $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$.

Here are our conventions regarding algebraic stacks:

When we say

*algebraic stack*we will mean an algebraic stacks over $S$, i.e., a category fibred in groupoids $p : \mathcal{X} \to (\mathit{Sch}/S)_{fppf}$ which satisfies the conditions of Algebraic Stacks, Definition 92.12.1.We will say $f : \mathcal{X} \to \mathcal{Y}$ is a

*morphism of algebraic stacks*to indicate a $1$-morphism of algebraic stacks over $S$, i.e., a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$, see Algebraic Stacks, Definition 92.12.3.A

*$2$-morphism*$\alpha : f \to g$ will indicate a $2$-morphism in the $2$-category of algebraic stacks over $S$, see Algebraic Stacks, Definition 92.12.3.Given morphisms $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ of algebraic stacks we abusively call the $2$-fibre product $\mathcal{X} \times _\mathcal {Z} \mathcal{Y}$ the

*fibre product*.We will write $\mathcal{X} \times _ S \mathcal{Y}$ for the product of the algebraic stacks $\mathcal{X}$, $\mathcal{Y}$.

We will often abuse notation and say two algebraic stacks $\mathcal{X}$ and $\mathcal{Y}$ are

*isomorphic*if they are equivalent in this $2$-category.

Here are our conventions regarding algebraic spaces.

If we say $X$ is an

*algebraic space*then we mean that $X$ is an algebraic space over $S$, i.e., $X$ is a presheaf on $(\mathit{Sch}/S)_{fppf}$ which satisfies the conditions of Spaces, Definition 63.6.1.A

*morphism of algebraic spaces*$f :X \to Y$ is a morphism of algebraic spaces over $S$ as defined in Spaces, Definition 63.6.3.We will

**not**distinguish between an algebraic space $X$ and the algebraic stack $\mathcal{S}_ X \to (\mathit{Sch}/S)_{fppf}$ it gives rise to, see Algebraic Stacks, Lemma 92.13.1.In particular, a

*morphism*$f : X \to \mathcal{Y}$ from $X$ to an algebraic stack $\mathcal{Y}$ means a morphism $f : \mathcal{S}_ X \to \mathcal{Y}$ of algebraic stacks. Similarly for morphisms $\mathcal{Y} \to X$.Moreover, given an algebraic stack $\mathcal{X}$ we say

*$\mathcal{X}$ is an algebraic space*to indicate that $\mathcal{X}$ is representable by an algebraic space, see Algebraic Stacks, Definition 92.8.1.We will use the following notational convention: If we indicate an algebraic stack by a roman capital (such as $X, Y, Z, A, B, \ldots $) then it will be the case that its inertia stack is trivial, and hence it is an algebraic space, see Algebraic Stacks, Proposition 92.13.3.

Here are our conventions regarding schemes.

If we say $X$ is a

*scheme*then we mean that $X$ is a scheme over $S$, i.e., $X$ is an object of $(\mathit{Sch}/S)_{fppf}$.By a

*morphism of schemes*we mean a morphism of schemes over $S$.We will

**not**distinguish between a scheme $X$ and the algebraic stack $\mathcal{S}_ X \to (\mathit{Sch}/S)_{fppf}$ it gives rise to, see Algebraic Stacks, Lemma 92.13.1.In particular, a

*morphism*$f : X \to \mathcal{Y}$ from a scheme $X$ to an algebraic stack $\mathcal{Y}$ means a morphism $f : \mathcal{S}_ X \to \mathcal{Y}$ of algebraic stacks. Similarly for morphisms $\mathcal{Y} \to X$.Moreover, given an algebraic stack $\mathcal{X}$ we say

*$\mathcal{X}$ is a scheme*to indicate that $\mathcal{X}$ is representable, see Algebraic Stacks, Section 92.4.

Here are our conventions regarding morphisms of algebraic stacks:

A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is

*representable*, or*representable by schemes*if for every scheme $T$ and morphism $T \to \mathcal{Y}$ the fibre product $T \times _\mathcal {Y} \mathcal{X}$ is a scheme. See Algebraic Stacks, Section 92.6.A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is

*representable by algebraic spaces*if for every scheme $T$ and morphism $T \to \mathcal{Y}$ the fibre product $T \times _\mathcal {Y} \mathcal{X}$ is an algebraic space. See Algebraic Stacks, Definition 92.9.1. In this case $Z \times _\mathcal {Y} \mathcal{X}$ is an algebraic space whenever $Z \to \mathcal{Y}$ is a morphism whose source is an algebraic space, see Algebraic Stacks, Lemma 92.9.8.

Note that every morphism $X \to \mathcal{Y}$ from an algebraic space to an algebraic stack is representable by algebraic spaces, see Algebraic Stacks, Lemma 92.10.11. We will use this basic result without further mention.

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