PROJECTIVE COVERS AND MINIMAL FREE RESOLUTIONS

Using a generalization of the definition of the projective cover of a module, a special type of surjective free resolution, known as the projective cover of a complex, may be defined. The projective cover is shown to be a direct summand of every surjective free resolution and to be the direct sum of the minimal free resolution and an exact complex. Necessary and sufficient conditions for the projective cover and minimal free resolution to be identical are discussed.


INTRODUCTION.
Let R be a commutative ring with identity.A complex C of R-modules is a sequence of R-module homomorphisms .... C, C,_, ..... C1 Co -*.. satisfying 6,-1 o 6, 0 for all E Z.The maps 6i are called the boundary maps of C. In this paper, we will assume that all complexes are bounded above, i.e.C, 0 for all < 0. If ker 6,-1 im 6 for all G Z then C is called an exact complex.
A map of complexes C D is a sequence of R-module homomorphisms C, 24 D, which commutes with the boundary maps of the complexes C and D. If each map , is surjeetive, we will say that is surjective and if each induced map H,(C) 0-H,(D) is an isomorphism, we will say that is a quasi-isomorphism.Generalizing the definition of the projective cover of a module, we define the projective cover of a complex C to be a complex P of projective modules and a map of complexes P C with the following two properties: can l)e c,npleted 1,v a nal, of complexes.(If P L C satisfies this conditmn alone, then the map is cdled a l)r()j('('live l)l ('('()v(,1 .)can only be completed bv maps of complexes which are automorphisms of P, in each degree.
It can be shown (see Goddard[1]) that every bounded above complex of finitely generated modules over a Noetherian local ring has a projective cover and that the projective cover is a unique surjective quasi-isomorphism.The proofs of these facts depend upon two key lemmas which we shall find useful later in this paper as well.The first lemma, which is proved by Roberts[2]   (pp.42-44.), is helpful in proving that the projective cover is a surjeetive map of complexes.
LEMMA 1.If G is a bounded above complex of finitely generated modules over a Noetherian local ring then there exists a map of complexes F -t C such that I.Each map F 2. C is surjective.
2. F is a free module for each .
Our second lemma, which along with the first is used to prove that the projective cover is a quasi-isomorphism, has been proven by Goddard[1].The proof is included here for completeness.
LEMMA 2. Let C be a bounded above complex.If P -' G is a surjective quasi-isomorphism then P is projective precover of (3 PROOF.Let Q be a complex of projective modules and let Q -C be a map of complexes. We need to findamapQ Psothateoh=.We leth,=0fori<0.For the=0case, we first need to note that for all > 0, e,(Z,(P)) Z,(G) {c 6 C,]6,(c) 0}.In order to do so, let c 6 Zi(C).Since is a quasi-isomorphism, there is a p 6 Z,(P) such that e,(p) c 6 B,(C).
Since eo(Zo(P)) Zo(C) and Qo is projective, the diagram Z0(P) o Zo(C) " can be completed by a map h0.We now proceed inductively to construct h.Assume that we have constructed h such that e h for all j _< and h commutes with the boundary maps.We now wish to construct h+. g-+ Snce oh, og+l(q) E B,(C) for all q E Q,+I and is a quasi-isomorphism, we have h, og,+l(q) B,(P) for all q Q,(R).Thus the diagram b,, B (P) P+ ht g+ \ can be completed by a map h' since O,+ is projective.For q ,+, we have (e,+l oh'-,+)(q) Z,+(C) so the diagram Z,+(P) q+' Qz+l can be completed by a map h" since the module Q,+, is projective and the map is surjeetive.If we let h,+l h'-h" then it follows eily that h commutes with the boundary and e,+ o h,+ ,+.
Since the existence of projective covers and the hypotheses of lemma both place the same requirements upon our complex C, we shall assume hereafter that C refers to a bounded above complex of finitely generated modules over a Noetherian local ring.Under these conditions, the projective cover P like the complex F constructed in lemma 1, turns out to be a complex of fr modules.Thus we define a fr resolution of a complex C to be a quasi-isomorphism F C from a complex of free modules into C.
The free resolution of a complex is not unique.The projective cover of is one example of a surjective free resolution.Another unique type of free resolution is given below: DEFINITION.A minimal free resolution of a complex C of R-modules is a free resolution F C such that each boundary map F, F,_ is defined by a matrix with entries in the mimal ideal m of the local ring R.
Roberts [2] shows that every complex has a unique minimal free resolution and that every free resolution is a direct sum of a minimal free resolution and an exact sequence.Later in th paper we will investigate the relationship betwn the projective cover and the minimal fr resolution in detail.
. ,S[IR.Ji'X'TIVI' FII'I' IESOI[ITiONS A<'cordinl4 to lemma 2, (,v<,rv sur.le<'tlve free resolution of a COml)lex C must be a projective 1)re<'ov<'r of C. The 1)r(,j(,<'ive <'ov(, tsclf is a unique examl,le of a surjective free resolution but lhcr(, mav le other suj(,('liv(, fiee resolutions as well.For example, if R Z/4 and C is the 1)ounde(l al)ove ('oml)lcx c o z/4 z/4 o .. then the projective cover of C is identical to C itself but if we utilize the construction used by Roberts[2] to prove lemma l, we obtain a different surjective fl resolution: o Notice in this case that the projective cover of C is identical to the minimal fr resolution of C. We will see in the next section how the structure of the complex C forces this to be true.While the projective cover and surjective free resolution are not identical, they are very closely related.The exact nature of this relationship is sn in the following theorem: THEOREM 1.The projective cover of a complex C is a direct summand of any surjeetive free resolution of the complex.
POOF.Let P C be the projective cover of C and let P C be a surjective free reso- lution of C. Since P is the projective cover of C, the diagram can be completed by a map of complexes N P. Also, since N is a projective preeover of C (by lemma 2), the diagram P can be completed by a map of complexes P N. Thus the map P P completes the diegrem P Since P & C is e projective cover, the map of complexes h o k must be an eutomorphism of P.
Therefore, h must be a surjeetion and we have the short exact sequence of complexes 0--,K' F---P--,0 To see that P s a dre<'l summand of F, we simply need to show that there exmts a map of complexes P -F s<'h that h o g zdp At lirst glance, this may seem obvmus snce P s a complex of 1)to.lecturemodules but t s not dflicult to find an example of a complex of projective modules whwh does not form a projective complex (see Goddard[1]).
To obtain the map g, we observe that both and @ nmst be quasi isomorphisms, so h s a quas-sonorphism as well.Snce h is also surjectve, by lemna 2, we know that h is a projective precover so the diagram can be completed by the desired map of complexes g. [] 3. MINIMAL FREE RESOLUTIONS Just as the projective cover is a direct summand of the surjectve free resolution, it is easy to see that the projective cover is a direct sum of the minimal free resolution and an exact complex.Although both the projective cover and the minimal free resolution are unique examples of free resolutions, they are not necessarily identical.For example, if R Z/4, the complex c ..... o z/4 z/4 o .... is equM to its projective cover (The map P C is simply the identity map) but is not equal to its minimal free resolution.In fact, the minimal free resolution of C is equal to the zero complex.This example and the example of the previous section suggest that the relationship between the projective cover and the minimal free resolution depends upon the boundary maps of C and the maximal ideal m of the ring R. Our next theorem will begin to formalize this relationship but we begin with a crucial lemma.
LEMMA 3. If P M is the projective cover of the finitely generated R-module M where R is a local ring with maximal ideal m, x P, and e(x) mM then x raP.
PROOF.The result follows easily from the construction of the projective cover (see Rot- man[3], pp 136.). []THEOREM 2. If C is a complex of modules over a local ring R with maximal ideal m, and C has boundary maps 6 such that 6,+1(C,+) _C taker(6,) for all z, then the projective cover P C is equal to the minimal free resolution of C.
PROOF.Let b be the boundary map of the complex P. To prove that P is the minimal free resolution, we must verify that b+(P+) c_ mP for all >_ 0. First, if p P then e0(bl(p)) 51(1(P)) mCo In the construction of the projective cover of a complex (see Goddard[1]), the map P0 Co is the (module) projective cover of Co.Thus by the previous lemma, we see that b(p) mPo.
Now suppose x P,+ for some _> 1.Once again invoking the construction of the projective cover we recall that C',+_[e,(p) 6,+(c)} where b,+ n o and e,+ r, o e.If e(x) (p, c) S,+, then we need to show that b,+(x) Since r,+(c) mker(/,) by hypothesis, we know ,+(c)-rc + +rc for some r, r,.r, m and c, c2 c, 6 ker(,).Since P, is the projective cover of S, {(p', c') ker(b,_) C, le,_(f 6,(d)}, (0, c) S, for < 3 _< n, and the projective cover is surjective, we can find elements p, ,p, ker(b,) such that e,(p3) c3 for < j <_ n.It follows that and e,(rp + + rnp.-p) 0 b,(rp + + r.p. p) 0 so the image of rp + + rnp, p in S, is (0, 0) mS,.Again invoking the previous lemma, we have rp -t-...r,p,, p mP, so p mP, as well. []One might wonder if we could weaken the hypotheses of our theorem slightly: If 6,+(C,+) C_ mC for all z, is the projective cover always equal to the minimal free resolution?As we shall see shortly in an example, the answer is no, but first we observe that while this slightly stronger conjecture is false, its converse is true.THEOREM 3. Let C be a complex of modules over a local ring R with maximal ideal m and boundary maps 6.If the projective cover of C is equal to the minimal free resolution of C then 6,+,(C,+t) c_ mC, for all i.
PROOF.Let c C+ and let P C be the projective cover of (3.Since the projective cover is surjective, there exists p Pi+ such that ei+(p) c.Since P (3 is also the minimal free resolution of (3, bi+(p) rnPi and e,(b,+(p)) rnC.Since e commutes with the boundary maps, we have ,+(c) mC,. [] At this point we have the following three statements which can be made about a complex C: 1. 6,+,(C,+,) _c mker(5,) 2. The projective cover of C is identical to the minimal free resolution of C 3. ,+l(Ci+l) C We have seen that (1) (2) (3) but in the following two examples, we will illustrate that statement (2)is equivalent to neither statement (]) nor statement (3).
First if we let R Z/4 with maximal ideal (2) and consider the complex w(, have the l)()]e('v(' ('()v('1 ald I11(, nnlal fie(' r(,s()lut()n of C l)oth equal () C tself where P C s the l(lCll[ltV nla l) "l'he ('()nl)h'x C does l()t, SatlSfV stat('nmnt (1) however so statements (1) and ( 2) ac n() e(lUlValent Next, if we le /t Z/S h naxmal (lcal (2) and let C which is it coml)lcx satisfying satement (3), we see that the projective cover P C is given by: (0, l) (0, 1) where the boundary real) b_o given by the matrix contains an entry not m the maximal ideal m.Thus P -L C s not the minimal free resolution of C and statements ( 2) and (3) are not equivalent.
In order to give necessary and sufficient conditions for the projective cover and minimal free resolution to be identical, we must define a new subcomplex D of C such that statement (2) holds if and only if 6,+ (C,+) c_ D, for all z.From our discussion up to this point, we certainly need mker(6,) C_ D, C_ for all i.With this in mind, given the complex C with projective cover P, we define the sub- module D of C, to be the set of elements c C, such that there exists elements r r in the maximal ideal m, elements c c C, and elements p, p ker(b,_) satisfying the following conditions: 1. c :rlC J-... + rkck 2. e,_(pa) 6,(%) for j k 3. rip + + rkpk 0 It is ey to see with D, defined in this manner, that D is a subcomplex of C and that each D, satisfies the subset inequality given above.
THEOREM 4. The projective cover P C of a complex C is equal to the minimal fr resolution of C if and only if 6,+(C+) D, for all k 0.
PROOF.First, let us assume that P C is the minimal fr resolution of C. In other words, if b is the boundary map of P then b,+(P,+) raP, for all z.