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In algebra, the notion of flat modules (due to GAGA) is a generalisation of the class of torsion-free modules over rings $R$ that are not PID.
An $R$-module $M$ is flat (from the French word “plat”) if it has “no torsion” in the sense that the Tor-functor vanishes for every $R$-module $X$:
(terminology and relation to fiber bundles)
Finitely presented flat modules are locally free [], hence under the dual geometric interpretation of modules as generalized vector bundles over the space on which $R$ is the ring of functions, flatness of a finitely presented module is essentially the local triviality of this bundle, hence in particular the fact that the fibers of the bundle do not change, up to isomorphism (cf. fiber bundle).
On the other hand there is no relation to “flat” as in flat connection on such a bundle.
We first state the definition and its equivalent reformulations over a commutative ring abstractly in In terms of exact functors and Tor-functors. Then we give an explicit element-wise characterization in Explicitly in terms of identities.
Much of this discussion also works in the more general case where the ring is not-necessarily taken to be commutative and not necessarily required to be equipped with a unit, this we indicate in For more general rings.
Let $R$ be a commutative ring.
An $R$-module $N$ is flat if tensoring with $N$ over $R$ as a functor from $R$Mod to itself
is an exact functor (sends short exact sequences to short exact sequences).
A module as above is faithfully flat if it is flat and tensoring in addition reflects exactness, hence if the tensored sequence is exact if and only if the original sequence was.
The condition in def. has the following immediate equivalent reformulations:
$N$ is flat precisely if $(-)\otimes_R N$ is a left exact functor,
because tensoring with any module is generally already a right exact functor;
$N$ is flat precisely if $(-)\otimes_R N$ sends monomorphisms (injections) to monomorphisms,
because for a right exact functor to also be left exact the only remaining condition is that it preserves the monomorphisms on the left of a short exact sequence;
$N$ is flat precisely if $(-)\otimes_R N$ is a flat functor,
because Mod is finitely complete;
$N$ is flat precisely if the degree-1 Tor-functor $Tor_1^{R Mod}(-,N)$ is zero,
because by the general properties of derived functors in homological algebra, $L_1 F$ is the obstruction to a right exact functor $F$ being left exact;
$N$ is flat precisely if all higher Tor functors $Tor_{\geq 1}(R Mod)(-,N)$ are zero,
because the higher derived functors of an exact functor vanish;
$N$ is flat precisely if $N$ is an acyclic object with respect to the tensor product functor;
because the Tor functor is symmetric in both arguments and an object is called tensor-acyclic object if all its positive-degree $Tor$-groups vanish.
The condition in def. also has a number of not so immediate equivalent reformulations. These we discuss in detail below in Equivalent characterizations. One of them gives an explicit characterization of flat modules in terms of relations beween their elements. An exposition of this we give now in In terms of identities.
There is a characterisation of flatness that says that a left $A$-module $M$ is flat if and only if “everything (that happens in $M$) happens for a reason (in $A$)”. We indicate now what this means. Below in prop. it is shown how this is equivalent to def. above.
The meaning of this is akin to the existence of bases in vector spaces. In a vector space, say $V$, if we have an identity of the form $\sum_i \alpha_i v_i = 0$ then we cannot necessarily assume that the $\alpha_i$ are all zero. However, if we choose a basis then we can write each $v_i$ in terms of the basis elements, say $v_i = \sum_j \beta_{i j} u_j$, and substitute in to get $\sum_{i j} \alpha_i \beta_{i j} u_j = 0$. Now as $\{u_j\}$ forms a basis, we can deduce from this that for each $j$, $\sum_i \alpha_i \beta_{i j} = 0$. These last identities happen in the coefficient field, which is standing in place of $A$ in the analogy.
When translating this into the language of modules we cannot use bases so we have to be a little more relaxed. The following statement is the right one.
Suppose there is some identity in $M$ of the form $\sum_i a_i m_i = 0$ with $m_i \in M$ and $a_i \in A$. Then there is a family $\{n_j\}$ in $M$ such that every $m_i$ can be written in the form $m_i = \sum_j b_{i j} n_j$ and the coefficients $b_{i j}$ have the property that $\sum_i a_i b_{i j} = 0$.
The module $M$ being flat is equivalent to being able always to do this.
There is an alternative way to phrase this which is less element-centric. The elements $m_i$ correspond to a morphism into $M$ from a free module, say $m \colon F \to M$. The $a_i$ correspond to a morphism $a \colon F \to F$, multiplying the $i$th term by $a_i$. That we have the identity $\sum_i a_i m_i = 0$ says that the composition $m a$ is zero, or that $m \colon F \to M$ factors through the coequaliser of $a$ and $0$.
Now we consider the elements $n_j$. These define another morphism from a free module, say $n \colon E \to M$. That the $m_i$ can be expressed in terms of the $n_j$ says that the morphism $m$ factors through $n$. That is, there is a morphism $b \colon F \to M$ such that $m = n b$. We therefore have two factorisations of $m$: one through $n$ and one through the cokernel $\coker a$. The question is as to whether these have any relation to each other. In particular, does $\coker a \to M$ factor through $n$? We can represent all of this in the following diagram.
Saying that $M$ is flat says that this lift always occurs.
A module $N$ is flat precisely if for every finite linear combination of zero, $\sum_i r_i n_i = 0\in N$ with $\{r_i \in R\}_i$, $\{n_i \in N\}$ there are elements $\{\tilde n_j \in N\}_j$ and linear combinations
with $\{b_{i j} \in R\}_{i,j}$ such that for all $j$ we have
A finite set $\{r_i \in R\}_i$ corresponds to the inclusion of a finitely generated ideal $I \hookrightarrow R$.
By theorem $N$ is flat precisely if $I \otimes_R N \to N$ is an injection. This in turn is the case precisely if the only element of the tensor product $I \otimes_R R$ that is 0 in $R \otimes_R N = N$ is already 0 on $I \otimes_R N$.
Now by definition of tensor product of modules an element of $I \otimes_R N$ is of the form $\sum_i (r_i ,n_i)$ for some $\{n_i \in N\}$. Under the inclusion $I \otimes_R N \to N$ this maps to the actual linear combination $\sum_i r_i n_i$. This map is injective if whenever this linear combination is 0, already $\sum_i (r_i, n_i)$ is 0.
But the latter is the case precisely if this is equal to a combination $\sum_j (\tilde r_j , \tilde n_j)$ where all the $\tilde r_j$ are 0. This implies the claim.
Taking this a step further, we consider the filtered family of all finite subsets of $M$. This generates a filtered family of finitely generated free modules with compatible morphisms to $M$. So there is a morphism from the colimit of this family to $M$. This morphism is surjective by construction. To show that it is injective, we need to show that any element in one of the terms in the family that dies by the time it reaches $M$ has actually died on the way. This is precisely what the above characterisation of flatness is saying: the element corresponding to $\sum_i a_i m_i$ that dies in $M$ is already dead by the time it reaches $E$.
We have thus arrived at the following result:
A module is flat if and only if it is a filtered colimit of finitely generated free modules.
This observation (Wraith, Blass) can be put into the more general context of modelling geometric theories by geometric morphisms from their classifying toposes, or equivalently, certain flat functors from sites for such topoi.
Even if the ring $R$ is not necessarily commutative and not necessarily unital, we can say:
A left $R$-module is flat precisely if the tensoring functor
from right $R$ modules to abelian groups is an exact functor.
The full subcategory of flat modules over a ring $R$ is stable under the following categorical constructions :
Every projective module is flat.
Every projective module is a retract of a free module, which in turn is a sum of copies of the base ring $R$, which is flat over itself.
Let $R$ be a commutative ring and $S \subset R$ be a multiplicative subset. Then the localisation $R[S^{-1}]$ is a flat $R$-module.
Regard the set $S$ as a poset with de divisibility relation. The fact that $S$ be multiplicative means that the poset is directed. One thus have that $R[S^{-1}] = \underset{\longrightarrow}{\lim}_s \;\frac{1}{s} \; R$ is a filtered colimit of free modules, hence it is flat.
(Stability under base change) Let $R \to S$ be a morphism of commutative rings and let $M$ be a flat $R$-module, then the $S$-module $M \otimes_R S$ is also flat.
Let $X \to Y$ be an injective morphism of $S$-modules, then the map $(M \otimes_R S) \otimes_S X \to (M \otimes_R S) \otimes_S Y$ is isomorphic to the map $M \otimes_R X \to M \otimes_R X$ which is injective because $M$ is flat.
(Flatness is a local property) Let $R$ be a commutative ring and let $M$ be an $R$-module. Then the following assertions are equivalent :
$1 \Rightarrow 2$. By base change.
$2 \Rightarrow 3$. Obvious.
$3 \Rightarrow 1$. Let $X \to Y$ be an injective morphism of $R$-modules and let $K$ denote the kernel of the induced map $M \otimes_R X \to M \otimes_R Y$. Since $R_\mathfrak{m}$ is a flat $R$-module it follows that $X_\mathfrak{m} \to Y_\mathfrak{m}$ is also injective ; and $K_\mathfrak{m}$ is the kernel of $M_\mathfrak{m} \otimes_{R_\mathfrak{m}} X_\mathfrak{m}\to M_\mathfrak{m} \otimes_{R_\mathfrak{m}} Y_\mathfrak{m}$. By the flatness assumption, we deduce that $K_\mathfrak{m} = 0$ for every maximal ideal $\mathfrak{m} \subset R$, so $K = 0$ and $M$ is flat.
For any module $M$ over a ring $R$, one has
For every regular element $p \in R$, the $p$-torsion submodule of $M$ is isomorphic to $\mathrm{Tor}^R_1(M, R/(p))$.
The converse holds for semi-hereditary rings.
(Reduction to cyclic modules) Let $R$ be a commutative ring and let $M$ be an $R$-module. If
for every finitely generated ideal $I \subset R$, then $M$ is a flat $R$-module.
Equivalently, if
is an isomorphism for every finitely generated ideal $I \subset R$, then $M$ is flat.
The proof goes in 3 steps.
which is exact in the middle, so $Tor_1(M, X) = 0$.
for every module $X$.
The equivalence with the second characterisation comes from the fact that using the exact sequence $0 \to I \to R \to R/I \to 0$ and tensoring it with $M$, one gets the long exact sequence
identifying $Tor_1(M, R/I)$ as the kernel of the always surjective map $I \otimes M \to IM \subset R$.
Thanks to the general structure theorem of flat modules, one gets
Given a finitely presented module $M$, one has
If $M$ is at the same time finitely presented and a filtered colimit of finitely generated free modules, the identity morphism $M \to M$ must factor as $M \to F \to M$ where $F$ is a finitely generated free module. This implies that $M$ is projective.
In many cases, “finite presentation” can be replaced with “finite generation”.
(When $R$ is local)
Let $M$ be a finitely generated module over a local ring $R$. Then one has
This is Matsumara, Theorem 7.10.
Thus we see that finitely generated flat modules are locally free in the weak sense. Let us recall the definition of locally free modules and their relation with projective modules.
An $R$-module $M$ is called a locally free module in the weak sense if for every prime ideal $\mathfrak{p} \hookrightarrow R$ the localised module $M_\mathfrak{p}$ is a free module over the localization $R_\mathfrak{p}$.
Let $M$ be a finitely generated module, then the following are equivalent :
Then one has the following characterisation of finitely generated flat modules.
Let $M$ be a finitely generated module, then
This characterisation can be strengthened in two very large cases
(When $R$ is Noetherian) If $R$ is a Noetherian ring and $M$ is a finitely generated module, then
Over a Noetherian ring all finitely generated modules are finitely presented. So this is a simple consequence of Corollary ()
(When $R$ is an integral domain) If $R$ is an integral domain and $M$ is a finitely generated module, then
Given two prime ideals $\mathfrak{p} \subset \mathfrak{q}$, one gets $\mathrm{r}(\mathfrak{p}) = \mathrm{r}(\mathfrak{q})$. Since $R$ is an integral domain $(0)$ is a prime ideal and $(0) \subset \mathfrak{p}$ for every prime ideal $\mathfrak{p}$. The function $\mathfrak{p} \mapsto \mathrm{r}(\mathfrak{p})$ is thus constant.
The proof of the above proposition shows that the same result is true for every ring $R$ with a unique minimal ideal.
In the general case, it is not true that all finitely generated flat modules are projective.
(counter-example) Let $R = \prod_{i = 0}^\infty k$ be an infinite product of fields. Let $I = \oplus_{i = 0}^\infty k \subset R$ its ideal.
Then $R/I$ is a finitely generated $R$-module, flat (all $R$-modules are flat), but not projective.
The class of commutative rings $R$ over which all finitely generated flat modules are projective has been determined in the 1970s.
Let $R$ be a commutative ring. The following two are equivalent :
One can see Gena Puninski; Philipp Rothmaler (2004). When every finitely generated flat module is projective. Journal of Algebra, 277(2), 542–558. doi:10.1016/j.jalgebra.2003.10.027
As we have seen finitely presented flat module are always projective ; their projective dimension is equal to zero. For countably presented flat modules, the dimension cannot exceed one.
A countably presented flat module has projective dimension? inferior or equal to $1$.
Let $M$ be a countably generated flat module. Being flat, it can be written as an inductive limit of finitely generated free modules $M = \underset{\longrightarrow}{\lim}_A M_\alpha$. Since $M$ is also countably presented, there exists a countable cofinal filtered subset $H \subset A$, so one can reduce to the case where $A = \mathbb{N}$.
Denoting by $g_n : M_n \to M_{n+1}$ the transition maps, one can build the short exact sequence
where the map $G$ restricted to $M_n$ is given by $1 - g_n$.
As we have seen, there is not much difference between flat modules and projective modules in the finitely generated case. The notion of flatness starts becoming really distinct in the case of infinitely generated modules. For example the $\mathbb{Z}$-module $\mathbb{Q}$ is flat (since it is torsion-free) but far from being free.
On the other side, for infinitely generated module, the notion of projectivity is less useful since for example the module $\mathbb{Z}[[X]]$ which is the infinite product of copies of $\mathbb{Z}$ is not free but still flat. It is common to encounter infinitely generated flat modules and uncommon to encounter infinitely generated projective modules.
To better understand the general notion of flatness from a geometric perspective, very early after the definition of the notion, two questions have been asked
For a commutative ring $R$, the following propositions are equivalent :
$1 \Rightarrow 2$. Let $\{Q_\alpha\}_{\alpha \in A}$ be a set of flat $R$-modules. By cyclic reduction, we only need to show that the map $I \otimes \prod_\alpha Q_\alpha \to \prod_\alpha Q_\alpha$ is injective for every finitely generated ideal $I \subset R$. Since $R$ is coherent every such ideal $I$ is a finitely presented module and thus the map $I \otimes \prod_\alpha Q_\alpha \to \prod_\alpha Q_\alpha$ is the product of the maps $I \otimes Q_\alpha \to Q_\alpha$. It is injective since each individual map is injective by flatness of each $Q_\alpha$.
$2 \Rightarrow 1$. Let $I \subset R$ be a finitely generated ideal. Using one of the equivalent characterisations of finitely presented modules, we shall show that the canonical map $I \otimes R^A \to I^A$ is an isomorphism, for every set $A$. By assumption, since $R^A$ is a flat module, the map $I \otimes R^A \to R^A$ is injective and since $I$ is finitely generated, its image is $I^A$.
We say that a ring $R$ is absolutely flat if every $R$-module is flat.
The following propositions are equivalent for a commutative ring $R$:
1 $\Rightarrow$ 2. Let $I \subset R$ be any ideal of $R$, then since $R$ is absolutely flat, the $R$-module $R/I$ must be flat. One has
the ideal $I$ must thus be idempotent;
2 $\Rightarrow$ 3. For every $x \in R$, the ideal $(x)$ must be idempotent, that is $(x)(x) = (x)$ which means that there must exist $y$ such that $x = x^2y$;
3 $\Rightarrow$ 4. Let $r \in \mathfrak{p}$. Let $u \in R$ such that $r u r = r$. Then $r ( 1 - u r ) = 0$. Since $u r \in \mathfrak{p}$ it follows that the image of $1- u r$ is invertible in $R_\mathfrak{p}$, so the image of $r$ is null in $R_\mathfrak{p}$. Thus the maximal ideal $R_\mathfrak{p}$ is trivial.
4 $\Rightarrow$ 5. Obvious.
5 $\Rightarrow$ 1. Because flatness is a local property.
Recall that a commutative ring $R$ is semi-hereditary? if every finitely generated ideal is projective.
For a commutative ring $R$, the following are equivalent :
First, if $R$ is a local ring, this follows from the equivalent characterisations of valuation rings. So we only need to take care of the global part of the theorem.
$1 \Rightarrow 2$. Every $R_\mathfrak{m}$ is a localisation of $R$, so it is semi-hereditary and hence a valuation ring. To show that $\mathrm{Q}(R)$ is absolutely flat, it shall be enough to show that for every $x \in R$, there exists a regular $r$ such that $x^2 = r x$. Let $x \in R$, then $(x)$ is projective. As a consequence the annihilator $Ann(x) \subset R$ is a direct summand, so there exists an idempotent $\pi \in R$ such that $Ann(x) = (\pi)$. Then $r \coloneqq x + \pi$ is a regular element such that $x^2 = r x$.
$2 \Rightarrow 3$. When $\mathrm{Q}(R)$ is absolutely flat, being torsion-free becomes a local property, as it is for flatness.
$3 \Rightarrow 1$. Torsion-free modules are stable under product, so it follows that flat modules are also stable under product which means that $R$ is coherent. Then every finitely generated ideal $I \subset R$ is finitely presented. Since they are also torsion-free, they are flat and thus projective.
Since $\mathbb{Z}$ is a PID, an abelian group is flat (regarded as a $\mathbb{Z}$-module) precisely if it is torsion-free.
In particular, the modules $\mathbb{Q}$ and $\mathbb{Z}[[X]]$ are flat.
The ring of power series $R[[X]]$ (which as an $R$-module is simply an infinite product of the base ring) is an interesting example.
For the flatness of $R[[X]]$ see Direct products of modules, Theorem 2.1.
Let $R$ be an integral domain, with field of fractions $K$. Then $K$ is a flat $R$-module, as it can be written as the filtered colimit
of free $R$-modules.
If $\pi$ is an idempotent element of a commutative ring $R$, then the cyclic module $R/(\pi)$ is projective, hence flat.
projective object, projective presentation, projective cover, projective resolution
injective object, injective presentation, injective envelope, injective resolution
flat object, flat resolution
free module$\Rightarrow$ projective module $\Rightarrow$ flat module $\Rightarrow$ torsion-free module
Flat modules (“modules plats” in French) where first defined in GAGA
Original articles include
Shizuo Endo, On flat modules over commutative rings, J. Math. Soc. Japan Volume 14, Number 3 (1962), 284-291. (EUCLID)
Shizuo Endo, On semi-hereditary rings, J. Math. Soc. Japan Vol. 13, No. 2, 1961
Michel Raynaud, Laurent Gruson, Critères de platitude et de projectivité, Techniques de “platification” d’un module. Invent. Math. 13 (1971), 1–89.
The characterization of flat modules as filtered colimits of projective modules is due to
For a general account see for instance section 3.2 of
For more details see
Lecture notes include
Further resources include
MO discussion flatness and local freeness
Stephen Chase, Direct product of modules
Last revised on July 29, 2023 at 09:21:02. See the history of this page for a list of all contributions to it.