## 29.25 Flat morphisms

Flatness is one of the most important technical tools in algebraic geometry. In this section we introduce this notion. We intentionally limit the discussion to straightforward observations, apart from Lemma 29.25.10. A very important class of results, namely criteria for flatness, are discussed in Algebra, Sections 10.99, 10.101, 10.128, and More on Morphisms, Section 37.16. There is a chapter dedicated to advanced material on flat morphisms of schemes, namely More on Flatness, Section 38.1.

Recall that a module $M$ over a ring $R$ is *flat* if the functor $-\otimes _ R M : \text{Mod}_ R \to \text{Mod}_ R$ is exact. A ring map $R \to A$ is said to be *flat* if $A$ is flat as an $R$-module. See Algebra, Definition 10.39.1.

Definition 29.25.1. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-modules.

We say $f$ is *flat at a point $x \in X$* if the local ring $\mathcal{O}_{X, x}$ is flat over the local ring $\mathcal{O}_{S, f(x)}$.

We say that $\mathcal{F}$ is *flat over $S$ at a point $x \in X$* if the stalk $\mathcal{F}_ x$ is a flat $\mathcal{O}_{S, f(x)}$-module.

We say $f$ is *flat* if $f$ is flat at every point of $X$.

We say that $\mathcal{F}$ is *flat over $S$* if $\mathcal{F}$ is flat over $S$ at every point $x$ of $X$.

Thus we see that $f$ is flat if and only if the structure sheaf $\mathcal{O}_ X$ is flat over $S$.

Lemma 29.25.2. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-modules. The following are equivalent

The sheaf $\mathcal{F}$ is flat over $S$.

For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the $\mathcal{O}_ S(V)$-module $\mathcal{F}(U)$ is flat.

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 modules $\mathcal{F}|_{U_ i}$ is flat over $V_ j$, for all $j\in J, i\in I_ j$.

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 $\mathcal{F}(U_ i)$ is a flat $\mathcal{O}_ S(V_ j)$-module, for all $j\in J, i\in I_ j$.

Moreover, if $\mathcal{F}$ is flat over $S$ then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $\mathcal{F}|_ U$ is flat over $V$.

**Proof.**
Let $R \to A$ be a ring map. Let $M$ be an $A$-module. If $M$ is $R$-flat, then for all primes $\mathfrak q$ the module $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$. Conversely, if $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ for all primes $\mathfrak q$ of $A$, then $M$ is flat over $R$. See Algebra, Lemma 10.39.18. This equivalence easily implies the statements of the lemma.
$\square$

Lemma 29.25.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

The morphism $f$ is flat.

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 flat.

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 flat.

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 $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is flat, for all $j\in J, i\in I_ j$.

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

**Proof.**
This is a special case of Lemma 29.25.2 above.
$\square$

Lemma 29.25.4. Let $f : X \to Y$ be an affine morphism of schemes over a base scheme $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then $\mathcal{F}$ is flat over $S$ if and only if $f_*\mathcal{F}$ is flat over $S$.

**Proof.**
By Lemma 29.25.2 and the fact that $f$ is an affine morphism, this reduces us to the affine case. Say $X \to Y \to S$ corresponds to the ring maps $C \leftarrow B \leftarrow A$. Let $N$ be the $C$-module corresponding to $\mathcal{F}$. Recall that $f_*\mathcal{F}$ corresponds to $N$ viewed as a $B$-module, see Schemes, Lemma 26.7.3. Thus the result is clear.
$\square$

Lemma 29.25.5. Let $X \to Y \to Z$ be morphisms of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $x \in X$ with image $y$ in $Y$. If $\mathcal{F}$ is flat over $Y$ at $x$, and $Y$ is flat over $Z$ at $y$, then $\mathcal{F}$ is flat over $Z$ at $x$.

**Proof.**
See Algebra, Lemma 10.39.4.
$\square$

Lemma 29.25.6. The composition of flat morphisms is flat.

**Proof.**
This is a special case of Lemma 29.25.5.
$\square$

Lemma 29.25.7. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-modules. Let $g : S' \to S$ be a morphism of schemes. Denote $g' : X' = X_{S'} \to X$ the projection. Let $x' \in X'$ be a point with image $x = g'(x') \in X$. If $\mathcal{F}$ is flat over $S$ at $x$, then $(g')^*\mathcal{F}$ is flat over $S'$ at $x'$. In particular, if $\mathcal{F}$ is flat over $S$, then $(g')^*\mathcal{F}$ is flat over $S'$.

**Proof.**
See Algebra, Lemma 10.39.7.
$\square$

Lemma 29.25.8. The base change of a flat morphism is flat.

**Proof.**
This is a special case of Lemma 29.25.7.
$\square$

Lemma 29.25.9. Let $f : X \to S$ be a flat morphism of schemes. Then generalizations lift along $f$, see Topology, Definition 5.19.4.

**Proof.**
See Algebra, Section 10.41.
$\square$

Lemma 29.25.10. A flat morphism locally of finite presentation is universally open.

**Proof.**
This follows from Lemmas 29.25.9 and Lemma 29.23.2 above. We can also argue directly as follows.

Let $f : X \to S$ be flat locally of finite presentation. To show $f$ is open it suffices to show that we may cover $X$ by open affines $X = \bigcup U_ i$ such that $U_ i \to S$ is open. By definition we may cover $X$ by affine opens $U_ i \subset X$ such that each $U_ i$ maps into an affine open $V_ i \subset S$ and such that the induced ring map $\mathcal{O}_ S(V_ i) \to \mathcal{O}_ X(U_ i)$ is of finite presentation. Thus $U_ i \to V_ i$ is open by Algebra, Proposition 10.41.8. The lemma follows.
$\square$

Lemma 29.25.11. Let $f : X \to Y$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume $f$ locally finite presentation, $\mathcal{F}$ of finite type, $X = \text{Supp}(\mathcal{F})$, and $\mathcal{F}$ flat over $Y$. Then $f$ is universally open.

**Proof.**
By Lemmas 29.25.7, 29.21.4, and 29.5.3 the assumptions are preserved under base change. By Lemma 29.23.2 it suffices to show that generalizations lift along $f$. This follows from Algebra, Lemma 10.41.12.
$\square$

reference
Lemma 29.25.12. Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism. A subset $T \subset Y$ is open (resp. closed) if and only $f^{-1}(T)$ is open (resp. closed). In other words, $f$ is a submersive morphism.

**Proof.**
The question is local on $Y$, hence we may assume that $Y$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ as a finite union of affine opens. Then $f' : X' = X_1 \amalg \ldots \amalg X_ n \to Y$ is a surjective flat morphism of affine schemes. Note that for $T \subset Y$ we have $(f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_ n$. Hence, $f^{-1}(T)$ is open if and only if $(f')^{-1}(T)$ is open. Thus we may assume both $X$ and $Y$ are affine.

Let $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be a surjective morphism of affine schemes corresponding to a flat ring map $A \to B$. Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(J)$ for $J \subset B$ an ideal. Then $T = f(f^{-1}(T)) = f(V(J))$ is the image of $\mathop{\mathrm{Spec}}(B/J) \to \mathop{\mathrm{Spec}}(A)$ (here we use that $f$ is surjective). On the other hand, generalizations lift along $f$ (Lemma 29.25.9). Hence by Topology, Lemma 5.19.6 we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under generalization. Hence $T$ is stable under specialization (Topology, Lemma 5.19.2). Thus $T$ is closed by Algebra, Lemma 10.41.5.
$\square$

Lemma 29.25.13. Let $h : X \to Y$ be a morphism of schemes over $S$. Let $\mathcal{G}$ be a quasi-coherent sheaf on $Y$. Let $x \in X$ with $y = h(x) \in Y$. If $h$ is flat at $x$, then

\[ \mathcal{G}\text{ flat over }S\text{ at }y \Leftrightarrow h^*\mathcal{G}\text{ flat over }S\text{ at }x. \]

In particular: If $h$ is surjective and flat, then $\mathcal{G}$ is flat over $S$, if and only if $h^*\mathcal{G}$ is flat over $S$. If $h$ is surjective and flat, and $X$ is flat over $S$, then $Y$ is flat over $S$.

**Proof.**
You can prove this by applying Algebra, Lemma 10.39.9. Here is a direct proof. Let $s \in S$ be the image of $y$. Consider the local ring maps $\mathcal{O}_{S, s} \to \mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$. By assumption the ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is faithfully flat, see Algebra, Lemma 10.39.17. Let $N = \mathcal{G}_ y$. Note that $h^*\mathcal{G}_ x = N \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x}$, see Sheaves, Lemma 6.26.4. Let $M' \to M$ be an injection of $\mathcal{O}_{S, s}$-modules. By the faithful flatness mentioned above we have

\begin{align*} \mathop{\mathrm{Ker}}( M' \otimes _{\mathcal{O}_{S, s}} N \to M \otimes _{\mathcal{O}_{S, s}} N) \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \\ = \mathop{\mathrm{Ker}}( M' \otimes _{\mathcal{O}_{S, s}} N \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x} \to M \otimes _{\mathcal{O}_{S, s}} N \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{X, x}) \end{align*}

Hence the equivalence of the lemma follows from the second characterization of flatness in Algebra, Lemma 10.39.5.
$\square$

Lemma 29.25.14. Let $f : Y \to X$ be a morphism of schemes. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module with scheme theoretic support $Z \subset X$. If $f$ is flat, then $f^{-1}(Z)$ is the scheme theoretic support of $f^*\mathcal{F}$.

**Proof.**
Using the characterization of scheme theoretic support on affines as given in Lemma 29.5.4 we reduce to Algebra, Lemma 10.40.4.
$\square$

Lemma 29.25.15. Let $f : X \to Y$ be a flat morphism of schemes. Let $V \subset Y$ be a retrocompact open which is scheme theoretically dense. Then $f^{-1}V$ is scheme theoretically dense in $X$.

**Proof.**
We will use the characterization of Lemma 29.7.5. We have to show that for any open $U \subset X$ the map $\mathcal{O}_ X(U) \to \mathcal{O}_ X(U \cap f^{-1}V)$ is injective. It suffices to prove this when $U$ is an affine open which maps into an affine open $W \subset Y$. Say $W = \mathop{\mathrm{Spec}}(A)$ and $U = \mathop{\mathrm{Spec}}(B)$. Then $V \cap W = D(f_1) \cup \ldots \cup D(f_ n)$ for some $f_ i \in A$, see Algebra, Lemma 10.29.1. Thus we have to show that $B \to B_{f_1} \times \ldots \times B_{f_ n}$ is injective. We are given that $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective and that $A \to B$ is flat. Since $B_{f_ i} = A_{f_ i} \otimes _ A B$ we win.
$\square$

slogan
Lemma 29.25.16. Let $f : X \to Y$ be a flat morphism of schemes. Let $g : V \to Y$ be a quasi-compact morphism of schemes. Let $Z \subset Y$ be the scheme theoretic image of $g$ and let $Z' \subset X$ be the scheme theoretic image of the base change $V \times _ Y X \to X$. Then $Z' = f^{-1}Z$.

**Proof.**
Recall that $Z$ is cut out by $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to g_*\mathcal{O}_ V)$ and $Z'$ is cut out by $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to (V \times _ Y X \to X)_*\mathcal{O}_{V \times _ Y X})$, see Lemma 29.6.3. Hence the question is local on $X$ and $Y$ and we may assume $X$ and $Y$ affine. Note that we may replace $V$ by $\coprod V_ i$ where $V = V_1 \cup \ldots \cup V_ n$ is a finite affine open covering. Hence we may assume $g$ is affine. In this case $(V \times _ Y X \to X)_*\mathcal{O}_{V \times _ Y X}$ is the pullback of $g_*\mathcal{O}_ V$ by $f$. Since $f$ is flat we conclude that $f^*\mathcal{I} = \mathcal{I}'$ and the lemma holds.
$\square$

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