Definition 100.25.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is flat if the equivalent conditions of Lemma 100.16.1 hold with $\mathcal{P} = \text{flat}$.
100.25 Flat morphisms
The property “being flat” of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark 73.20.5. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma 66.30.4 and Descent on Spaces, Lemma 73.11.13. Hence, by Lemma 100.16.1 above, we may define what it means for a morphism of algebraic spaces to be flat as follows and it agrees with the already existing notion defined in Properties of Stacks, Section 99.3 when the morphism is representable by algebraic spaces.
Lemma 100.25.2. The composition of flat morphisms is flat.
Proof. Combine Remark 100.16.3 with Morphisms of Spaces, Lemma 66.30.3. $\square$
Lemma 100.25.3. A base change of a flat morphism is flat.
Proof. Combine Remark 100.16.4 with Morphisms of Spaces, Lemma 66.30.4. $\square$
Lemma 100.25.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a surjective flat morphism of algebraic stacks. If the base change $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is flat, then $f$ is flat.
Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Then $W \to \mathcal{Z}$ is surjective and flat (Morphisms of Spaces, Lemma 66.37.7) hence $W \to \mathcal{Y}$ is surjective and flat (by Properties of Stacks, Lemma 99.5.2 and Lemma 100.25.2). Since the base change of $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ by $W \to \mathcal{Z}$ is a flat morphism (Lemma 100.25.3) we may replace $\mathcal{Z}$ by $W$.
Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We have to show that $U \to V$ is flat. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times _\mathcal {Y} U$, $V' = W \times _\mathcal {Y} V$, $\mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}$, and $\mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times _{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by our definition of flat morphisms of algebraic stacks and the assumption that $\mathcal{X}' \to \mathcal{Y}'$ is flat, we see that $U' \to V'$ is flat. Then, since $V' \to V$ is surjective as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ is flat by Morphisms of Spaces, Lemma 66.31.3 (2) and we win. $\square$
Lemma 100.25.5. Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{Z}$ is flat and $\mathcal{X} \to \mathcal{Y}$ is surjective and flat, then $\mathcal{Y} \to \mathcal{Z}$ is flat.
Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times _\mathcal {Z} \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We know that $U \to V$ is flat and that $U \to W$ is flat. Also, as $\mathcal{X} \to \mathcal{Y}$ is surjective we see that $U \to V$ is surjective (as a composition of surjective morphisms). Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Morphisms of Spaces, Lemma 66.31.5. $\square$
Lemma 100.25.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a flat morphism of algebraic stacks. Let $\mathop{\mathrm{Spec}}(A) \to \mathcal{Y}$ be a morphism where $A$ is a valuation ring. If the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to a point of $|\mathcal{Y}|$ in the image of $|\mathcal{X|} \to |\mathcal{Y}|$, then there exists a commutative diagram where $A \to A'$ is an extension of valuation rings (More on Algebra, Definition 15.123.1).
\[ \xymatrix{ \mathop{\mathrm{Spec}}(A') \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathcal{Y} } \]
Proof. The base change $\mathcal{X}_ A \to \mathop{\mathrm{Spec}}(A)$ is flat (Lemma 100.25.3) and the closed point of $\mathop{\mathrm{Spec}}(A)$ is in the image of $|\mathcal{X}_ A| \to |\mathop{\mathrm{Spec}}(A)|$ (Properties of Stacks, Lemma 99.4.3). Thus we may assume $\mathcal{Y} = \mathop{\mathrm{Spec}}(A)$. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. Then we can apply Morphisms of Spaces, Lemma 66.42.4 to the morphism $U \to \mathop{\mathrm{Spec}}(A)$ to conclude. $\square$
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