AMERICAN JOURNAL OF UNDERGRADUATE RESEARCH VOL. 3 NO. 3 (2004)
Exactness, Tor and Flat Modules Over a Commutative Ring
Abhishek Banerjee Indian Statistical Institute 203 Barrackpore Trunk Road Kolkata 700108 INDIA
Received: June 4, 2004 Accepted: November 20, 2004
ABSTRACT
In this paper, we principally explore flat modules over a commutative ring with identity. We do this in relation to projective and injective modules with the help of derived functors like Tor and Ext. We also consider an extension of the property of flatness and induce analogies with the “special cases” occurring in flat modules. We obtain some results on flatness in the context of a noetherian ring. We also characterize flat modules generated by one element and obtain a necessary condition for flatness of finitely generated modules.
I. DEFINITIONS • Flat modules: Those modules F for which the functor _⊗F is exact are Let R be a commutative ring with termed flat modules. identity and consider the modules over R. • Derived functors: Let T be an additive • Exact sequence: A sequence of maps functor and N be an R-module. If the A→B→C is said to be exact at B if the Ci’s are all projective modules and the image of the map “entering” B is equal following sequence is exact: to the kernel of the map “leaving” B. …. →…→C1→C0→N→0, then we have • Functor: It is a map from the set of R- a projective resolution of N. The ith- modules to itself. derived functor with respect to T is the • Exact functor: A functor T which when homology module at T(Ci ), i.e., the applied to all the terms of an exact quotient of the kernel of the map leaving sequence induces another exact T(Ci) to the image of the map entering it sequence is said to be exact. The from T(Ci-1). It can be proved that the functor T in question should be additive, derived functor is determined uniquely i.e. if f is a map from M to N, there up to isomorphism by any projective should exist an induced map T(f) from resolution [1-7]. T(M) to T(N). • Projective modules: A module P having the following property: If p is a map from II. FLATNESS M onto N and f is a map from P to N, there exists a map g from P to M such Let R be a commutative ring with that pog = f. In other words, P is such identity [2]. Flatness may be defined through that the functor Hom(P,_) is exact. either of the following equivalent conditions: • Injective modules: A module Q having the following property: If i is a one to one (i) If P→Q is a monomorphism of R- map from K to M and f is a map from K modules, the induced map from M⊗P → to Q, there exists a map g from M to Q M⊗Q is also a monomorphism. such that goI = f. In other words, the (ii) The functor _ ⊗M is exact. functor Hom(_,Q) is exact.
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We note the following interesting hence we have Tor(M,E)=0 for all modules isomorphism: M. Thus, E is flat. Hom(E,Hom(F,G)) ≅ Hom(E⊗F,G).1 From these we obtain the following results: III. DEFINITION OF THE EXT Result (a): Suppose that G is injective FUNCTOR and we have the exact sequence: 0→Hom(M,G)→Hom(N,G) →Hom(P,G) Suppose that M is an arbitrary R- →0. If E is flat, the functor Hom(E,_) module and that the following is a projective
preserves the exactness of this resolution of M:...→C2 →C1 →C0 →M →0. sequence. Thus, the functor Hom(_,G) We consider the nth right derived functor of for injective G enables a flat module to Hom(_,N). This is denoted by Extn(M,N). If do what a projective module does. 0 →M’ →M →M’’ →0 is exact, we have the This follows if we apply the functors E⊗_ exact sequence, 0 → Hom(M’’,N)→ and Hom(_,G) one after the other to the Hom(M,N) → Hom(M’,N) → Ext1(M’’,N) sequence 0→P→N→M→0. Each of →Ext1(M,N) →Ext1(M’,N) →…. these functors preserves the exactness From this, we obtain the following of the sequence. set of equivalent statements:
Result (b): The condition that (i) M is projective. Hom(E⊗_,G) is exact can be replaced (ii) Extn(M,N)=0 for all N and all n>=1. by the condition that Hom(E,G) is (iii) Ext1(M,N)=0 for all N. injective. We will now prove that if
Hom(E,G) is injective for all injective We first determine as to when there exist modules G, then E is flat. projective modules that are direct
summands of R. In this respect, we obtain Result (c): If E and F are projective, so the result that follows. is E⊗F. This is because the successive use of the two functors Hom(F,_) and Result: There exist projective modules Hom(E,_) can be replaced by the that are direct summands of R if and functor Hom(E⊗F,_). only if R contains idempotents other than 0 and 1. Result (a) can also be used to characterize flat modules. If we assume that the functor Suppose that R can be written as a direct Hom(E⊗_,G) is exact for all injective sum I⊕J, where I and J are submodules modules G, we can prove that E is flat. (and hence ideals of R). Then 1 can be Let M be an arbitrary R-module. We written as a sum i+(1-i), where i is in I and 1- represent M as the quotient of a free module i is in J. Consider the element i(1-i). This F. Thus we have the exact sequence lies in both I and J and hence must be 0. 0→K→F→M→0. Tensoring with E, we have Thus i is an idempotent in R. If i is 0 or 1, the exact sequence 0→Tor(M,E)→E⊗K→ one of the submodules I and J will contain 1 E⊗F→E⊗M→0. and hence will be equal to R. Thus, R Let G be any injective module. contains an idempotent not equal to 0 or 1. Then 0 → Hom(E⊗M,G) → Hom(E⊗F,G) → Conversely, if we assume that R Hom(E⊗K,G) → Hom(Tor(M,E),G) → 0 is contains an idempotent j ≠ 0,1 we always exact. But, if we assume that Hom(E⊗_,G) have a nontrivial projective module which is is exact, we have Hom(Tor(M,E),G)=0 for all a direct summand of R. Consider the ideals injective G. Now, Tor(M,E) can be (j) and (1-j). None of these modules is zero embedded in some injective module G and and hence it is enough to show that their intersection is 0. Say that jx = (1-j)y. Then y = j(x+y). Then jx = (1-j)j(x+y). But j(1-j) = 0 and hence jx = 0. Thus, the intersection is 1 We note that the RHS is symmetric in E (0). and F, while the LHS is apparently not so. We have Hom(E,Hom(F,G)) ≅ Hom(F,Hom(E,G)).
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0 (i) M is a flat module. ↓ (ii) Tori(N,M)=0 for all R-modules N and
0 Tor(N,M) all i>0. ↓ ↓ (iii) Tor1(N,M)=0 for all R-modules N. M’⊗N’→ F ⊗N’ → M⊗N’→ 0 M b. Isomorphisms ↓ ↓ ↓ 0 → M’⊗FN → FM⊗FN → M⊗FN → 0 Now that we have defined both Ext ↓ ↓ ↓ and Tor, we obtain the following M’⊗N → FM⊗N → M⊗N → 0 isomorphisms: ↓ ↓ ↓ (a) For any projective module P: 0 0 0 Hom(P,Ext(F,G))≈Ext(P⊗F,G) for all F and G. Figure 1. A ‘tensor product’ of the (b) For any injective module Q: sequences in (1). Hom(Tor(E,F),Q)≈Ext(F,Hom(E,Q)) for all F and finitely generated E. To prove (a) we start with the exact IV. THE TOR FUNCTOR AND sequence: 0 →G →In →I →0, where In is FLATNESS an injective module. (n.b., Any module can be embedded in an injective module.) We a. Symmetry of the Tor Functor obtain: 0 → Hom(F,G) → Hom(F,In) → Hom(F,I) → Ext(F,G) → 0. Since P is The Tor functor directly measures projective, we have: 0→Hom(P,Hom(F,G)) the degree to which a module is flat. As the →Hom(P, Hom(F,In)) → Hom(P, Hom(F,I)) tor is part of a family of functors, it lends →Hom(P,Ext(F,G)) →0. But we can directly itself to defining flatness in a more general obtain: 0 → Hom(P⊗F,G) → Hom(P⊗F,In) setting. First we show that the tor functor is →Hom(P⊗F,I) → Ext(P⊗F,G) → 0. Since symmetric: Tor(M,N) ≅ Tor(N,M). the first three terms of the two exact Represent M and N as quotients of sequences are isomorphic, we have the flat modules, result of part (a).
0 → M’ → FM → M → 0 To prove part (b), we start with the 0 → N’ → FN → N → 0 (1) exact sequence: 0 → I → Fr → E → 0, where Fr is a free module. (i.e. We write E We now ‘tensor the sequences with each as the quotient of a free module.) We tensor other’—as shown in Figure 1, above. The the sequence with F to obtain: 0 → Tor(E,F) snake lemma yields the exact sequence, → I ⊗F → Fr⊗F → E⊗F → 0. 0 → Tor(N,M) → M’⊗N → F ⊗N. But we M Applying the functor Hom(_,Q) we already have, obtain 0 → Hom(E⊗F,Q) → Hom(Fr⊗F,Q)
0 → Tor(M,N) → M’⊗N → FM ⊗N, →Hom(I⊗F,Q)→Hom(Tor(E,F),Q) → 0. But, we also have: 0 → Hom(E,Q) → Hom(Fr,Q) and hence Tor(M,N) ≅ Tor(N,M). An R- → Hom(I,Q) →0. And we get: 0→ module M is flat if and only if Tor(N,M) = 0 Hom(F,Hom(E,Q)) → Hom(F,Hom(Fr,Q)) → for all R-modules N. If we have a projective Hom(F, Hom(I,Q)) → Ext(F,Hom(E,Q)) → 0. resolution of N, say, Hom(Fr,Q) is isomorphic to a direct sum of
finitely many copies of Q and hence is … →C →C →C →N→0, 2 1 0 injective (since E is finitely generated, Fr
may be taken to be of finite basis). From the then Tor is the ‘1st’ derived functor of the isomorphism of the first three terms in the complex derived by tensoring the above two exact sequences, we obtain part (b). resolution by M. Similarly Torn is the nth derived functor of the complex. Derived functors (of additive functors) are c. Result of the Long Exact Sequence independent of the resolution chosen and First, we note that the long exact hence we have the following set of sequence of homology gives us the equivalent statements: following: If 0→M’→M→M’’→0 is an exact
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sequence, then the following is exact: …→ Clearly, Tork(N,M)=0 iff Tork-1(I,M) = 0. Tor2(M’,N) → Tor2(M,N) →Tor2(M’’,N) → Hence, M is k-flat if and only if Tork-1(I,M) = 0 Tor1(M’,N)→Tor1(M,N)→Tor1(M’’,N)→ M’⊗N for all submodules I of free modules. → M⊗N→ M’’⊗N→0. If N is an R-module We can also take another look at the result: If Hom(F,Q) is injective for each such that Torj(M,N)=0 for all modules M, it is injective module Q, then F is flat. We can easy to see that Tork(M,N)=0 for all modules M and all k>j. This follows easily from the extend this to the following: above long exact sequence. Any module M Result: If F is finitely generated F is 2- can be written as the image of a free module flat if and only if Hom(F,Q) is 2-injective F and if we tensor that exact sequence by N for each injective module Q. to obtain its long exact sequence, we have the result. If F is 2-flat, it must be the quotient of a free module Fr with respect to a flat module I:0 d. Follow-on Definitions and Observations →I→Fr→F→0. If Q is injective, we obtain the sequence: We therefore define the following: 0→Hom(F,Q)→Hom(Fr,Q) →Hom(I,Q) →0 From this we obtain: Ext(E,Hom(I,Q))→ • A module M is k-flat if Tor (N,M)=0 for k Ext2(E,Hom(F,Q)) → Ext2(E, Hom(Fr,Q)) → all R-modules N. (And we define k- Ext2(E, Hom(I,Q)). projective and k-injective likewise with k The first and last terms are zero because Ext ) Hom(I,Q) is injective (since I is flat). Also
Hom(Fr,Q) is isomorphic to a direct sum of Analogously, we have the following set of finitely many copies of Q and hence is equivalent statements: injective. Thus Ext2(E,Hom(F,Q)=0 for any
E. Thus Hom(F,Q) is 2-injective. (i) M is k-flat. Conversely, suppose that Hom(F,Q) (ii) Tor (N,M)=0 for all modules N. k is 2-injective for any injective module Q. Let (iii) Tor (N,M)=0 for all N and all j>=k. j I and Fr be defined as in the previous part.
We can obtain the exact sequence: Thus, if 0→N’→N→N’’→0 is an exact sequence and M is a 2-flat module, it is 0→Hom(F,Q)→Hom(Fr,Q)→Hom(I,Q)→0. carried to the exact sequence: Applying the functor Hom(E,_), we get: 0→Tor(N’,M)→Tor(N,M)→Tor(N’’,M)→N’⊗M Ext(E,Hom(Fr,Q)) → Ext(E,Hom(I,Q)) → →N⊗M→N’’⊗M→0 Ext2(E,Hom(F,Q)) → Ext2(E,Hom(Fr,Q)). The first and last terms are zero. Since For 3-flat modules the sequence will have Hom(F,Q) is 2-injective, the third term is also the Tor terms; for 4-flat modules the 2 zero. Thus Hom(I,Q) is injective. Hence I is sequence will have Tor terms and so on. 3 flat. Thus, F is 2-flat. If a module M is 2-flat, we want to e. Results for k-flat Modules find conditions under which it becomes flat (or 1-flat). Accordingly, we have the Now we have an if and only if following result: condition for k-flat modules. Result: The following statements are Result: “ A module M is k-flat if and only equivalent: if Tork-1(I,M)=0 for all submodules I of (ii) M is flat. free modules.” (iii) M is 2-flat and Tor(Q,M)=0 for all injective modules Q. In order to demonstrate this, we take an (i)⇒(ii) is obvious. arbitrary module N and write it as the quotient of a flat (free) module F: To prove the second part, we use the fact 0→I→F→N→0. We tensor the sequence by that every module can be embedded into an M and write out the long exact sequence: injective module. Let M be 2-flat and N be
→…→0→Tork(N,M)→Tork-1(I,M)→0→…→ an arbitrary R-module. Then there exists an Tor1(N,M)→I⊗M→F⊗M→N⊗M→0. injective module Q such that N can be embedded in Q. Thus, there is an exact
10 AMERICAN JOURNAL OF UNDERGRADUATE RESEARCH VOL. 3 NO. 3 (2004) sequence 0→N→Q→Q/N→0. Since M is 2- which are cyclic modules and hence flat, we have the sequence: isomorphic to some R/I. Thus, Tor(R/I,M)=0 for each I implies that M is flat. 0→Tor1(N,M)→Tor1(Q,M)→Tor1(Q/N,M)→N ⊗M→… VI. FLATNESS OF PROJECTIVE If we allow Tor(Q,M) = 0 for all injective MODULES modules Q, we have Tor(N,M) = 0 for all modules N. Hence, M is flat. The above theorem shows that in order to check whether a module is flat, we Each of the properties (flatness, 2- need to consider the tensor products I⊗M, flatness and so on is a local property: where I is an ideal. Suppose that x is a non- zero element of M and a is an element such Result: The following statements are that ax=0. Clearly, a is a nonunit and hence equivalent: there exists a maximal ideal m containing x. (i) M is k-flat. Since the map from m⊗M to M must be (ii) Mp is k-flat for each prime ideal p. injective, we must have a⊗x=0 in m⊗M. In (iii) Mm is k-flat for each maximal ideal general, we can say that if I is an ideal and m. ∑aixi=0 with ai∈I we must have ∑ai⊗xi=0. In this respect, we state the following criterion
Let …→C2→C1→C0→N→0 be a projective (without proof) of when an element of the resolution of an R-mdoule N. We form the tensor product of two modules M and N is complex M⊗C. Now, Tor (N,M) is the nth zero. k homology module of this complex and its Criterion: Let N be generated by a set of localizations are the homology modules of elements {ni}. Every element of M⊗N may the complex after localization. Since Q = 0 be written as a finite sum ∑mi⊗ni, where the ⇔ each Qp = 0 ⇔ each Qm = 0 for any mi lie in M. Such an expression is 0 iff there module Q, and we have our result. exist elements mj’ of M and elements aij of R such that ∑aijmj’=mi for each i (sum is taken V. THEOREM over j) and ∑aijni=0 for each j (sum is taken over i). Theorem: Let M be an R-mdoule. If I is an ideal of R, then the map I⊗M→M is an Projective modules are always flat. injection if and only if Tor(R/I,M)=0. The This follows from the fact that a direct sum module M is flat if and only if this is so for (finite or infinite) of R-modules is flat if and every ideal I. only if each of its direct summands is finite. This result can be easily generalized:
Proof: Consider the exact sequence: Result: A direct sum of R-modules is k- 0→I→R→R/I→0. We obtain the exact flat if and only each of its direct sequence: 0→Tor(R/I,M)→I⊗M→R⊗M. summands is k-flat. Since R⊗M=M, Tor(R/I,M) is the kernel of We will actually prove that Tor (P,M⊕N)=0 iff the map from I⊗M to M. Thus the map is an k Tor (P,M)=0 and Tor (P,N)=0. This is true injection if and only if Tor(R/I,M)=0. k k for k=1. We assume it to be true up to k-1. Now, assume that Tor(R/I,M)=0 for Express P as a quotient of a free module: all ideals I. Suppose that P→Q is an 0→I→F→P→0 and consider the exact injection of R-modules M⊗P→M⊗Q is not sequence an injection. Then there exists a non-zero 0→Tor (P,M⊕N)→Tor (I,M⊕N)→0. element m⊗p that goes to zero. If we restrict k k-1 If Tor (P,M⊕N)=0, we have Tor (I,M⊕N)=0 the map to the module generated by the k k-1 which means that Tork-1(I,M)=0 and so also finitely many elements required to take m⊗p for N. From the sequence obtained by to 0, we obtain a finitely generated module substituting M for M⊕N in the above for which this map is not an injection. But sequence, we have Tor (P,M)=0. The every finitely generated module can be k converse is proved similarly. decomposed into a finite chain of We note that properties of flat submodules, the successive quotients of modules tend to carry over to k-flat modules
11 AMERICAN JOURNAL OF UNDERGRADUATE RESEARCH VOL. 3 NO. 3 (2004) in an analogous manner. Let us generalize (ii) M is 2-flat and Tor(R/q,M)=0 for the statement “Projective modules are flat”. each primary ideal q of R.
Let M be a non-zero cyclic module over a Result: k-projective modules are k-flat. local ring (R,M). Then M is isomorphic to
some R/I. Let us assume that I is non-zero. Let M be k-projective, i.e. Extk(M,N)=0 for all Suppose that M is flat. Take an element a in N. Let k>1. We write M as the quotient of a I such that ax=0, where x is the generator of free module F: 0 →I →F →M →0. We the cyclic module M. Since M is flat, there derive the exact sequence: k-1 k exist elements b1, b2…bk in R and elements 0→ Ext (I,N) → Ext (M,N) →0 for any y ,y …y in M such that b a=0 for each module N. Thus Extk-1(I,N)=0 for all N and 1 2 k j 0
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an1+bn2=0. If this ideal is contained in the maximal ideal of R, the coefficient of m1 in INDIAN the 1st equation becomes a unit and hence {m1,m2} is not a minimal generating set. STATISTICAL Thus, we can say that I=R. The same holds for the second “co-ordinate”. This process INSTITUTE can be extended to higher dimensions. Thus, we can develop a necessary condition Calcutta, New Delhi, Bangalore for flatness of a finitely generated module and Hyderabad over a local ring. This may be extended to a general R.
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