nLab flat module




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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 RR that are not PID.

An RR-module MM is flat (from the French word “plat”) if it has “no torsion” in the sense that the Tor-functor vanishes for every RR-module XX:

Mis flatXRModTor 1 R(M,X)=0. M\;\text{is flat} \;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\; \underset{X \in R Mod}{\forall} \;\;\; \mathrm{Tor}_1^R(M, X) \,=\, 0 \,.


(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 RR 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.

In terms of exact functors and Tor-functors

Let RR be a commutative ring.


An RR-module NN is flat if tensoring with NN over RR as a functor from RRMod to itself

() RN:RModRMod (-)\otimes_R N : R Mod \to R Mod

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. Equivalently, one says a module is faithfully flat is it is flat and a module MM is nonzero if and only if the module MNM\otimes N is nonzero.


The condition in def. has the following immediate equivalent reformulations:

  1. NN is flat precisely if () RN(-)\otimes_R N is a left exact functor,

    because tensoring with any module is generally already a right exact functor;

  2. NN is flat precisely if () RN(-)\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;

  3. NN is flat precisely if () RN(-)\otimes_R N is a flat functor,

    because Mod is finitely complete;

  4. NN is flat precisely if the degree-1 Tor-functor Tor 1 RMod(,N)Tor_1^{R Mod}(-,N) is zero,

    because by the general properties of derived functors in homological algebra, L 1FL_1 F is the obstruction to a right exact functor FF being left exact;

  5. NN is flat precisely if all higher Tor functors Tor 1(RMod)(,N)Tor_{\geq 1}(R Mod)(-,N) are zero,

    because the higher derived functors of an exact functor vanish;

  6. NN is flat precisely if NN 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 TorTor-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.

Explicitly in terms of identities

There is a characterisation of flatness that says that a left AA-module MM is flat if and only if “everything (that happens in MM) happens for a reason (in AA)”. 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 VV, if we have an identity of the form iα iv i=0\sum_i \alpha_i v_i = 0 then we cannot necessarily assume that the α i\alpha_i are all zero. However, if we choose a basis then we can write each v iv_i in terms of the basis elements, say v i= jβ iju jv_i = \sum_j \beta_{i j} u_j, and substitute in to get ijα iβ iju j=0\sum_{i j} \alpha_i \beta_{i j} u_j = 0. Now as {u j}\{u_j\} forms a basis, we can deduce from this that for each jj, iα iβ ij=0\sum_i \alpha_i \beta_{i j} = 0. These last identities happen in the coefficient field, which is standing in place of AA 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 MM of the form ia im i=0\sum_i a_i m_i = 0 with m iMm_i \in M and a iAa_i \in A. Then there is a family {n j}\{n_j\} in MM such that every m im_i can be written in the form m i= jb ijn jm_i = \sum_j b_{i j} n_j and the coefficients b ijb_{i j} have the property that ia ib ij=0\sum_i a_i b_{i j} = 0.

The module MM 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 im_i correspond to a morphism into MM from a free module, say m:FMm \colon F \to M. The a ia_i correspond to a morphism a:RFa \colon R \to F, where 1 maps to the tuple whose iith term is a ia_i. That we have the identity ia im i=0\sum_i a_i m_i = 0 says that the composition mam a is zero, or that m:FMm \colon F \to M factors through the coequaliser of aa and 00.

Now we consider the elements n jn_j. These define another morphism from a free module, say n:EMn \colon E \to M. That the m im_i can be expressed in terms of the n jn_j says that the morphism mm factors through nn. That is, there is a morphism b:FMb \colon F \to M such that m=nbm = n b. We therefore have two factorisations of mm: one through nn and one through the cokernel cokera\coker a. The question is as to whether these have any relation to each other. In particular, does cokeraM\coker a \to M factor through nn? We can represent all of this in the following diagram.

Layer 1 R R F F M M coker a \coker a E E a a b b n n m m

Saying that MM is flat says that this lift always occurs.


A module NN is flat precisely if for every finite linear combination of zero, ir in i=0N\sum_i r_i n_i = 0\in N with {r iR} i\{r_i \in R\}_i, {n iN}\{n_i \in N\} there are elements {n˜ jN} j\{\tilde n_j \in N\}_j and linear combinations

n i= jb ijn˜ jN n_i = \sum_j b_{i j} \tilde n_j \;\;\in N

with {b ijR} i,j\{b_{i j} \in R\}_{i,j} such that for all jj we have

ir ib ij=0R. \sum_i r_i b_{i j} = 0\;\;\; \in R \,.

A finite set {r iR} i\{r_i \in R\}_i corresponds to the inclusion of a finitely generated ideal IRI \hookrightarrow R.

By theorem NN is flat precisely if I RNNI \otimes_R N \to N is an injection. This in turn is the case precisely if the only element of the tensor product I RRI \otimes_R R that is 0 in R RN=NR \otimes_R N = N is already 0 on I RNI \otimes_R N.

Now by definition of tensor product of modules an element of I RNI \otimes_R N is of the form i(r i,n i)\sum_i (r_i ,n_i) for some {n iN}\{n_i \in N\}. Under the inclusion I RNNI \otimes_R N \to N this maps to the actual linear combination ir in i\sum_i r_i n_i. This map is injective if whenever this linear combination is 0, already i(r i,n i)\sum_i (r_i, n_i) is 0.

But the latter is the case precisely if this is equal to a combination j(r˜ j,n˜ j)\sum_j (\tilde r_j , \tilde n_j) where all the r˜ j\tilde r_j are 0. This implies the claim.

Taking this a step further, we consider the filtered family of all finite subsets of MM. This generates a filtered family of finitely generated free modules with compatible morphisms to MM. So there is a morphism from the colimit of this family to MM. 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 MM has actually died on the way. This is precisely what the above characterisation of flatness is saying: the element corresponding to ia im i\sum_i a_i m_i that dies in MM is already dead by the time it reaches EE.

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.

For more general rings

Even if the ring RR is not necessarily commutative and not necessarily unital, we can say:

A left RR-module is flat precisely if the tensoring functor

() R:Mod RAb (-)\otimes_R \colon Mod_R \to Ab

from right RR modules to abelian groups is an exact functor.

General properties

Properties of the category of flat modules


The full subcategory of flat modules over a ring RR is stable under the following categorical constructions :

  1. (infinite) sums ;
  2. filtered colimits ;
  3. retracts;
  4. tensor product.


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 RR, which is flat over itself.


Let RR be a commutative ring and SRS \subset R be a multiplicative subset. Then the localisation R[S 1]R[S^{-1}] is a flat RR-module.


Regard the set SS as a poset with de divisibility relation. The fact that SS be multiplicative means that the poset is directed. One thus have that R[S 1]=lim s1sRR[S^{-1}] = \underset{\longrightarrow}{\lim}_s \;\frac{1}{s} \; R is a filtered colimit of free modules, hence it is flat.

Flatness is a local property


(Stability under base change) Let RSR \to S be a morphism of commutative rings and let MM be a flat RR-module, then the SS-module M RSM \otimes_R S is also flat.


Let XYX \to Y be an injective morphism of SS-modules, then the map (M RS) SX(M RS) SY(M \otimes_R S) \otimes_S X \to (M \otimes_R S) \otimes_S Y is isomorphic to the map M RXM RXM \otimes_R X \to M \otimes_R X which is injective because MM is flat.


(Flatness is a local property) Let RR be a commutative ring and let MM be an RR-module. Then the following assertions are equivalent :

  1. MM is flat;
  2. M 𝔭M_\mathfrak{p} is a flat R 𝔭R_\mathfrak{p}-module, for every prime ideal 𝔭R\mathfrak{p} \subset R;
  3. M 𝔪M_\mathfrak{m} is a flat R 𝔪R_\mathfrak{m}-module, for every maximal ideal 𝔪R\mathfrak{m} \subset R.


121 \Rightarrow 2. By base change.

232 \Rightarrow 3. Obvious.

313 \Rightarrow 1. Let XYX \to Y be an injective morphism of RR-modules and let KK denote the kernel of the induced map M RXM RYM \otimes_R X \to M \otimes_R Y. Since R 𝔪R_\mathfrak{m} is a flat RR-module it follows that X 𝔪Y 𝔪X_\mathfrak{m} \to Y_\mathfrak{m} is also injective ; and K 𝔪K_\mathfrak{m} is the kernel of M 𝔪 R 𝔪X 𝔪M 𝔪 R 𝔪Y 𝔪M_\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 𝔪=0K_\mathfrak{m} = 0 for every maximal ideal 𝔪R\mathfrak{m} \subset R, so K=0K = 0 and MM is flat.

Flat modules are torsion-free


For any module MM over a ring RR, one has

Mis flatMis torsion-free. M\;\text{is flat} \;\;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\;\; M\;\text{is torsion-free} \,.


For every regular element pRp \in R, the pp-torsion submodule of MM is isomorphic to Tor 1 R(M,R/(p))\mathrm{Tor}^R_1(M, R/(p)).

The converse holds for semi-hereditary rings.

Reduction to cyclic modules


(Reduction to cyclic modules) Let RR be a commutative ring and let MM be an RR-module. If

Tor 1 R(M,R/I)=0 Tor_1^R(M, R/I) = 0

for every finitely generated ideal IRI \subset R, then MM is a flat RR-module.

Equivalently, if

I RMIM I \otimes_R M \to IM

is an isomorphism for every finitely generated ideal IRI \subset R, then MM is flat.


The proof goes in 3 steps.

  1. Tor 1 R(M,R/I)=0Tor_1^R(M, R/I) = 0 for every ideal IRI \subset R. Indeed, write II as the union of the finitely generated ideals I finII_\mathrm{fin} \subset I, then R/I=lim I finIR/I finR/I = \underset{\longrightarrow}{\lim}_{I_\mathrm{fin} \subset I} R/I_\mathrm{fin}. Then because Tor 1Tor_1 commutes with filtered colimits
    Tor 1(M,R/I)=Tor 1(M,lim I finIR/I fin)=lim I finITor 1(M,R/I fin)=0 Tor_1(M, R/I) = Tor_1(M, \underset{\longrightarrow}{\lim}_{I_fin \subset I} R/I_fin) = \underset{\longrightarrow}{\lim}_{I_fin \subset I} Tor_1(M, R/I_fin) = 0
  2. Tor 1 R(M,X)=0Tor_1^R(M, X) = 0 for every finitely generated module XX. We can show this by induction on the number of generators of the module. The case of n=1n = 1 generator is covered by the previous step. Assume that Tor 1(M,X)=0Tor_1(M, X) = 0 for every module generated by nn elements. Let XX be generated by n+1n+1 elements x 0,,x nx_0, \dots, x_n. Let us call YY the submodule of XX generated by x 1,,x nx_1, \dots, x_n, then one has a short exact sequence 0YXX/Y00 \to Y \to X \to X/Y \to 0 leading to the sequence
    0=Tor 1(M,Y)Tor 1(M,X)Tor 1(M,X/Y)=0 0 = Tor_1(M, Y) \to Tor_1(M, X) \to Tor_1(M, X/Y) = 0

    which is exact in the middle, so Tor 1(M,X)=0Tor_1(M, X) = 0.

  3. For the last step, we use the fact that every module is the union of its finitely generated submodules and that Tor 1(M,)Tor_1(M,-) commutes with filtered colimits.
    Tor 1(M,X)=Tor 1(M, X finXX fin)=lim X finXTor 1(M,X fin)=0 Tor_1(M, X) = Tor_1\left(M, \bigcup_{X_fin \subset X} X_fin\right) = \underset{\longrightarrow}{\lim}_{X_fin \subset X} Tor_1(M, X_fin) = 0

    for every module XX.

The equivalence with the second characterisation comes from the fact that using the exact sequence 0IRR/I00 \to I \to R \to R/I \to 0 and tensoring it with MM, one gets the long exact sequence

0Tor 1(M,R/I)IRRR/I0 0 \to Tor_1(M, R/I) \to I \otimes R \to R \to R/I \to 0

identifying Tor 1(M,R/I)Tor_1(M, R/I) as the kernel of the always surjective map IMIMRI \otimes M \to IM \subset R.

Finitely generated flat modules

Thanks to the general structure theorem of flat modules, one gets


Given a finitely presented module MM, one has

Mis flatMis projective. M\;\text{is flat} \;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\; M\;\text{is projective} \,.


If MM is at the same time finitely presented and a filtered colimit of finitely generated free modules, the identity morphism MMM \to M must factor as MFMM \to F \to M where FF is a finitely generated free module. This implies that MM is projective.

In many cases, “finite presentation” can be replaced with “finite generation”.


(When RR is local)

Let MM be a finitely generated module over a local ring RR. Then one has

Mis flatMis free. M\;\text{is flat} \;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\; M\;\text{is free} \,.

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 RR-module MM is called a locally free module in the weak sense if for every prime ideal 𝔭R\mathfrak{p} \hookrightarrow R the localised module M 𝔭M_\mathfrak{p} is a free module over the localization R 𝔭R_\mathfrak{p}.


Let MM be a finitely generated module, then the following are equivalent :

  1. MM is projective ;
  2. MM is locally free in the weak sense and the rank function 𝔭r(𝔭)\mathfrak{p} \mapsto \mathrm{r}(\mathfrak{p}) is locally constant.

Then one has the following characterisation of finitely generated flat modules.


Let MM be a finitely generated module, then

Mis flatMis locally free in the weak sense. M\;\text{is flat} \;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\; M\;\text{is locally free in the weak sense} \,.

This characterisation can be strengthened in two very large cases


(When RR is Noetherian) If RR is a Noetherian ring and MM is a finitely generated module, then

Mis flatMis projective. M\;\text{is flat} \;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\; M\;\text{is projective} \,.


Over a Noetherian ring all finitely generated modules are finitely presented. So this is a simple consequence of Corollary ()


(When RR is an integral domain) If RR is an integral domain and MM is a finitely generated module, then

Mis flatMis projective. M\;\text{is flat} \;\;\;\;\;\;\;\; \Leftrightarrow \;\;\;\;\;\;\;\; M\;\text{is projective} \,.


Given two prime ideals 𝔭𝔮\mathfrak{p} \subset \mathfrak{q}, one gets r(𝔭)=r(𝔮)\mathrm{r}(\mathfrak{p}) = \mathrm{r}(\mathfrak{q}). Since RR is an integral domain (0)(0) is a prime ideal and (0)𝔭(0) \subset \mathfrak{p} for every prime ideal 𝔭\mathfrak{p}. The function 𝔭r(𝔭)\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 RR 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= i=0 kR = \prod_{i = 0}^\infty k be an infinite product of fields. Let I= i=0 kRI = \oplus_{i = 0}^\infty k \subset R its ideal.

Then R/IR/I is a finitely generated RR-module, flat (all RR-modules are flat), but not projective.

The class of commutative rings RR over which all finitely generated flat modules are projective has been determined in the 1970s.


Let RR be a commutative ring. The following two are equivalent :

  1. Every finitely generated flat RR-module is projective ;
  2. Every sequence (a 0,a 1,)(a_0, a_1, \dots) of elements of RR such that a n+1a n=a na_{n+1} a_n = a_n for every 0n0 \leq n, is eventually constant.

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 

Countably presented flat modules

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 11.


Let MM be a countably generated flat module. Being flat, it can be written as an inductive limit of finitely generated free modules M=lim AM αM = \underset{\longrightarrow}{\lim}_A M_\alpha. Since MM is also countably presented, there exists a countable cofinal filtered subset HAH \subset A, so one can reduce to the case where A=A = \mathbb{N}.

Denoting by g n:M nM n+1g_n : M_n \to M_{n+1} the transition maps, one can build the short exact sequence

0M nGM nM0 0 \to \oplus M_n \overset{G}{\to} \oplus M_n \to M \to 0

where the map GG restricted to M nM_n is given by 1g n1 - 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 [[X]]\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

  1. A product of torsion-free modules is always torsion-free, is it true for flat modules ?
  2. What type of rings RR have all their modules flat?
  3. For what type of rings RR is it true that all torsion-free modules are flat?
  4. For what type of rings can flat modules be the filtered union of their finitely generated projective submodules?

Coherent rings


For a commutative ring RR, the following propositions are equivalent :

  1. RR is a coherent ring ;
  2. the product of a family of flat RR-modules is flat.


121 \Rightarrow 2. Let {Q α} αA\{Q_\alpha\}_{\alpha \in A} be a set of flat RR-modules. By cyclic reduction, we only need to show that the map I αQ α αQ αI \otimes \prod_\alpha Q_\alpha \to \prod_\alpha Q_\alpha is injective for every finitely generated ideal IRI \subset R. Since RR is coherent every such ideal II is a finitely presented module and thus the map I αQ α αQ αI \otimes \prod_\alpha Q_\alpha \to \prod_\alpha Q_\alpha is the product of the maps IQ αQ αI \otimes Q_\alpha \to Q_\alpha. It is injective since each individual map is injective by flatness of each Q αQ_\alpha.

212 \Rightarrow 1. Let IRI \subset R be a finitely generated ideal. Using one of the equivalent characterisations of finitely presented modules, we shall show that the canonical map IR AI AI \otimes R^A \to I^A is an isomorphism, for every set AA. By assumption, since R AR^A is a flat module, the map IR AR AI \otimes R^A \to R^A is injective and since II is finitely generated, its image is I AI^A.

Absolutely flat rings


We say that a ring RR is absolutely flat if every RR-module is flat.


The following propositions are equivalent for a commutative ring RR:

  1. RR is absolutely flat;
  2. every ideal IRI \subset R is idempotent, i.e. I 2=II^2 = I;
  3. for every xRx \in R there exists an element yRy \in R such that x=x 2yx = x^2 y ;
  4. R 𝔭R_\mathfrak{p} is a field for every prime ideal 𝔭R\mathfrak{p} \subset R;
  5. R 𝔪R_\mathfrak{m} is a field for every maximal ideal 𝔪R\mathfrak{m} \subset R.


1 \Rightarrow 2. Let IRI \subset R be any ideal of RR, then since RR is absolutely flat, the RR-module R/IR/I must be flat. One has

Tor 1(R/I,R/I)I/I 2 Tor_1(R/I, R/I) \simeq I/{I^2}

the ideal II must thus be idempotent;

2 \Rightarrow 3. For every xRx \in R, the ideal (x)(x) must be idempotent, that is (x)(x)=(x)(x)(x) = (x) which means that there must exist yy such that x=x 2yx = x^2y;

3 \Rightarrow 4. Let r𝔭r \in \mathfrak{p}. Let uRu \in R such that rur=rr u r = r. Then r(1ur)=0r ( 1 - u r ) = 0. Since ur𝔭u r \in \mathfrak{p} it follows that the image of 1ur1- u r is invertible in R 𝔭R_\mathfrak{p}, so the image of rr is null in R 𝔭R_\mathfrak{p}. Thus the maximal ideal R 𝔭R_\mathfrak{p} is trivial.

4 \Rightarrow 5. Obvious.

5 \Rightarrow 1. Because flatness is a local property.

Semi-hereditary rings


Recall that a commutative ring RR is semi-hereditary? if every finitely generated ideal is projective.


For a commutative ring RR, the following are equivalent :

  1. RR is semi-hereditary? ;
  2. The total ring of fractions Q(R)\mathrm{Q}(R) of RR is absolutely flat and for every maximal ideal 𝔪\mathfrak{m} of RR, the local ring R 𝔪R_\mathfrak{m} is a valuation ring ;
  3. Every torsion-free RR-module is flat.


First, if RR 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.

121 \Rightarrow 2. Every R 𝔪R_\mathfrak{m} is a localisation of RR, so it is semi-hereditary and hence a valuation ring. To show that Q(R)\mathrm{Q}(R) is absolutely flat, it shall be enough to show that for every xRx \in R, there exists a regular rr such that x 2=rxx^2 = r x. Let xRx \in R, then (x)(x) is projective. As a consequence the annihilator Ann(x)RAnn(x) \subset R is a direct summand, so there exists an idempotent πR\pi \in R such that Ann(x)=(π)Ann(x) = (\pi). Then rx+πr \coloneqq x + \pi is a regular element such that x 2=rxx^2 = r x.

232 \Rightarrow 3. When Q(R)\mathrm{Q}(R) is absolutely flat, being torsion-free becomes a local property, as it is for flatness.

313 \Rightarrow 1. Torsion-free modules are stable under product, so it follows that flat modules are also stable under product which means that RR is coherent. Then every finitely generated ideal IRI \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 [[X]]\mathbb{Z}[[X]] are flat.


The ring of power series R[[X]]R[[X]] (which as an RR-module is simply an infinite product of the base ring) is an interesting example.

  1. It is always torsion-free;
  2. it is a flat RR-module if and only if RR is a coherent ring;
  3. it is almost never free. In particular [[X]]\mathbb{Z}[[X]] is not free.

For the flatness of R[[X]]R[[X]] see Direct products of modules, Theorem 2.1.


Let RR be an integral domain, with field of fractions KK. Then KK is a flat RR-module, as it can be written as the filtered colimit

K=lim r1rR K = \underset{\longrightarrow}{\lim}_r \;\frac{1}{r}\;R

of free RR-modules.


If π\pi is an idempotent element of a commutative ring RR, then the cyclic module R/(π)R/(\pi) is projective, hence flat.


Flat modules (“modules plats” in French) where first defined in GAGA

  • Jean-Pierre Serre, Géométrie Algébrique et Géométrie Analytique (pdf)

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.

  • S. Jondrup, Flat and projective modules, Math, Scand. 43 (1978) (pdf)

The characterization of flat modules as filtered colimits of projective modules is due to

  • Daniel Lazard, Sur les modules plats C. R. Acad. Sci. Paris 258, 6313–6316 (1964)

For a general account see for instance section 3.2 of

For more details see

  • Matsumura’s CRT book

Lecture notes include

  • Arthur Ogus, Flatness – a brief overview (pdf)

Further resources include

Last revised on March 25, 2024 at 19:16:08. See the history of this page for a list of all contributions to it.