# Flat Morphisms and Flatness

In this post we will review some of the basic properties of flat/faithuflly flat modules, define flat morphisms of schemes, and discuss some of the nice properties that these morphisms have.

## Motivation

Flatness is a property of modules that is familiar to anyone who has taken algebra. In particular, $M$ being flat tells us that if things are put together in a certain way (i.e. they fit into an exact sequence) then their respective tensor products with $M$ also fit together in a predictable way (i.e. the resulting sequence is still exact). Or, said differently, the functor $-\otimes M$ is exact–it takes exact sequences to exact sequences.

In some ways, this is precisely the role that flatness plays in algebraic geometry. But now, instead of flat modules, we talk about flat morphisms. While it seems difficult to immediately see how to define a flat morphism, given only the definition of a flat module, it’s simple if one thinks functorially. Given two schemes $X$ and $Y$, a morphism $\pi:X\to Y$ induces a functor $\pi^\ast:\mathsf{Qcoh}(Y)\to\mathsf{Qcoh}(X)$ between the categories of quasicoherent sheaves on the respective schemes. In general, this functor will not be exact, and it is precisely flatness of the morphism $\pi$ which does guarantees exactness. This should not be all that surprising since, for a map of affine schemes $\text{Spec}(B)\to\text{Spec}(A)$, the pull-back functor corresponds precisely to the tensor functor $-\otimes_A B:A\text{-}\mathsf{mod}\to B\text{-}\mathsf{mod}$.

Of course, when we think about what the pull-back functor does, this tells us geometrically what a flat morphism is. Pull-back should be thought of, in some sense, as generalized restriction. Thus, flat morphisms are precisely describing morphisms for which restricting along them should respect the way that (quasicoherent) sheaves are put-together. So, intuitively, one might expect that open embeddings should be flat (you’re just leaving off some “small” set of points), but that closed embedding shouldn’t be (you’re leaving off a huge set of points, and possibly ‘crushing’ the structure sheaf along the way). This corresponds, roughly, to the fact that localizations are flat, but quotients aren’t (most of the time!).

It will turn out that adapting the notion of flatness to schemes will, unsurprisingly, allow us to highlight some of the more geometric properties of flatness that were absent when we were thinking purely module theoretically. We will see that flat morphisms, in reasonable circumstances, have unexpectedly nice topological properties. For example, flat morphisms will, in tame cases, be open mappings. We will also see that fibers of surjective flat (faithfully flat!) morphisms of irreducible varieties have the dimension we’d expect (the difference in the dimension of the ambient schemes). In fact, it is true that in nice situations a property closely related to this “expected fiber dimension property” (something close to “expected fiber codimension of points”) characterizes flat maps.

This last remark, is related to an intuitively pleasing property of flat morphisms which I, unfortunately, won’t have the time to discuss. One of the most commonly cited reasons that flat morphisms are “useful” is that they describe “continuously/smoothly varying families of varieties”. To try and understand what this means, suppose that $\pi:X\to Y$ is of finite type, and $X$ is reduced. Then, we can think of $\pi$ as describing a method of piecing together the family of varieties $\left\{X_y:=\pi^{-1}(y)\to\text{Spec}(k(y))\right\}_{y\in Y}$. When should we say that this piecing together is such that the $X_y$ vary “nicely” with $y$? In particular, what should it mean to say that $\pi$ pieces the $X_y$ “continuously”, or even “smoothly”.

While answering this question in a precise manner is not in this blog’s cards (for now), one property that suggests itself is that $\dim X_y$ is constant, which, as mentioned above, holds for faithfully flat morphisms of irreducible varieties. Pinning down the notion of “continuously varying fibers”, especially in a way that actually fits out intuition, is the goal in the relevant part of The Geometry of Schemes by Eisenbud and Harris, and should be read by anyone interested in developing an even more geometric view of flatness.

## Algebra

To start let’s do a bit of review about flatness. It is assumed that the reader has had previous exposure to flatness and is comfortable with most of its  properties, and some of the more advanced machinery associated with them (e.g. the $\text{Tor}$ functor). What we will discuss in much more detail, as it is not nearly as common of a topic in a beginning algebra course, is the notion of faithful flatness.

### Flatness

Let $A$ be a commutative, unital ring, and let $M$ be an $A$-module. We say that $M$ is flat over $A$, if for all injections

$0\to N\to L$

the sequence resulting by tensoring this sequence with $M$ remains exact or, in symbols,

$0\to N\otimes_A M\to L\otimes_A M$

is exact. It is a common fact that since the functor $-\otimes_A M$ is right exact, this is equivalent to the statement that for any short exact sequence

$X\to Y\to Z$

of $A$-modules, the associated sequence

$X\otimes_A M\to Y\otimes_A M\to Z\otimes_A M$

is exact.  We say that a ring map $A\to B$ is flat if $B$ is flat as an $A$-module, where the $A$-module structure is endowed through the ring map. We often say, in this case, that $B$ is a flat $A$-algebra.

There are many basic properties pertaining to flat modules which are commonly proved in a standard course on algebra. The only one for which there is a possibility the reader is unaware of is the following:

Theorem 1: Let $A\to B$ be a ring map, and let $N$ be a $B$-module. Then, $N$ is a flat $A$-module (with the obvious structure) if and only if for all maximal ideals $\mathfrak{n}$ of $B$

Proof: This is proved in the standard way, by showing that $\text{Tor}_1^A(N,-)$ is locally zero (i.e. it’s localization at maximal ideals is zero) considered as a $B$-module, using the flat base change property, and thus zero itself. $\blacksquare$

In particular, by taking the identity map $A\to A$, we see that an $A$-module $M$ is flat if and only if $M_\mathfrak{m}$ is flat over $A$ for every maximal ideal $\mathfrak{m}$ of $A$. But, since the map $A\to A_\mathfrak{m}$ is flat, it actually only suffices to show that $M_\mathfrak{m}$ is a flat $A_\mathfrak{m}$-module for every $\mathfrak{m}$. This is often paraphrased as “flatness is a local condition”.

One nice theorem that comes out of this theorem, is a satisfying classification of flat modules over Dedekind domains.

Theorem 2: Let $A$ be a Dedekind domain. Then, an $A$-module $M$ is flat if and only if it’s torsion free.

Recall that a Noetherian domain $A$ (which is not a field) is Dedekind if for every non-zero prime $\mathfrak{p}$ of $A$, the localization $A_\mathfrak{p}$ is a PID (necessarily a DVR). Of course, there are many equivalent definitions, familiar to any who have studied any amount of algebraic number theory. For example, a domain $A$ is Dedekind if and only if it is Noetherian, dimension $1$, and normal (integrally closed). They are also the domains for which unique factorization of ideals into prime ideals holds. The most common examples of Dedekind domains are the ring of integers $\mathcal{O}_K$ inside of some number field $K$.

For those that think more geometrically, one usually thinks about Dedekind domains as being one-dimensional regular affine schemes. Thus, the cuspidal cubic $\text{Spec}(k[x,y]/(y^2-x^3))$ is a non-example. In particular, one can check that $k[x,y]/(y^2-x^3)$ is not normal (the problem prime/point being $(x,y)$).

Proof(Theorem 2): It is trivial, by considering the sequence

$0\to A\xrightarrow{\text{mult. by }a}A\to A/a\to 0$

for $a\in A$, that for every flat module $M$ (over any domain!) multiplication by $a$ is an injective map $M\to M$, and thus $M$ is torsion free.

Conversely, suppose that $M$ is torsion free. Since every module is a direct limit of finitely generated submodules, and direct limit of flat modules are flat, we may assume that $M$ is finitely generated over $A$. By the comment following the previous theorem, to prove that $M$ is flat, it suffices to show that for every maximal $\mathfrak{m}$ of $A$, the $A_\mathfrak{m}$-module $M_\mathfrak{m}$ is flat. But, now since $A$ is Dedekind, we know that $A_\mathfrak{m}$ is a PID. So, by the structure theorem for PIDs, since $M_\mathfrak{m}$ is a finitely generated torsion free $A_\mathfrak{m}$ module, we know that $M_\mathfrak{m}$ is actually a free $A_\mathfrak{m}$-module and so trivially flat. $\blacksquare$

This theorem is really nice, because it will save us a lot of headache when it comes to verifying that some given example is a flat extension (since many of our examples will involve Dedekind domains). It immediately tells us, for example, that if $L/K$ is an extension of number fields, that the associated ring map $\mathcal{O}_K\to\mathcal{O}_L$ is flat.

A sophisticated example of non-flatness comes from normalization. Namely, suppose that $A$ is a non-normal domain, and $B$ is its normalization (e.g. the integral closure of $B$ in $\text{Frac}(A)$), then $A\to B$ is never flat. While there are geometric proofs of this, the quickest proof uses Serre’s criterion for normality, and the fact that $R_i$ and $S_i$ are preserved under flat extensions (see Theorem 23.9 in Matsumura’s Commutative Ring Theory).

### Faithful Flatness

Now, while flatness is often times good enough to get one through a basic algebra course, it is actually a much stronger condition that will mainly preoccupy us in scheme land. Namely, we say an $A$-module $M$ is faithfully flat if a sequence of $A$-modules

$X\to Y\to Z$

is exact if and only if the associated sequence

$X\otimes_A M\to Y\otimes_A M\to Z\otimes_A M$

is exact. We say a ring map $A\to B$ is faithfully flat, or that $B$ is a faithfully flat $A$-algebra, if $B$ is faithfully flat as an $A$-module

There is a nice way to understand, categorically, faithfully flat modules.

To this end, recall that a functor $F:\mathscr{C}\to\mathscr{D}$ is faithful if for all objects $X$ and $Y$ of $\mathscr{C}$, the natural map

$\text{Hom}_\mathscr{C}(X,Y)\to\text{Hom}_\mathscr{D}(F(X),F(Y))$

is injective. We then have the following nice characterization of faithful exact additive functors between abelian categories:

Theorem 3: Let $\mathscr{A}$ and $\mathscr{B}$ be abelian categories, and $F:\mathscr{A}\to\mathscr{B}$ be an additive functor. Then, $F$ is exact and faithful if and only if the following holds: a sequence in $\mathscr{A}$

$X\to Y\to Z$

is exact if and only if the sequence in $\mathscr{B}$

$F(X)\to F(Y)\to F(Z)$

is exact.

Proof: Suppose first that $F$ is faithful and exact. Then, we know that

$X\xrightarrow{\alpha} Y\xrightarrow{\beta} Z$

implies

$F(X)\xrightarrow{F(\alpha)} F(Y)\xrightarrow{F(\beta)} F(Z)$

is exact by definition. Conversely, suppose that

$F(X)\xrightarrow{F(\alpha)} F(Y)\xrightarrow{F(\beta)} F(Z)$

is exact. We know that $\text{im}(\alpha)\subseteq\ker\beta$, and (by exactness)

$F(\ker\beta/\text{im}(\alpha))\cong F(\ker\beta)/F(\text{im}(\alpha))=0$

Now, all that is left is to observe that $F$ can’t take non-zero objects to zero objects. For, if this were true, then we could conclude that $\ker\beta/\text{im}(\alpha)=0$ as desired. Now, if $W\ne 0$ were such that $F(W)=0$, then the map

$\text{Hom}_\mathscr{C}(W,W)\to\text{Hom}_\mathscr{D}(F(W),F(W))=0$

wouldn’t be injective, since $\text{Hom}_\mathscr{C}(W,W)$ contains at least two elements (the identity and zero morphism).

Conversely, let’s assume that a sequence is exact if and only if its image is exact. This implies that $F$ is exact by definition. Conversely, suppose that $X$ and $Y$ are objects of $\mathscr{C}$. We want to show that the map

$\text{Hom}_\mathscr{C}(X,Y)\to\text{Hom}_\mathscr{D}(F(X),F(Y))$

is injective. Suppose not, and that $f$ is in the kernel of this map. Then, the sequence

$X\xrightarrow{\text{id}}X\xrightarrow{f}Y$

is not exact (since $f\ne 0$), but after applying $f$ we see that

$F(X)\xrightarrow{\text{id}}F(X)\xrightarrow{F(f)=0}F(Y)$

is exact. But, this is a contradiction. $\blacksquare$

Thus, from this proposition, a module $M$ is faithfully flat if and only if the functor $-\otimes M$ is both exact and faithful. Thus, the word “faithfully” in faithfully flat.

Now, let us focus back on modules. Faithful flatness is a MUCH stronger condition than pure flatness, and excludes many “classic” examples of flat modules. For example, it is trivial that $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is flat (since it’s just a localization), but it’s not faithfully flat. The sequence

$0\xrightarrow{}\mathbb{Z}/2\mathbb{Z}\xrightarrow{}0$

is certainly not exact, but after tensoring with $\mathbb{Q}$ we obtain the sequence

$0\to 0\to 0$

which is exact.

So, what is an example of a faithfully flat module? Well, it’s pretty obvious that any free module is faithfully flat, but what’s a non-trivial example? Before we write one down, let us prove the following helpful theorem which says something like faithfully flat modules are those modules $M$ such that the associated functor $-\otimes M$ is exact, and has “trivial kernel” (really it’s talking about faithfulness, as above, but we’ll prove it differently for variety):

Theorem 4: Let $A$ be a ring, and $M$ an $A$-module. Then, the following are equivalent

1. $M$ is faithfully flat.
2. $M$ is flat and $N\otimes_A M\ne 0$ for every $N\ne 0$.
3. $M$ is flat and $A/\mathfrak{m}\otimes_A M\ne 0$ for all maximal ideals $\mathfrak{m}$ of $A$.

Proof: To see 1. implies 2.,  we can generalize the counterexample listed above for $\mathbb{Q}$. Namely, if $N\otimes_A M=0$ for some non-zero module $N$, then the sequence

$0\xrightarrow{}N\xrightarrow{}0$

would be non-exact, but after tensoring with $M$, we obtain the exact sequence

$0\to0\to0$

Conversely, let’s show that 2. implies 1. To see this, we must show that if

$X\xrightarrow{f}Y\xrightarrow{g}Z$

is a sequence, such that

$X\otimes_A M\xrightarrow{f\otimes 1} Y\otimes_A M\xrightarrow{g\otimes 1} Z\otimes_A M$

is exact, then so was the original sequence. So, we need to show that $\text{im }f=\ker g$. Now, we know that $\text{im }f\subseteq\ker g$, and so we must only show that $\ker g/\text{im }f$ is zero. But,

$(\ker g/\text{im }f)\otimes_A M=(\ker g\otimes_A M)/(\text{im }f\otimes_A M)\cong \ker(g\otimes 1)/\text{im }(f\otimes 1)=0$

where the second to last equality follows from flatness (since flat modules preserve kernels). By assumption this implies that $\ker g=\text{im }f$ as desired.

Now, it’s trivial that 2. implies 3. Finally, to see that 3. implies 1., let $N\ne 0$. Choose some $n\ne 0$ in $N$, and consider the map $A\to M:a\mapsto an$, and let its kernel be $I$. We then obtain an injection $A/I\to N$. Let $\mathfrak{m}$ be any maximal ideal of $A$ lying above $I$, and consider the surjection $A/I\to A/\mathfrak{m}$. Note then that since $M$ is flat, we obtain an injection $A/I\otimes_A M\to N\otimes_A M$ and a surjection $A/I\otimes_A M\to A/\mathfrak{m}\otimes_A M$. Now, by assumption $A/\mathfrak{m}\otimes_A M$ is non-zero, and since $A/I\otimes_A M$ surjects onto this, we have that $A/I\otimes_A M$ is non-zero. But, since $A/I\otimes_A M$ injects into $N\otimes_A M$, we obtain that $N\otimes_A M\ne 0$ as desired. $\blacksquare$

This theorem is most useful in its application to the following, seemingly inane, but incredibly useful corollary:

Corollary 5: Let $\varphi:(A,\mathfrak{m})\to (B,\mathfrak{n})$ be a local ring, and let $M$ be a finitely generated, non-zero $B$-module. Then, $M$ is faithfully flat if and only if it’s flat.

Proof: Obviously if $M$ is faithfully flat, then it is flat. Conversely, if $M$ is flat, to check faithful flatness we need only show that for the any maximal ideal of $A$, namely $\mathfrak{m}$, we have that $M\otimes_A A/\mathfrak{m}\ne 0$. But, since the morphism $\varphi$ is local, we have that

$M\otimes_A A/\mathfrak{m}=M/\varphi(\mathfrak{m})M=M/\mathfrak{n}M\ne 0$

where the last equality holds by Nakayama’s lemma, since $M\ne 0$, and $M$ is a finitely generated $B$-module. $\blacksquare$

So, this corollary allows us to create tons of non-trivial examples of faithfully flat modules. Namely, take any flat ring map $\varphi:A\to B$, let $\mathfrak{q}$ be a prime of $B$, and let $\varphi^{-1}(\mathfrak{q})=\mathfrak{p}$. We then obtain a flat ring map $A_\mathfrak{p}\to B_\mathfrak{q}$. Since $B_\mathfrak{q}$ is a finitely generated non-zero $B_\mathfrak{q}$-module, the previous corollary tells us that $B_\mathfrak{q}$ is a flat $A_\mathfrak{p}$-module, and so the ring map $A_\mathfrak{p}\to B_\mathfrak{q}$ is faithfully flat!

For example, take any extension of number fields $L/K$, and consider the associated ring map $\mathcal{O}_K\to\mathcal{O}_L$–we have already commented that this ring map is flat. Let $\mathfrak{p}$ be a prime of $K$, and $\mathfrak{q}$ a prime of $L$ lying over $\mathfrak{p}$. Then, the ring map $(\mathcal{O}_K)_\mathfrak{p}\to(\mathcal{O}_L)_\mathfrak{q}$ is faithfully flat. To have a concrete example, the ring map $\mathbb{Z}_{(2)}\to \mathbb{Z}[i]_{(1+i)}$ is faithfully flat.

Now, as in most of the examples we’ve written down, in the future we will care most about faithful flatness of algebras opposed to just modules (since this will be the geometric setting), and so it would be nice if we could use the extra information present in a ring map $A\to B$ to decide when it’s a faithfully flat map. It turns out, there is such a nice (geometric!) condition. In fact, there are a few that are important:

Theorem 6: Let $\varphi:A\to B$ be a ring map. Then, then following are equivalent

1. $\varphi$ is faithfully flat.
2. $\varphi$ is flat, and the induced map $\text{Spec}(B)\to\text{Spec}(A)$ is surjective.
3. $\varphi$ is injective, and $\text{coker}(\varphi)$ is flat as an $A$-module

Proof: To prove 1. implies 2., let us prove something slightly stronger. Namely, if there exists a non-zero $B$-module $M$ such that $M$ is a faithfully flat $A$-module, then the induced map on spectra is surjective (the actual problem will follow by taking $M=B$). For this, we use one of the sneakiest tricks in the book. Namely, to show that $\text{Spec}(B)\to\text{Spec}(A)$ is surjective, we need to show that the fiber over $\mathfrak{p}\in\text{Spec}(A)$ is non-empty for all $\mathfrak{p}$. But, as a scheme, the fiber is merely $\text{Spec}(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})$. We use then the deep theorem that if $R$ is a ring, and there exists a non-zero $R$-module, then $R\ne 0$. In our case, we are trying to show that $B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is non-zero, but the module $M\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is non-zero since $M$ is faithfully flat, and $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is nonzero.

To prove 1. implies 3. we, again, use a very sneaky trick (one that comes up often in descent theory). Namely, we want to show first that the map $A\to B$ is injective. To do this, it suffices, by faithful flatness, to show that $B=B\otimes_A A \to B\otimes_A B$ is injective. But, this map has an explicit left inverse–the multiplication map $\mu:B\otimes_A B\to B$ given on simple tensors by the equation $\mu(b_1\otimes b_2)=b_1b_2$.

To prove 2. implies 1., since we are assuming that $\varphi$ is flat, it suffices to show (by Theorem 4) that for all maximal ideals $\mathfrak{m}$ of $A$ that $B\otimes_A A/\mathfrak{m}$ is non-zero. But, $\text{Spec}(B\otimes_A A/\mathfrak{m})$ is precisely the fiber over $\mathfrak{m}$ of the map $\text{Spec}(B)\to\text{Spec}(A)$ which, by assumption, is non-empty, and so $B\otimes_A A/\mathfrak{m}$ is non-zero.

Finally, to prove that 3. implies 1., we proceed as follows. Let $M$ be any $A$-module. Note that, by assumption, we have the short exact sequence

$0\to A\to B\to\text{coker}(\varphi)\to 0$

of $A$-modules. We then obtain a long-exact sequence containing

$0\to\text{Tor}_1^A(M,B)\to0\to A\otimes_A M\to B\otimes_A M\to \text{coker}(\varphi)\otimes_A M\to 0$

where we used the fact that $A$ and $\text{coker}(\varphi)$ are flat over $A$. In particular, we see two things. First, this says that $\text{Tor}^A_1(M,B)=0$ for all $A$-modules $M$, and so $B$ is flat over $A$. Second, since we assumed that $\text{coker}(\varphi)$ is flat, we know that $\text{Tor}^1_A(M,\text{coker}(\varphi))=0$, and so we have an injection $M=A\otimes_A M\to B\otimes_A M$. Thus, we see that if $M$ is non-zero, then neither is $M\otimes_A B$$\blacksquare$

So, for example, we see that $\mathbb{Z}\to\mathbb{Q}$ had no chance of being faithfully flat, since the induced map on spectra $\text{Spec}(\mathbb{Q})\to\text{Spec}(\mathbb{Z})$ was infinitely far from being surjective.

Another distinct advantage of faithfully flat maps is that they have a certain “cancellation property”. If we know that the composition of two maps is flat, and the second is faithfully flat, we can conclude that the first is flat. More precisely:

Theorem 7: Let $A\to B$ and $B\to C$ be ring maps with $B\to C$ faithfully flat. Then, the composition $A\to C$ is flat if and only if $A\to B$ is flat.

Proof: Certainly if $A\to B$ is flat, then so is $A\to C$. Conversely, suppose that $A\to C$ is flat. Let

$0\to M\to N$

be an exact sequence of $A$-modules. We need to show that

$0\to M\otimes_A B\to N\otimes_A B$

is exact. But, since $B\to C$ is faithfully flat, it suffices to show that

$0\to (M\otimes_A B)\otimes_B C\to (N\otimes_A B)\otimes_B C$

is exact. But, this is precisely the sequence

$0\to M\otimes_A C\to N\otimes_A C$

which is exact since $A\to C$ is flat. $\blacksquare$

Note that this can fail without the assumption of faithfulness for $B\to C$. For example, consider the normalization of the cuspidal cubic: $k[t^2,t^3]\to k[t]$. This is not flat since it’s a normalization and, as we stated above, normalizations (of non-normal things) are never flat. More down to earth, one can check that $0\to (t^2,t^3)\to k[t^2,t^3]$ doesn’t remain injective after tensoring with $k[t]$. That said, the ring maps $k[t^2,t^3]\to k(t)$ and $k[t]\to k(t)$ are both flat since, in both cases, we are dealing with a localization. Thus, $k[t^2,t^3]\to k[t]\to k(t)$ produces a counterexample to the previous theorem above if we replace “$B\to C$ is faithfully flat” with just “$B\to C$ is flat”.

Of course, this is not a counterexample to Theorem 7 as is, since $k[t]\to k(t)$ is not faithfully flat. Indeed, the induced map $\text{Spec}(k(t))\to\text{Spec}(k[t])$ is far from surjective.

## Geometry

### Definition and Basic Properties

Now that we have developed sufficient algebraic machinery, we can now discuss how it relates to geometry. We start with definitions which, at this point, shouldn’t be unexpected. Namely, we call a map of schemes $f:X\to Y$ flat at $x\in X$ if the induced map $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ is flat. We call $f$ flat if it is flat at all $x\in X$. We call a map of schemes faithfully flat if it is both flat and surjective (imitating the conclusion of Theorem 6).

Before we discuss flatness and further, let us first remark that this definition is consistent with flatness of ring maps/algebras. By this I mean that $A\to B$ is flat if and only if $\text{Spec}(B)\to\text{Spec}(A)$ is. Evidently if $A\to B$ is flat, then for all $\mathfrak{q}\in\text{Spec}(B)$ we have that $A_{f^{-1}(\mathfrak{q})}\to B_\mathfrak{q}$ is flat, and so $\text{Spec}(B)\to\text{Spec}(A)$ is a flat map of schemes. To prove the converse, we note that by definition we have that $A_{f^{-1}(\mathfrak{q})}\to B_\mathfrak{q}$ is flat for all primes $\mathfrak{q}\in\text{Spec}(B)$. Since $A\to A_{f^{-1}(\mathfrak{q})}$ is flat, we then see that $A\to B_\mathfrak{q}$ is flat for all $\mathfrak{q}\in\text{Spec}(B)$. In light of Theorem 1, we may then conclude that $A\to B$ is indeed flat. Of course, we also see that faithful flatness is a consistent notion as well.

As is true about all “reasonable classes of morphisms” (a la Vakil’s text Foundations of Algebraic Geometry) one can immediately check that flatness and faithful flatness are properties preserved under all of the standard operations: composition, base change, products. It is also clear that all open embeddings are flat maps since the map on stalks is an isomorphism

As was stated in the motivation, purely by inspection (recalling that exactness is a stalk-local condition!) we see that flatness may be characterized as follows:

Theorem 8: Let $f:X\to Y$ be a morphism of schemes. Then, $f$ is flat if and only if the pull-back functor $f^\ast:\mathsf{Qcoh}(Y)\to\mathsf{Qcoh}(X)$ is exact.

And, if we assume that $f$ is flat and surjective (i.e. faithfully flat!) then a sequence in $\mathsf{Qcoh}(Y)$ is exact if and only if it’s image is. Thus, using Theorem 3 (since $f^\ast$ is additive) we may conclude:

Theorem 9: Let $f:X\to Y$ be a morphism of schemes. Then, $f$ is faithfully flat if and only if the pull-back functor $f^\ast:\mathsf{Qcoh}(Y)\to\mathsf{Qcoh}(X)$ is exact and faithful.

Purely in terms of practicality, one can often reduce checking flatness at primes of $A$. To make this exact, let us suppose that $f:A\to B$ is a ring map, and $\mathfrak{q}\in\text{Spec}(B)$, and $\mathfrak{p}=f^{-1}(\mathfrak{q})$. Then $B_{\mathfrak{p}}\to B_\mathfrak{q}$ is a localization, and thus flat. But, since both rings are local, we can conclude by Corollary 5 that, in fact, $B_{\mathfrak{p}}\to B_\mathfrak{q}$ is faithfully flat. Thus, using Theorem 7, we see that $A_\mathfrak{p}\to B_{\mathfrak{q}}$ is flat if and only if $A_\mathfrak{p}\to B_\mathfrak{p}$ is flat. This often times allows us to reduce checking flatness at a point to a purely $A$-focused (i.e. working with primes of $A$) task.

### Flat Locus

While one can think of many examples of scheme maps which are flat since, as remarked above, flatness of algebras coincides with flatness of affine schemes, there is one big difference between the purely ring-theoretic case and the geometric one. Usually in ring land we think of flatness as a black and white notion. Either a ring map is flat, or it is not. Now, we can ask a more specific question: for what points is a ring map flat? Let us call the set of points $x\in X$ such that $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ the flat locus of $f$.

For example, consider the cuspidal cubic $C=\text{Spec}(k[x,y]/(y^2-x^3))=\text{Spec}(k[t^2,t^3])$, and the normalization $\varphi:\mathbb{A}^1_k\to C$ given by $k[t^2,t^3]\hookrightarrow k[t]$. We claim that the flat locus of this morphism is $\mathbb{A}^1_k-\{0\}$. To see this, let us consider the map $\mathbb{A}^1_k-\{0\}\to C-\{(0,0)\}$ given by $k[t^2,t^3]_{t^2}\to k[t]_{t^2}=k[t]_t$, it’s easy to see that this is an isomorphism. This should be expected since normalizations of curves are isomorphisms off of the singular points. This in particular tells us though that $\mathcal{O}_{C,\varphi(p)}\to \mathcal{O}_{p,\mathbb{A}^1_k}$ is an isomorphism, and in particular flat, for $p\ne (t)$. Or, in other words, the flat locus of $\varphi$ contains $\mathbb{A}^1_k-\{(0,0)\}$. But, since $k[t]$ is not a flat $k[t^2,t^3]$-module (as remarked above), this allows us to conclude that the flat locus of $\varphi$ is precisely $\mathbb{A}^1-\{0\}$, else $\varphi$ would actually be flat.

The fact that the flat locus in this example is open is no mistake. In fact, we actually have the following:

Theorem 10: Let $f:X\to Y$ be a morphism locally of finite type. Then, the flat locus of $f$ is open.

The proof of this statement is surprisingly hard. We will discuss a proof later in this post, of a slightly weaker version. The full proof can be found in EGA IV as Theorem 11.3.1. A very rough heuristic for why one might expect this to be true is as follows. Flatness is determined by the vanishing of some module (namely $\text{Tor}$), and since the vanishing set of modules is open (at least in the coherent case) we might expect the flat locus to be open. This is flawed on several levels though (at least in rigor), since it only works if the $\text{Tor}$ is finitely generated, and realistically any attempt at a proof will likely involve intersecting infinitely many of these vanishing sets.

### Significant Properties

Let us now point out some interesting, non-trivial properties of flat morphisms.

The first is the seemingly trivial, but very powerful observation that if $f:X\to Y$ is flat, then for all $x\in X$ one has that $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ is faithfully flat. Indeed, it’s a flat local map and so faithfully flat by Corollary 5. A somewhat surprising implication of this is that, by Theorem 6, the map $\mathcal{O}_{Y,f(x)}\to\mathcal{O}_{X,x}$ is injective. This forces, at least in the case of faithful flatness, something quite strong. Namely, in this case $f$ is surjective, and for all $x\in X$ we have that $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ is injective. But, since $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ factors as $\mathcal{O}_{Y,f(x)}\to (f_\ast\mathcal{O}_X)_{f(x)}\to \mathcal{O}_{X,x}$ we see that $\mathcal{O}_{Y,f(x)}\to (f_\ast\mathcal{O}_X)_{f(x)}$ is injective. Thus, we may conclude that $f$ is surjective and the sheaf map $\mathcal{O}_Y\to f_\ast\mathcal{O}_X$ is injective and thus:

Theorem 11: Let $f:X\to Y$ be faithfully flat. Then, $f$ is an epimorphism in the category $\mathsf{Sch}$ of schemes.

This is somewhat surprising. It tells us that if have a surjective map of schemes which plays nicely with restriction along it (i.e. exact sequences restricted along it stay exact), then the morphism can’t equalize two different morphisms. At first glance the properties seem unrelated.

Next let us prove that flat morphisms of finite presentation are open mappings. This is a somewhat surprising topological property for a notion which, ostensibly, is purely algebraic.

Theorem 12: Let $f:X\to Y$ be a locally finitely presented flat morphism. Then, $f$ is a universally open mapping.

The proof of Theorem 12 is much easier (or at least much easier if one assumes some commonly known theorems) if one assumes that $X$ and $Y$ are Noetherian, and so this is the only case that we will prove. One can deduce the general case from the Noetherian case by appealing to Grothendieck’s “passage to the limit” technique, where one can show that our finitely presented flat morphism is the base change of a flat morphism of finite type between Noetherian schemes. The general result can be found here.

Our proof for the Noetherian case will rely heavily upon the classic result of Chevalley:

Theorem(Chevalley): Let $X$ and $Y$ be Noetherian schemes, and $f:X\to Y$ a finite morphism. Then, the image of a constructible (i.e. finite union of locally closed) set under $f$ is constructible.

along with the following point-set fact:

Lemma: Let $X$ be a Noetherian space. Then, a constructible subset of $X$ is open if and only if it is closed under generalization.

The first of these is a difficult theorem, but one which is usually discussed in a first course in algebraic geometry (a guided exercise to the proof can be found in Vakil’s Foundations of Algebraic Geometry in section 7.4.1) . The second is much less difficult, albeit annoying, and is just a standard point-set proof.

Proof(Theorem 12 assuming $X,Y$ Noetherian): Since the base change of a morphism locally of finite presentation and flat map is locally of finite presentation and flat, it suffices to prove that $f$ itself open.

Let $U\subseteq X$ be open. Then, by Chevalley’s theorem we know that $f(U)\subseteq Y$ is constructible, and thus by the previous lemma it suffices to show that $f(U)$ is closed under generalization. To see this, let $f(x)\in f(U)$ be arbitrary. Note by Theorem 5 that since $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ is faithfully flat, the induced morphism $\text{Spec}(\mathcal{O}_{X,x})\to\text{Spec}(\mathcal{O}_{Y,f(x)})$ is surjective. But, recall that in general the image of $\text{Spec}(\mathcal{O}_{Z,z})\to Z$ is precisely the points of $Z$ generalizing $z$. So, since $U\subseteq X$ is open, and so closed under generlization, we see that $\text{Spec}(\mathcal{O}_{X,x})\to X$ has image in $U$. But, the following diagram (of set maps) commutes:

$\begin{matrix} X & \to & Y\\ \uparrow & & \uparrow\\ \text{Spec}(\mathcal{O}_{X,x}) & \to & \text{Spec}(\mathcal{O}_{Y,f(x)}\end{matrix}$

and thus we see that $\text{Spec}(\mathcal{O}_{Y,y})\to Y$ lands in $f(U)$, and so $f(U)$ is closed under generalization. The conclusion follows. $\blacksquare$

As a quick, but useful, application of Theorem 12 we immediately see that the structure morphism $X\to\text{Spec}(k)$ of any finite type $k$-scheme is universally open. Since $X\to\text{Spec}(k)$ is surjective, this implies that the base change $X_K\to X$ is actually a quotient map for all extensions $K/k$.

Let us now show that the fibers of a flat map vary continuously, in so much as all points of $X$ have predictable codimension in their associated fiber (the fiber associated to $x\in X$ is $f^{-1}(f(x))$).

Theorem 13: Let $X$ and $Y$ be locally Noetherian, and $f:X\to Y$ flat. Then, if $x\in X$ and $y=f(x)$ one has that $\text{codim}_{X_y} \overline{\{x\}}=\text{codim}_X \overline{\{x\}}-\text{codim}_Y \overline{\{y\}}$.

The proof of this can be found in Qing Liu’s Algebraic Geometry and Arithmetic Curves as Theorem 3.12. The rough idea is to induct on $\dim Y$, and use the fact that a non-zero divisor germ $f\in\mathcal{O}_{Y,y}$ (e.g. an element of a system of parameters) passes to a non-zero divisor in $\mathcal{O}_{X,x}$ by flatness (otherwise $\mathcal{O}_{X,x}$ would be a torsion $\mathcal{O}_{Y,y}$-module, and so not flat!). This allows one to effectively reduce dimensions by $1$ by reducing modulo $f$.

By taking $x$ be a closed point of $X_y$, and using the fact that $\dim Z=\text{tr.deg}_k K(Z)$ (and its corollary that $\text{codim}_Z C=\dim Z-\dim C$ for any irreducible closed subset $C\subseteq Z$) for an irreducible $k$-variety $Z$, one deduces the following from the previous theorem:

Theorem 14: Let $X$ and $Y$ be irreducible varieties, and $f:X\to Y$ faithfully flat. Then,

$\dim X_y=\dim X-\dim Y$

for all $y\in Y$.

It is a standard result that if $f:X\to Y$ is a surjective morphism of irreducible varieties, then there exists some non-empty open subset $U\subseteq Y$ such that $\dim X_y=\dim X-\dim Y$ for all $y\in U$ (cf. Foundations of Algebraic Geometry 11.4.1), and one might have wondered under what conditions on $f$ one is guaranteed that we can take $U=Y$–of course, the above says that this condition is flatness of $f$.

Now, as was mentioned in the motivation, under fairly general conditions one can actual phrase Theorem 13 as an if and only if. Namely:

Theorem(Miracle Flatness): Let $X$ be a Cohen-Macaulay scheme and $Y$ a a regular scheme. A morphism $f:X\to Y$ is flat if and only if for all $x\in X$, with $y=f(x)$, one has that

$\text{codim}_{X_y}\overline{\{x\}}=\text{codim}_X\overline{\{x\}}-\text{codim}_Y\overline{\{y\}}$

This follows from Theorem 23.1 in Matsumura’s Commutative Ring Theory–this theorem is often times called miracle flatness. This theorem is not particularly useful, at least in my experience. But it does tell you that under fairly mild conditions flatness is equivalent to continuously varying fibers, assuming you define this to mean that codimensions vary as expected, which is intuitively helpful.

Of course, the properties in the Miracle Flatness theorem cannot be tweaked. For example, it’s pivotal that $Y$ is regular. Consider, once again, the normalization of the cuspidal cubic $\varphi:\mathbb{A}^1\to \text{Spec}(k[t^2,t^3])$. Both of these are irreducible varieties, and so the condition of Miracle Flatness is equivalent to the dimension of the fibers all being $\dim k[t]-\dim k[t^2,t^3]=1-1=0$. But, this is certainly true since every fiber of $\varphi$ is a singleton! That said, $\varphi$ is not flat, as previously mentioned. Of course, what goes wrong in this case is that $\text{Spec}(k[t^2,t^3])$ is not regular.

Generic Freeness  and the Flat Locus Revisited

Now that we have worked a little bit with flat morphisms, let us return to the proof of Theorem 10, or an ever so slightly weaker version thereof. The key ingredient, as it turns out, to the proof is the solution to a less difficult problem. Namely, instead of asking whether or not the flat locus of a morphism $f:X\to Y$ is open, one could ask if the flat locus of $f$ contains a non-empty open subset of $X$. The answer to this is yes if we assume that $Y$ is integral and locally Noetherian. The proof follows immediately from a famous theorem of Grothendieck:

Theorem(Grothendieck’s Generic Freeness): Let $A$ be a Noetherian integral domain, and $B$ a finite $A$-algebra. For any finitely generated $B$-module $M$, there exists a non-zero $f\in A$ such that $M_f$ is a free $A_f$-module.

The proof of Generic Freeness can be found in many texts. For example, there is a guided exercise in Foundations of Algebraic Geometry (starting with exercise 7.4.F), or a full proof can be found in EGA IV Lemma 6.9.2, or in Eisenbud’s Commutative Algebra as Theorem 14.4.

Then we prove the following weaker version of Theorem 10, but in the more general context of the flat locus of a coherent module. In particular, if $f:X\to Y$ is a map of schemes, and $\mathscr{F}$ a quasicoherent $\mathcal{O}_X$-module, we define the flat locus of $\mathscr{F}$ to be the set of $x\in X$ such that $\mathscr{F}_x$ is a flat $\mathcal{O}_{Y,f(x)}$-module.

Theorem 10(Weak Version): Let $f:X\to Y$ be a finite type morphism between Noetherian schemes, and let $\mathscr{F}$ be a coherent $\mathcal{O}_X$-module. Then, the flat locus of $\mathscr{F}$ is open.

The proof requires not only generic freeness, but a very difficult technical commutative algebra lemma:

Lemma(Gross Technical): Let $A\to B$ be a morphism of Noetherian rings. Suppose that $I$ is an ideal of $A$ such that $IB$ is contained in the Jacobson radical of $B$, and that $M$ is a finitely generated $B$-module. Then, the following are equivalent

1. $M$ is flat over $A$
2. $M/(IB)M$ is flat over $A/I$ and $\text{Tor}_1^A(M,A/I)=0$

The proof of this, I will admit, is really horrendous. Or, at least the one I know is. That said, it does show up somewhat often in certain circles. Parts of the proof can be found scattered over Matsumura’s Commutative Algebra.

Now, we only need one more result before we actually prove the main theorem, but one which is much easier than the above two results

Lemma 15: Let $A$ be a Noetherian ring, $B$ a finite type $A$-algebra, $M$ a finitely generated $B$-module, $\mathfrak{q}\in\text{Spec}(B)$, and $\mathfrak{p}$ the image of $\mathfrak{q}$ under $\text{Spec}(B)\to\text{Spec}(A)$. Suppose that $M_\mathfrak{q}$ is flat over $A_\mathfrak{p}$, then there is some $g\in B-\mathfrak{q}$ such that $(M/\mathfrak{p}M)_g$ is flat over $A/\mathfrak{p}$ and $\text{Tor}_1^A(M,A/\mathfrak{p})_g=0$.

Proof: Since $A/\mathfrak{p}$ is a Noetherian domain, and $M/\mathfrak{p}M$ a finitely generated $B$-module, we can use Generic Freeness to produce some element $f\in A-\mathfrak{p}$ such that $(M/\mathfrak{p}M)_f$ is a free $A_f$-module.

Now, since $M_\mathfrak{q}$ is flat over $A$, since by assumption $M_\mathfrak{q}$ is flat over $A_\mathfrak{p}$ and $A\to A_\mathfrak{p}$ is flat, we deduce

$\text{Tor}_1^A(M,A/\mathfrak{p})_\mathfrak{q}\cong\text{Tor}_1^A(M_\mathfrak{q},A/\mathfrak{p})=0$

We’d be done if we knew $\text{Tor}_1^A(M,A/\mathfrak{p})$ were a finitely generated $B$-module, since we could clearly then find $g_1,\ldots,g_r\in B-\mathfrak{q}$ annihilating the generators of $\text{Tor}_1^A(M,A/\mathfrak{p})$ and then take $g=fg_1\cdots g_r$. To see that $\text{Tor}_1^A(M,A/\mathfrak{p})$ is a finitely generated we merely use the short exact sequence

$0\to\mathfrak{p}\to A\to A/\mathfrak{p}\to 0$

to get the long exact sequence

$\cdots\to\underbrace{\text{Tor}_1^A(M,A)}_{=0}\to\text{Tor}_1^A(M,A/\mathfrak{p})\to M\otimes_A\mathfrak{p}\to M\otimes_A \to M\otimes_A A/\mathfrak{p}\to 0$

to see that $\text{Tor}_1^A(M,A/\mathfrak{p})$ is a submodule of the finitely generated $B$-module $M\otimes_A\mathfrak{p}$ which, since $B$ is Noetherian, implies that $\text{Tor}_1^A(M,A/\mathfrak{p})$ is finitely generated as desired. $\blacksquare$

With this, we have the following corollary:

Corollary 16: Let us have the same hypotheses as in the previous theorem. Then, for every prime $\mathfrak{Q}$ of $B$ which contains $\mathfrak{p}$ but which does not contain $g$ has the property that $M_\mathfrak{Q}$ is flat over $A$.

Proof: Consider the map $A\to B_\mathfrak{Q}$. Then apply the Gross Technical lemma to the ideal $I=\mathfrak{p}$ and the $B_\mathfrak{Q}$ module $M_\mathfrak{Q}$, in conjunction with the previous theorem. $\blacksquare$

Now, we can finally prove our desired theorem:

Proof(Theorem 10 Weak Version): It’s clear that this is a local property, and so we may as well assume that we are in affine land, say $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$, and that $\mathscr{F}$ is $\widetilde{M}$ for some finitely generated $B$-module $M$. Let $U$ denote the flat locus of $f$.

Note first that $U$ is clearly closed under generalization. Then, since our spaces are Noetherian, it suffices to show that $U$ is constructible. But, one form of constructibility is that if $C$ is an irreducible closed subset of $X$, then if $C\cap U$ is dense (i.e. non-empty), then there exists some non-empty open subset $O\subseteq C$ such that $C\cap U\supseteq O$.

To see this holds, suppose that $U\cap V(\mathfrak{p})$ is non-empty. Then, by definition, we know that $M_\mathfrak{q}$ is a flat $A_\mathfrak{p}$-module. Then, by the corollary to the Gross Technical lemma we know that $U\cap V(\mathfrak{p})$ contains $V(\mathfrak{p})\cap D(g)$.

The conclusion follows. $\blacksquare$