COMMUTATIVE ALGEBRA Contents - Columbia University

COMMUTATIVE ALGEBRA8(1) We say M is a finite R-module, or a finitely generated R-module if thereexist n N and x1 , . . . , xn M such that every element of M is a R-linearcombination of the xi . Equivalently, this means there exists a surjectionR n M for some n N.(2) We say M is a finitely presented R-module or an R-module of finite presentation if there exist integers n, m N and an exact sequenceR m R n M 0Informally, M is a finitely presented R-module if and only if it is finitely generatedand the module of relations among these generators is finitely generated as well. Achoice of an exact sequence as in the definition is called a presentation of M .07JXLemma 5.2. Let R be a ring. Let α : R n M and β : N M be modulemaps. If Im(α) Im(β), then there exists an R-module map γ : R n N suchthat α β γ.Proof. Let ei (0, . . . , 0, 1, 0, . . . , 0) be the ith basis vector of R n . Let xi Nbe an element with α(ei ) β(xi ) which exists by assumption. Set γ(a1 , . . . , an ) Pai xi . By construction α β γ. 0519Lemma 5.3. Let R be a ring. Let0 M1 M2 M 3 0be a short exact sequence of R-modules.(1) If M1 and M3 are finite R-modules, then M2 is a finite R-module.(2) If M1 and M3 are finitely presented R-modules, then M2 is a finitely presented R-module.(3) If M2 is a finite R-module, then M3 is a finite R-module.(4) If M2 is a finitely presented R-module and M1 is a finite R-module, thenM3 is a finitely presented R-module.(5) If M3 is a finitely presented R-module and M2 is a finite R-module, thenM1 is a finite R-module.Proof. Proof of (1). If x1 , . . . , xn are generators of M1 and y1 , . . . , ym M2 areelements whose images in M3 are generators of M3 , then x1 , . . . , xn , y1 , . . . , ymgenerate M2 .Part (3) is immediate from the definition.Proof of (5). Assume M3 is finitely presented and M2 finite. Choose a presentationR m R n M3 0By Lemma 5.2 there exists a map R n M2 such that the solid diagram0R m/ R n/ M3 / M1 / M2 / M3/0id/0commutes. This produces the dotted arrow. By the snake lemma (Lemma 4.1) wesee that we get an isomorphismCoker(R m M1 ) Coker(R n M2 )

COMMUTATIVE ALGEBRA9In particular we conclude that Coker(R m M1 ) is a finite R-module. SinceIm(R m M1 ) is finite by (3), we see that M1 is finite by part (1).Proof of (4). Assume M2 is finitely presented and M1 is finite. Choose a presentation R m R n M2 0. Choose a surjection R k M1 . ByLemma 5.2 there exists a factorization R k R n M2 of the compositionR k M1 M2 . Then R k m R n M3 0 is a presentation.Proof of (2). Assume that M1 and M3 are finitely presented. The argument in theproof of part (1) produces a commutative diagram0/ R n/ R n m/ R m/00 / M1 / M2 / M3/0with surjective vertical arrows. By the snake lemma we obtain a short exact sequence0 Ker(R n M1 ) Ker(R n m M2 ) Ker(R m M3 ) 0By part (5) we see that the outer two modules are finite. Hence the middle one isfinite too. By (4) we see that M2 is of finite presentation. 00KZLemma 5.4. Let R be a ring, and let M be a finite R-module. There exists afiltration by R-submodules0 M0 M1 . . . Mn Msuch that each quotient Mi /Mi 1 is isomorphic to R/Ii for some ideal Ii of R.Proof. By induction on the number of generators of M . Let x1 , . . . , xr M bea minimal number of generators. Let M 0 Rx1 M . Then M/M 0 has r 1generators and the induction hypothesis applies. And clearly M 0 R/I1 withI1 {f R f x1 0}. 0560Lemma 5.5. Let R S be a ring map. Let M be an S-module. If M is finite asan R-module, then M is finite as an S-module.Proof. In fact, any R-generating set of M is also an S-generating set of M , sincethe R-module structure is induced by the image of R in S. 6. Ring maps of finite type and of finite presentation00F200F3Definition 6.1. Let R S be a ring map.(1) We say R S is of finite type, or that S is a finite type R-algebra if thereexist an n N and an surjection of R-algebras R[x1 , . . . , xn ] S.(2) We say R S is of finite presentation if there exist integers n, m N andpolynomials f1 , . . . , fm R[x1 , . . . , xn ] and an isomorphism of R-algebrasR[x1 , . . . , xn ]/(f1 , . . . , fm ) S.Informally, R S is of finite presentation if and only if S is finitely generated asan R-algebra and the ideal of relations among the generators is finitely generated.A choice of a surjection R[x1 , . . . , xn ] S as in the definition is sometimes calleda presentation of S.

COMMUTATIVE ALGEBRA00F410Lemma 6.2. The notions finite type and finite presentation have the followingpermanence properties.(1) A composition of ring maps of finite type is of finite type.(2) A composition of ring maps of finite presentation is of finite presentation.(3) Given R S 0 S with R S of finite type, then S 0 S is of finite type.(4) Given R S 0 S, with R S of finite presentation, and R S 0 offinite type, then S 0 S is of finite presentation.Proof. We only prove the last assertion. Write S R[x1 , . . . , xn ]/(f1 , . . . , fm ) andS 0 R[y1 , . . . , ya ]/I. Say that the class ȳi of yi maps to hi mod (f1 , . . . , fm ) in S.Then it is clear that S S 0 [x1 , . . . , xn ]/(f1 , . . . , fm , h1 ȳ1 , . . . , ha ȳa ). 00R2Lemma 6.3. Let R S be a ring map of finite presentation. For any surjectionα : R[x1 , . . . , xn ] S the kernel of α is a finitely generated ideal in R[x1 , . . . , xn ].Proof. Write S R[y1 , . . . , ym ]/(f1 , . . . , fk ). Choose gi R[y1 , . . . , ym ] whichare lifts of α(xi ). Then we see that S R[xi , yj ]/(fl , xi gi ). Choose hj R[x1 , . . . , xn ] such that α(hj ) corresponds to yj mod (f1 , . . . , fk ). Consider themap ψ : R[xi , yj ] R[xi ], xi 7 xi , yj 7 hj . Then the kernel of α is the image of(fl , xi gi ) under ψ and we win. 0561Lemma 6.4. Let R S be a ring map. Let M be an S-module. Assume R Sis of finite type and M is finitely presented as an R-module. Then M is finitelypresented as an S-module.Proof. This is similar to the proof of part (4) of Lemma 6.2. We may assume S R[x1 , . . . , xn ]/J. PChoose y1 , . . . , ym M which generate M as an R-module andchoose relationsaij yj 0, i 1, . . . , t which generate the kernel of R m M .For any i 1, . . . , n and j 1, . . . , m writeXxi yj aijk ykfor some aijk P R. Consider the S-module N generatedP by y1 , . . . , ym subject tothe relationsaij yj 0, i 1, . . . , t and xi yj aijk yk , i 1, . . . , n andj 1, . . . , m. Then N has a presentationS nm t S m N 0By construction there is a surjectivemap ϕ : N M . To finish the proof we showPϕ is injective. Suppose z bj yj N for some bj S. We may think of bjas a polynomial Pin x1 , . . . , xn with coefficients in R. By applying the relations ofthe form xi yj aijkPyk we can inductively lower the degree of the polynomials.Hence we see that z cj yj for some cj R. Hence if ϕ(z) 0 then the vector(c1 , . . . , cm ) is an R-linear combination of the vectors (ai1 , . . . , aim ) and we concludethat z 0 as desired. 7. Finite ring maps0562Here is the definition.0563Definition 7.1. Let ϕ : R S be a ring map. We say ϕ : R S is finite if S isfinite as an R-module.00GJLemma 7.2. Let R S be a finite ring map. Let M be an S-module. Then Mis finite as an R-module if and only if M is finite as an S-module.

COMMUTATIVE ALGEBRA11Proof. One of the implications follows from Lemma 5.5. To see the other assumethat M is finite as an S-module. Pick x1 , . . . , xn S which generate S as anR-module. Pick y1 , . . . , ym M which generate M as an S-module. Then xi yjgenerate M as an R-module. 00GLLemma 7.3. Suppose that R S and S T are finite ring maps. Then R Tis finite.Proof. If ti generate T as an S-module and sj generate S as an R-module, thenti sj generate T as an R-module. (Also follows from Lemma 7.2.) 0D46Lemma 7.4. Let ϕ : R S be a ring map.(1) If ϕ is finite, then ϕ is of finite type.(2) If S is of finite presentation as an R-module, then ϕ is of finite presentation.Proof. For (1) if x1 , . . . , xn S generate S as anthen x1 , . . . , xn genPR-module,erate S as an R-algebra. For (2), suppose thatrji xi 0, j 1, . . . , m is a setof generators of the relations Pamong the xi when viewed as R-moduleP k generatorsof S. Furthermore, write 1 ri xi for some ri R and xi xj rij xk for somekrij R. ThenXXXkS R[t1 , . . . , tn ]/(rji ti , 1 ri ti , ti tj rijtk )as an R-algebra which proves (2). For more information on finite ring maps, please see Section 36.8. Colimits07N7Some of the material in this section overlaps with the general discussion on colimits in Categories, Sections 14 – 21. The notion of a preordered set is defined inCategories, Definition 21.1. It is a slightly weaker notion than a partially orderedset.00D4Definition 8.1. Let (I, ) be a preordered set. A system (Mi , µij ) of R-modulesover I consists of a family of R-modules {Mi }i I indexed by I and a family ofR-module maps {µij : Mi Mj }i j such that for all i j kµii idMiµik µjk µijWe say (Mi , µij ) is a directed system if I is a directed set.This is the same as the notion defined in Categories, Definition 21.2 and Section21. We refer to Categories, Definition 14.2 for the definition of a colimit of adiagram/system in any category.00D5Lemma 8.2. Let (Mi , µij ) be a system of R-modules over theLpreordered set I.The colimit of the system (Mi , µij ) is the quotient R-module ( i I Mi )/Q whereQ is the R-submodule generated by all elementsιi (xi ) ιj (µij (xi ))Lwhere ιi : Mi i is the natural inclusion. We denote the colimit M i I MLcolimi Mi . We denote π : i I Mi M the projection map and φi π ιi : Mi M.

COMMUTATIVE ALGEBRA12Proof. This lemma is a special case of Categories, Lemma 14.12 but we will alsoprove it directly in this case. Namely, note that φi φj µij in the above construction. To show the pair (M, φi ) is the colimit we have to show it satisfies theuniversal property: for any other such pair (Y, ψi ) with ψi : Mi Y , ψi ψj µij ,there is a unique R-module homomorphism g : M Y such that the followingdiagram commutes:µij/ MjMiφjφiψiM}ψjg YAnd this is clear becausewecandefineg by taking the map ψi on the summandLMi in the direct sumMi . 00D6Lemma 8.3. Let (Mi , µij ) be a system of R-modules over the preordered set I.Assume that I is directed. The colimit of the system (Mi , µij ) is canonically isomorphic to the module M defined as follows:(1) as a set let a M Mi / i Iwhere for m Mi and m0 Mi0 we havem m0 µij (m) µi0 j (m0 ) for some j i, i0(2) as an abelian group for m Mi and m0 Mi0 we define the sum of theclasses of m and m0 in M to be the class of µij (m) µi0 j (m0 ) where j Iis any index with i j and i0 j, and(3) as an R-module define for m Mi and x R the product of x and theclass of m in M to be the class of xm in M .The canonical maps φi : Mi M are induced by the canonical maps Mi i I Mi .Proof. Omitted. Compare with Categories, Section 19.00D7 Lemma 8.4. Let (Mi , µij ) be a directed system. Let M colim Mi with µi :Mi M . Then, µi (xi ) 0 for xi Mi if and only if there exists j i such thatµij (xi ) 0.Proof. This is clear from the description of the directed colimit in Lemma 8.3. 00D8Example 8.5. Consider the partially ordered set I {a, b, c} with a b and a cand no other strict inequalities. A system (Ma , Mb , Mc , µab , µac ) over I consists ofthree R-modules Ma , Mb , Mc and two R-module homomorphisms µab : Ma Mband µac : Ma Mc . The colimit of the system is justM : colimi I Mi Coker(Ma Mb Mc )where the map is µab µac . Thus the kernel of the canonical map Ma M isKer(µab ) Ker(µac ). And the kernel of the canonical map Mb M is the imageof Ker(µac ) under the map µab . Hence clearly the result of Lemma 8.4 is false forgeneral systems.

COMMUTATIVE ALGEBRA00D913Definition 8.6. Let (Mi , µij ), (Ni , νij ) be systems of R-modules over the samepreordered set I. A homomorphism of systems Φ from (Mi , µij ) to (Ni , νij ) is bydefinition a family of R-module homomorphisms φi : Mi Ni such that φj µij νij φi for all i j.This is the same notion as a transformation of functors between the associateddiagrams M : I ModR and N : I ModR , in the language of categories. Thefollowing lemma is a special case of Categories, Lemma 14.8.00DALemma 8.7. Let (Mi , µij ), (Ni , νij ) be systems of R-modules over the same preordered set. A morphism of systems Φ (φi ) from (Mi , µij ) to (Ni , νij ) induces aunique homomorphismcolim φi : colim Mi colim Nisuch that/ colim MiMiφicolim φi / colim Ni Nicommutes for all i I.Proof. Write M colim Mi and N colim Ni and φ colim φi (as yet to beconstructed). We will use the explicit description of M and N in Lemma 8.2without further mention. The condition of the lemma is equivalent to the conditionthatL/Mi I MiLφiL i Iφ /NNicommutes. Hence it is clear that if φ exists, then it is unique. To see that φ exists,Lit suffices to show that the kernel of the upper horizontal arrow is mapped byφito the kernel of the lower horizontal arrow. To see this, let j k and xj Mj .ThenM(φi )(xj µjk (xj )) φj (xj ) φk (µjk (xj )) φj (xj ) νjk (φj (xj ))which is in the kernel of the lower horizontal arrow as required.00DB Lemma 8.8. Let I be a directed set. Let (Li , λij ), (Mi , µij ), and (Ni , νij ) besystems of R-modules over I. Let ϕi : Li Mi and ψi : Mi Ni be morphismsof systems over I. Assume that for all i I the sequence of R-modulesLiϕi/ Mi/ Niψiis a complex with homology Hi . Then the R-modules Hi form a system over I, thesequence of R-modulescolimi Liϕ/ colimi Miψ/ colimi Niis a complex as well, and denoting H its homology we haveH colimi Hi .

COMMUTATIVE ALGEBRA14/ colimi Ni is a complex./ colimi MiProof. It is clear that colimi LiFor each i I, there is a canonical R-module morphism Hi H (sending each[m] Hi Ker(ψi )/ Im(ϕi ) to the residue class in H Ker(ψ)/ Im(ϕ) of the imageof m in colimi Mi ). These give rise to a morphism colimi Hi H. It remains toshow that this morphism is surjective and injective.ψϕWe are going to repeatedly use the description of colimits over I as in Lemma 8.3without further mention. Let h H. Since H Ker(ψ)/ Im(ϕ) we see that h isthe class mod Im(ϕ) of an element [m] in Ker(ψ) colimi Mi . Choose an i suchthat [m] comes from an element m Mi . Choose a j i such that νij (ψi (m)) 0which is possible since [m] Ker(ψ). After replacing i by j and m by µij (m) wesee that we may assume m Ker(ψi ). This shows that the map colimi Hi H issurjective.Suppose that hi Hi has image zero on H. Since Hi Ker(ψi )/ Im(ϕi ) we mayrepresent hi by an element m Ker(ψi ) Mi . The assumption on the vanishing ofhi in H means that the class of m in colimi Mi lies in the image of ϕ. Hence thereexists a j i and an l Lj such that ϕj (l) µij (m). Clearly this shows that theimage of hi in Hj is zero. This proves the injectivity of colimi Hi H. 00DCExample 8.9. Taking colimits is not exact in general. Consider the partiallyordered set I {a, b, c} with a b and a c and no other strict inequalities, as inExample 8.5. Consider the map of systems (0, Z, Z, 0, 0) (Z, Z, Z, 1, 1). From thedescription of the colimit in Example 8.5 we see that the associated map of colimitsis not injective, even though the map of systems is injective on each object. Hencethe result of Lemma 8.8 is false for general systems.04B0Lemma 8.10. Let I be an index category satisfying the assumptions of Categories,Lemma 19.8. Then taking colimits of diagrams of abelian groups over I is exact(i.e., the analogue of Lemma 8.8 holds in this situation). Proof. By Categories, Lemma 19.8 we may write I j J Ij with each Ij afiltered category, and J possibly empty. By Categories, Lemma 21.5 taking colimitsover the index categories Ij is the same as taking the colimit over some directed set.Hence Lemma 8.8 applies to these colimits. This reduces the problem to showingthat coproducts in the category of R-modules over the set J are exact. In otherwords, exact sequences Lj Mj Nj of R modules we have to show thatMMMLj Mj Njj Jj Jj Jis exact. This can be verified by hand, and holds even if J is empty. 9. Localization00CM00CNDefinition 9.1. Let R be a ring, S a subset of R. We say S is a multiplicativesubset of R if 1 S and S is closed under multiplication, i.e., s, s0 S ss0 S.Given a ring A and a multiplicative subset S, we define a relation on A S asfollows:(x, s) (y, t) u S such that (xt ys)u 0

COMMUTATIVE ALGEBRA15It is easily checked that this is an equivalence relation. Let x/s (or xs ) be theequivalence class of (x, s) and S 1 A be the set of all equivalence classes. Defineaddition and multiplication in S 1 A as follows:x/s y/t (xt ys)/st,x/s · y/t xy/stOne can check that S 1 A becomes a ring under these operations.00CODefinition 9.2. This ring is called the localization of A with respect to S.We have a natural ring map from A to its localization S 1 A,A S 1 A,x 7 x/1which is sometimes called the localization map. In general the localization map isnot injective, unless S contains no zerodivisors. For, if x/1 0, then there is au S such

00AP Basic commutative algebra will be explained in this document. A reference is [Mat70]. 2. Conventions 00AQ A ring is commutative with 1. The zero ring is a ring. In fact it is the only ring thatdoesnothaveaprimeideal. TheKroneckersymbolδ ijwillbeused. IfR S isaring