Realization of graded monomial ideal rings modulo torsion
aa r X i v : . [ m a t h . A T ] A ug REALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULOTORSION
TSELEUNG SO AND DONALD STANLEY
Abstract.
Let A be the quotient of a graded polynomial ring Z [ x , · · · , x m ] ⊗ Λ[ y , · · · , y n ]by an ideal generated by monomials with leading coefficients 1. Then we constructed aspace X A such that A is isomorphic to H ∗ ( X A ) modulo torsion elements. Introduction
A classical problem in algebraic topology asks: which commutative graded R -algebra A areisomorphic to H ∗ ( X A ; R ) for some space X A ? The space X A , if exists, is called a realizationof A . According to Aguad´e [1] the problem goes back to at least Hopf, and was later explicitlystated by Steenrod [14]. To solve the problem in general is probably too ambitious, but manyspecial cases have been proven.One of Quillen’s motivations for his seminal work on rational homotopy theory [13] wasto solve this problem over Q . He showed that all simply connected graded Q -algebras havea realization. The problem of which polynomial algebras over Z have realizations has a longhistory and a complete solution was given by Anderson and Grodal [2], (see also [12]). Morerecently Trevisan [15] and later Bahri-Bendersky-Cohen-Gitler [4] constructed realizationsof Z [ x , · · · , x m ] /I , where | x i | = 2 and I is an ideal generated by monomials with leadingcoefficient 1.We want to consider a related problem that lies between the solved realization problemover Q and the very difficult realization problem over Z . We do this by modding out torsion. Problem 1.1.
Which commutative graded R -algebra A are isomorphic to H ∗ ( X A ; R ) / torsionfor some space X A ? Such an X A is called a realization modulo torsion of A . For example, a polynomial ring Z [ x ]has a realization modulo torsion given by Eilenberg-MacLane space K ( Z , | x | ) if | x | is even,while Z [ x ] has a realization if and only if | x | = 2 or 4 [14]. Here we ask: do all finite typeconnected commutative graded Z-algebra have a realization modulo torsion?Notice that modding out by torsion is different from taking rational coefficients. Forexample, both H ∗ (Ω S n +1 ; Q ) and H ∗ ( K ( Z , n ); Q ) are Q [ x ] generated by x of degree 2 n .But H ∗ ( K ( Z , n )) / torsion is Z [ x ], while H ∗ (Ω S n +1 ) ∼ = Γ[ x ] is free and is the divided poly-nomial algebra generated by x .In this paper, we construct realizations modulo torsion of graded monomial ideal rings A which are of the following form. Let P = Z [ x , · · · , x m ] ⊗ Λ[ y , · · · , y n ] be a graded poly-nomial ring where x i ’s have arbitrary positive even degrees and y j ’s have arbitrary positiveodd degrees, and let I = ( M , · · · , M r ) be an ideal generated by r monomials M j = x a j x a j · · · x a mj m ⊗ y b j · · · y b nj n , ≤ j ≤ r, Mathematics Subject Classification.
Primary 55N10 Secondary 13F55, 55P99, 55T20.
Key words and phrases. cohomology realization problem, polyhedral product. where indices a ij are non-negative integers and b ij are either 0 or 1. Then the quotientalgebra A = P/I is called a graded monomial ideal ring . Theorem 1.2 (Main Theorem) . Let A be a graded monomial ideal ring. Then there exists aspace X A such that H ∗ ( X A ) /T is isomorphic to A , where T is the ideal consisting of torsionelements in H ∗ ( X A ) . Moreover, there is a ring morphism A → H ∗ ( X A ) that is right inverseto the quotient map H ∗ ( X A ) → H ∗ ( X A ) /T ∼ = A . If all of the even degree generators are in degree 2, then we do not need to mod out bytorsion and so we get a generalization (Proposition 4.5) of the results of Bahri-Bendersky-Cohen-Gitler [4, Theorem 2.2] and Trevisan [15, Theorem 3.6].The structure of the paper is as follows. Section 2 contains preliminaries, algebraictools and lemmas that are used in later sections. In Section 3 we recall the definitionof polyhedral products and modify a result of Bahri-Bendersky-Cohen-Gitler [3] to com-pute H ∗ (( X, ∗ ) K ) /T . In Section 4 we prove the Main Theorem in several steps. First, weconsider the special case of A = P/I where I is square-free and construct a polyhedral prod-uct X A as a realization modulo torsion of A . Next, for the general case, we polarize A to getanothers graded monomial ideal ring A ′ = P ′ /I ′ , where I ′ is square-free, and construct X A from X A ′ via a fibration sequence. Then we show H ∗ ( X A ) /T ∼ = A using Eilenberg-Moorespectral sequence. In Section 5 we demonstrate how to construct X A for an easy exampleof A . 2. Preliminaries
Quotients of algebras by torsion elements.
It is natural to study an algebra A byfactoring out the torsion elements since the quotient algebra is torsion-free and has a simplerstructure. Driven by this, we start investigating the quotients of cohomology rings of spacesby their torsion elements. Since we cannot find related reference in topological literature,here we fix the notations and develop lemmas for our purpose.A graded module A = { A i } i ∈ S is a family of indexed modules A i . Since we are interestedin cochain complexes and cohomology rings of connected, finite type CW-complexes, weassume A to be a connected, finite type graded module with non-positive degrees. Thatis S = N ≤ , A = Z and each component A i is finitely generated. We follow the conventionand denote A i by A − i . Remark 2.1.
Algebraists define a graded module to be a module with a grading structure,that is the direct sum A = L i ∈ S A i of a family of indexed modules. This definition is slightlydifferent from the definition aforementioned. In fact, we will shift to algebraists’ definitionlatter on as we need some algebraic results for our arguments. An element x ∈ A is torsion if cx = 0 for some non-zero integer c , and is torsion-free otherwise. The torsion submodule A t of A is the graded submodule consisting of torsionelements and the torsion-free quotient module A f = A/A t is their quotient. If B is anothergraded module and g : A → B is a morphism, then it induces a morphism g f : A f → B f sending a + A t ∈ A f to g ( a ) + B t ∈ B f . This kind of structure is important in abeliancategories and was formalized with Dixon’s notion of a torsion theory [6], but in this paperwe only use the structure in a naive way. EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 3
Lemma 2.2. If → A g → B h → C → is a short exact sequence of graded modules, then wehave C f ∼ = ( B f /A f ) f . Furthermore, if the sequence is split exact, then so is −→ A f g f −→ B f h f −→ C f −→ . Proof.
Consider a commutative diagram0 (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) (cid:15) / / A t (cid:15) (cid:15) g t / / B t (cid:15) (cid:15) p / / B t /A tu (cid:15) (cid:15) / / / / A (cid:15) (cid:15) g / / B (cid:15) (cid:15) h / / C v (cid:15) (cid:15) / / / / A f (cid:15) (cid:15) g f / / B f (cid:15) (cid:15) q / / B f /A f (cid:15) (cid:15) / /
00 0 0where g t is the restriction of g to A t , p and q are the quotient maps, and u and v are theinduced maps. By construction all rows and columns are exact sequences except for theright column. A diagram chase implies that u is injective and v is surjective. We claimthat the column is exact at C . Obviously v ◦ u is trivial. Take an element c ∈ ker ( v ) andits preimage b ∈ B . A diagram chase implies b = g ( a ) + b ′ for some a ∈ A and b ′ ∈ B t .So c = h ( b ′ ) = u ◦ p ( b ′ ) is in Im ( u ) and the right column 0 → B t /A t u → C v → B f /A f → v f : C f → ( B f /A f ) f is an isomorphism.Since v is surjective, so is v f . Take c ′ ∈ ker ( v f ) and its preimage ˜ c ′ ∈ C . Then v (˜ c ′ ) is atorsion element in B f /A f and mv (˜ c ′ ) = 0 for some non-zeor integer m . So m ˜ c ′ ∈ ker ( v ).As ker ( v ) = Im ( u ) consists of torsion elements, m ˜ c ′ is a torsion and so is ˜ c ′ . Therefore c ′ = 0in C f and v f is injective.Notice that an exact sequence being split is equivalent to B ∼ = A ⊕ C . So B f ∼ = A f ⊕ C f and 0 → A f g f → B f h f → C f → (cid:3) A graded algebra A is a graded module equipped with an associative bilinear multiplication m = { m i,j : A i ⊗ A j → A i + j } such that 1 ∈ A is the multiplicative identity. Since A t is an ideal, A f is a quotient algebra.Before describing the multiplication on A f , we check that taking tensor products and takingquotients by torsion elements are compatible. Lemma 2.3. If A is a graded module (not necessarily of finite type), then there is an iso-morphism ( A ⊗ A ) f → A f ⊗ A f . TSELEUNG SO AND DONALD STANLEY
Proof.
It suffices to show that ( A i ⊗ A j ) f ∼ = A if ⊗ A jf for any positive integers i and j .Consider commutative diagram0 / / A it ⊗ A j ⊕ A i ⊗ A jt ı / / a (cid:15) (cid:15) A i ⊗ A j π / / A if ⊗ A jfb (cid:15) (cid:15) / / / / ( A i ⊗ A j ) t ı / / A i ⊗ A j π / / ( A i ⊗ A j ) f / / a, ı and ı are inclusions, π and π are quotient maps and b is the induced map.We want to show that b is an isomorphism, which is equivalent to showing that a is anisomorphism. If A is of finite type, then a is an isomorphism since A i and A j are finitelygenerated abelian groups. In the general case, a is an isomorphism by [9, Theorem 61.5]. (cid:3) By Lemma 2.3 the multiplication on A f is the induced morphism m f : A f ⊗ A f ∼ = ( A ⊗ A ) f −→ A f . Let π : A → A f be the quotient map. Then there is a commutative diagram A ⊗ A m / / π ⊗ π (cid:15) (cid:15) A π (cid:15) (cid:15) A f ⊗ A f m f / / A f A differential graded algebra (
A, d ) is a couple consisting of a graded algebra A and adifferential d = { d i : A i → A i +1 } satisfying d ( ab ) = d ( a ) b + ( − | a | d ( b ). Let d t be therestriction of d to A t and d f = { d if : A if → A i +1 f } be the induced differential on A f .Then ( A t , d t ) is a cochain complex and ( A f , d f ) is a differential graded algebra.Given ( A, d ), the union of their cohomology groups { H i ( A, d ) } i ≥ forms a graded algebra.Cohomology ring H ∗ ( A, d ) = L i ≥ H i ( A, d ) is their direct sum and is a ring with a gradingstructure. Denote its torsion-free quotient ( H ∗ ( A, d )) f by H ∗ f ( A, d ). The following lemmacompares H ∗ f ( A, d ) and H ∗ f ( A f , d f ). Lemma 2.4.
Let ( A, d ) be a differential graded algebra. Then there is a ring monomorphism π ′ : H ∗ f ( A, d ) −→ H ∗ f ( A f , d f ) . Moreover, if H i +1 ( A t , d t ) = 0 , then π ′ : H if ( A, d ) → H if ( A f , d f ) is an isomorphism.Proof. Let ı : ( A t , d t ) → ( A, d ) be the inclusion. Then the short exact sequence of cochaincomplexes 0 → ( A t , d t ) ı → ( A, d ) π → ( A f , d f ) → · · · → H i − ( A f , d f ) → H i ( A t , d t ) ı ∗ → H i ( A, d ) π ∗ → H i ( A f , d f ) → H i +1 ( A t , d t ) → · · · Take π ′ : H ∗ f ( A, d ) → H ∗ f ( A f , d f ) to be the morphism induced by π ∗ : H ∗ ( A, d ) → H ∗ ( A f , d f ).We show that it has the asserted properties.To show the injectivity of π ′ , take a cocycle class [ a ] ∈ ker ( π ′ ) and represent it by atorsion-free cocycle class [˜ a ] ∈ H i ( A, d ). Then π ∗ [˜ a ] is torsion and π ∗ ( c [˜ a ]) = 0 for somenon-zero number c . Since H i ( A t , d t ) is torsion, so is Im ( ı ∗ ) = ker ( π ∗ ) and c [˜ a ] is a torsion.Therefore [˜ a ] is a torsion. By definition, ˜ a is zero. So π ′ is injective. EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 5
Suppose A i +1 has no torsion elements. Then A i +1 t = 0 and H i +1 ( A t , d t ) = 0. So π ∗ issurjective. By definition we have commutative diagram H i ( A, d ) π ∗ / / (cid:15) (cid:15) H i ( A f , d f ) (cid:15) (cid:15) H if ( A, d ) π ′ / / H if ( A f , d f ) , where vertical arrows are quotient maps and are surjective. So π ′ : H if ( A, d ) → H if ( A f , d f )is surjective and hence isomorphic. (cid:3) Example:
The surjectivity of π ′ : H if ( A, d ) → H if ( A f , d f ) may fail if A i +1 contains torsionelements. Let ( A, d ) be a cochain complex where A i = Z i = 0 Z / Z i = 10 otherwise , and d i are trivial for all i except for d : Z → Z / Z being the quotient map. Then H ( A )and H ( A f ) are Z while π ′ : H ( A ) → H f ( A ) is multiplication 2 : Z → Z .2.2. Eilenberg-Moore spectral sequence.
Let F → E → B be a fibration sequencewhere all spaces are connected CW-complexes and B is simply connected. The E page ofEilenberg-Moore spectral sequence is given by E − p, ∗ = Tor − pH ∗ ( B ) ( Z , H ∗ ( E ))which strongly converges to H ∗ ( F ) ∼ = Tor C ∗ ( B ) ( Z , C ∗ ( E )). In our case, H ∗ ( E ) and H ∗ ( B )are hard to describe, while H ∗ f ( E ) and H ∗ f ( B ) are much simpler. So we want to study E − p, ∗ by Tor − pH ∗ f ( B ) ( Z , H ∗ f ( E )).Recall the bar construction. Given ( A, d ) a differential graded algebra, let (
M, d M )be a left A -module and let ( N, d N ) be a right A -modules. For any positive integer i ,let B − i ( M, A, N ) = M ⊗ ( ¯ A ) ⊗ i ⊗ N and let B − i,j ( M, A, N ) be the degree j componentin B − i ( M, A, N ). Here we follow the notation in [11] and denote an element in B − i ( M, A, N )by x [ a | · · · | a i ] y for x ∈ M , a i ∈ ¯ A and y ∈ N . The internal and external differentials d I : B − i,j ( M, A, N ) → B − i,j +1 ( M, A, N ) and d E : B − i,j ( M, A, N ) → B − i +1 ,j ( M, A, N )are given by d I ( x [ a | · · · | a i ] y ) = ( d M x )[ a | · · · | a i ] y + i X j =1 ( − ǫ j − x [ a | · · · | a j − | d A a j | a j +1 | · · · | a i ] y +( − ǫ i x [ a | · · · | a i ]( d N y ) ,d E ( x [ a | · · · | a i ] y ) = ( − | x | ( xa )[ a | · · · | a i ] y + i − X j =1 ( − ǫ j x [ a | · · · | a j − | a j · a j +1 | · · · | a i ] y +( − ǫ i − x [ a | · · · | a i − ]( a i y ) , TSELEUNG SO AND DONALD STANLEY where ǫ k = k + | x | + P kj =1 | a j | . The total complex (Tot(B( M, A, N )) , d Tot ) is a differentialgraded module where Tot(B(
M, A, N )) n = M − i + j = n B − i,j ( M, A, N ) and d Tot = M − i + j = n ( d I + d E )for n ≥ Lemma 2.5. If A be simply connected, then there is a spectral sequence { E ∗ , ∗ r } that stronglyconverges to H ( Tot ( B ( M, A, N ))) , and there is a monomorphism of modules π ′ : ( E − p,q ) f −→ (cid:16) Tor − p,qH f ( A ) ( H f ( M ) , H f ( N )) (cid:17) f . Moreover, π ′ is isomorphic for p = 0 .Proof. Take the filtration F − p = L ≤ k ≤ p B − k ( M, A, N ). Since A is simply connected, theinduced spectral sequence strongly converges to H (Tot(B( M, A, N ))). The E -page is givenby E − p, ∗ = F − p / F − p +1 = M ⊗ ( ¯ A ⊗ p ) ⊗ N and d = d I . By K¨unneth Theorem the E -page is given by E − p, ∗ ∼ = H ( M ) ⊗ ( ˜ H ( A ) ⊗ p ) ⊗ H ( N ) ⊕ T where T is a torsion term, and d is induced by d E . Denote H ( M ), H ( A ) and H ( N ) by M ′ , A ′ and N ′ for short. In E − p, ∗ , the summand M ′ ⊗ ( ¯ A ′ ) ⊗ p ⊗ N ′ is B − p ( M ′ , A ′ , N ′ ). By Lemma 2.3,(B − p ( M ′ , A ′ , N ′ )) f ∼ = B − p ( M ′ f , A ′ f , N ′ f ) . Since inclusion B ∗ ( M ′ , A ′ , N ′ ) → E ∗ , ∗ is multiplicative, the induced differential ( d ) f isthe external differential of B ∗ ( M ′ f , A ′ f , N ′ f ). Denote ( d ) f by d ′ . By Lemma 2.4 there is amonomorphism π ′ : ( E − p,q ) f −→ H − pf (B ∗ ,q ( M ′ f , A ′ f , N ′ f ) , d ′ ).Notice that B ∗ ( M ′ f , A ′ f , N ′ f ) ∼ = M ′ f ⊗ A ′ f B ∗ ( A ′ f , A ′ f , N ′ f ) and d ′ = ⊗ A ′ f d ′′ , where d ′′ is theexternal differential of B ∗ ( A ′ f , A ′ f , N ′ f ). Since M ′ , A ′ and N ′ have trivial differentials and · · · −→ B − ( A ′ f , A ′ f , N ′ f ) d ′′ −→ B ( A ′ f , A ′ f , N ′ f ) −→ N ′ f −→ N ′ f over A ′ f , the monomorphism becomes π ′ : ( E − p,q ) f −→ (cid:16) Tor − p,qA ′ f ( M ′ f , N ′ f ) (cid:17) f . Since B ( M ′ , A ′ , N ′ ) = 0, π ′ is isomorphic for p = 0 by Lemma 2.4. (cid:3) Apply Lemma 2.5 to calculate H ∗ ( F ). Lemma 2.6.
Let F → E → B be a fibration sequence such that E and B are simply connectedspaces, and let { E − p,q } be the E -page of Eilenberg-Moore spectral sequence. Then there isa monomorphism π ′ : ( E − p,q ) f → (cid:16) Tor − p,qH ∗ f ( B ) ( Z , H ∗ f ( E )) (cid:17) f as modules such that π ′ is anisomorphism for p = 0 .Proof. Since E is simply connected, Universal Coefficient Theorem implies that H ( E ) isfree. By [7, Proposition 2.8] C ∗ ( E ) is quasi-isomorphic to a finite type differential gradedalgebra. Since the inclusion C ∗ ( E ) → ( C ∗ ( E )) ∨ are quasi-isomorphisms of differential gradedcoalgebras, we may assume C ∗ ( E ) to be a finite type differential graded coalgebra. Simi-larly we assume C ∗ ( B ) to be of finite type. The dual statement of [7, Theorem II] impliesthat C ∗ ( F ) is quasi-isomorphic to the total complex Tot(B( C ∗ ( E ) , C ∗ ( B ) , Z )) as modules of EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 7 differential graded coalgebras. Take M = C ∗ ( E ) , A = C ∗ ( B ) and N = Z and use Lemma 2.5to obtain the morphism π ′ . (cid:3) Regular sequences and freeness.
From now on we adopt algebraists’ definitionsof graded algebras and graded modules. A commutative graded algebra A = L i ≥ A i isan algebra with a grading structure such that ab = ( − ij ba for a ∈ A i and b ∈ A j ,and a graded A -module M = L j ≥ M j is the direct sum of a family of A -modules. Aset { r , · · · , r n } of elements in M is called an M -regular sequence if ideal ( r , · · · , r n ) M isnot equal to M and the multiplication r i : M/ ( r , · · · , r i − ) M → M/ ( r , · · · , r i − ) M is injective for 1 ≤ i ≤ n . In the special case where M is a K [ x , · · · , x n ]-module forsome field K and the grading of M has a lower bound, M is a free K [ x , · · · , x n ]-moduleif { x i } ni =1 is a regular sequence in M . We want to extend this fact to the case where M isa Z [ x , · · · , x n ]-module. Recall a corollary of the graded Nakayama Lemma. Lemma 2.7.
Let A be a graded ring and let M be an A -module. Suppose A and M arenon-negatively graded, and I = ( r , · · · , r n ) ⊂ A is an ideal generated by homogeneouselements r i of positive degrees. If { m α } α ∈ S is a set of homogeneous elements in M whoseimages generate M/IM , then { m α } α ∈ S generates M . Lemma 2.8.
Let M be a Z [ x , · · · , x n ] -module with non-negative degrees. If M/ ( x , · · · , x n ) M is a free Z -module and { x , · · · , x n } is a regular sequence on M , then M is a free Z [ x , · · · , x n ] -module.Proof. Denote I = ( x , · · · , x n ). By assumption there is a set { m α } α ∈ S of homogeneous ele-ments in M such that their quotient images form a basis in M/IM . By Lemma 2.7 { m α } α ∈ S generates M . We need to show that { m α } α ∈ S is linear independent over Z [ x , · · · , x n ].For 0 ≤ i ≤ n , let M i = M/ ( x , · · · , x n − i ) M , A i = Z [ x n +1 − i , · · · , x n ] and m α,i be thequotient image of m α in M i . Prove that M i is a free A i -module with a basis { m α,i } α ∈ S byinduction on i . For i = 0, M = M/IM and A = Z . The statement is true since { m α, } α ∈ S is a basis by construction. Assume the statement holds for i ≤ k . For i = k + 1, if there isa collection { f α } α ∈ S of polynomials satisfying(1) X α ∈ S f α · m α,k +1 = 0 , we show that all f α ’s are zero.If not, then there are finitely many non-zero polynomials f j , · · · , f j r . Quotient M k +1 and A k +1 by ideal ( x n − k ) and let ¯ f j i be the image of f j i in A k . Then (1) becomes r X i =1 ¯ f j i · m j i ,k = 0 . By inductive assumption, { m α,k } is a basis in M k . So ¯ f j i = 0 and f j i = x n − k g j i for somepolynomial g j i ∈ A k +1 . Since x n − k is not a zero-divisor, putting f j i = x n − k g j i to (1) and get r X i =1 g j i · m j i ,k +1 = 0 . TSELEUNG SO AND DONALD STANLEY So g j , · · · , g j r are non-zero polynomials satisfying (1) and | g j i | < | f j i | for 1 ≤ i ≤ r . Iteratingthis argument implies that | f j i | ’s are arbitrarily large, but this is impossible. So f j i ’s mustbe zero and { m α,k +1 } is linear independent. It follows that M k +1 is a free A k +1 -module. (cid:3) Cohomology rings of polyhedral products
Let [ m ] = { , · · · , m } , let K be a simplicial complex on [ m ] and let ( X, A ) = { ( X i , A i ) } mi =1 be a sequence of pairs of relative CW-complexes. For any simplex σ ∈ K define( X, A ) σ = ( ( x , · · · , x m ) ∈ m Y i =1 X i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x i ∈ A i for i / ∈ σ ) as a subspace of Q mi =1 X i , and define the polyhedral product ( X, A ) K = [ σ ∈ K ( X, ∗ ) σ to be the colimit of ( X, A ) σ over K .If X i = CP ∞ and A i = ∗ for all i , then ( CP ∞ , ∗ ) K is homotopy equivalent to Davis-Januszkiewicz space [5, Theorem 4.3.2]. For any principal ideal domain R , H ∗ (( CP ∞ , ∗ ) K ; R )is isomorphic to the Stanley-Reisner ring R [ x , · · · , x m ] /I K . Here I K is the ideal generatedby x j · · · x j k for x j i ∈ ˜ H ∗ ( X j i ; R ) and { j , · · · , j k } / ∈ K , and is called the Stanley-Reisnerideal of K . In general, a similar formula holds for H ∗ (( X, ∗ ) K ) whenever X i ’s are any spaceswith free cohomology. Theorem 3.1 ([3]) . Let R be a principal ideal domain, let K be a simplicial complex on [ m ] and let X = { X i } mi =1 be a sequence of CW-complexes. If H ∗ ( X i ; R ) is a free R -module forall i , then H ∗ (( X, ∗ ) K ; R ) ∼ = m O i =1 H ∗ ( X i ; R ) /I K , where I K is generated by x j ⊗ · · · ⊗ x j k for x j i ∈ ˜ H ∗ ( X j i ; R ) and { j , · · · , j k } / ∈ K and iscalled the generalized Stanley-Reisner ideal of K . The proof of Theorem 3.1 uses the strong form of K¨unneth Theorem, which says that µ : m O i =1 H ∗ ( X i ; R ) → H ∗ ( m Y i =1 X i ; R ) , x ⊗ · · · ⊗ x m π ∗ ( x ) ∪ · · · ∪ π ∗ m ( x m ) , where π ∗ j is induced by projection π j : Q mi =1 X i → X j , is isomorphic if all H ∗ ( X i ; R )’s arefree. In the reduced version of K¨unneth Theorem, ¯ µ : N mi =1 ˜ H ∗ ( X i ) → ˜ H ∗ ( V mi =1 X i ) is alsoisomorphic if all ˜ H ∗ ( X i ; R )’s are free. The goal of this section is to modify Theorem 3.1by removing the freeness assumption on H ∗ ( X i )’s. As a trade-off, we need to mod out thetorsion elements of H ∗ ( X i )’s. First let us refine K¨unneth Theorem. Lemma 3.2.
Let X = { X i } mi =1 be a sequence of spaces X i . Then the induced morphisms µ f : N mi =1 H ∗ f ( X i ) → H ∗ f ( Q mi =1 X i ) and ¯ µ f : N mi =1 ˜ H ∗ f ( X i ) → ˜ H ∗ f ( V mi =1 X i ) . EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 9 are isomorphisms as algebras, and there is a commutative diagram N mi =1 ˜ H ∗ f ( X i ) ¯ µ f / / (cid:15) (cid:15) ˜ H ∗ f ( V mi =1 X i ) q ∗ f (cid:15) (cid:15) N mi =1 H ∗ f ( X i ) µ f / / H ∗ f ( Q mi =1 X i ) where q ∗ f is induced by the quotient map q : Q mi =1 X i → V mi =1 X i .Proof. It suffices to show the m = 2 case. Let ( X, A ) and (
Y, B ) be pairs of relative CW-complexes and let π X : ( X × Y, A × Y ) → ( X, A ) and π Y : ( X × Y, X × B ) → ( Y, B ) beprojections. By the generalized version of K¨unneth Theorem [10, Chapter XIII, Theorem11.2], the sequence0 → M i + j = n H i ( X, A ) ⊗ H j ( Y, B ) µ ′ → H n ( X × Y, X × B ∪ A × Y ) → T → T is a torsion term and µ ′ sends u ⊗ v ∈ H i ( X, A ) ⊗ H j ( Y, B ) to π ∗ X ( u ) ∪ π ∗ Y ( v ), is splitexact. By Lemma 2.3 ( H ∗ ( X, A ) ⊗ H ∗ ( Y, B )) f ∼ = H ∗ f ( X, A ) ⊗ H ∗ f ( Y, B ) and by Lemma 2.2 µ ′ f : H ∗ f ( X, A ) ⊗ H ∗ f ( Y, B ) −→ H ∗ f ( X × Y, X × B ∪ A × Y )is an isomorphism. Since µ ′ is multiplicative, so is µ ′ f . Letting A and B be the basepointsof X and Y or be the empty set gives the isomorphisms µ f : H ∗ f ( X ) ⊗ H ∗ f ( Y ) ∼ = H ∗ f ( X × Y ) and ¯ µ f : ˜ H ∗ f ( X ) ⊗ ˜ H ∗ f ( Y ) ∼ = ˜ H ∗ f ( X ∧ Y ) . The asserted commutative diagram follows from commutative diagram N mi =1 ˜ H ∗ ( X i ) ¯ µ / / (cid:15) (cid:15) ˜ H ∗ ( V mi =1 X i ) q ∗ (cid:15) (cid:15) N mi =1 H ∗ ( X i ) µ / / H ∗ ( Q mi =1 X i ) (cid:3) Proposition 3.3.
Let X = { X i } mi =1 be a sequence of spaces X i , and let K be a simplicialcomplex on [ m ] . Then the inclusion ı : ( X, ∗ ) K → Q mi =1 X i induces a ring isomorphism H ∗ f (( X, ∗ ) K ) ∼ = m O i =1 H ∗ f ( X i ) ! /I K where I K is generated by x j ⊗ · · · ⊗ x j k for x j i ∈ ˜ H ∗ f ( X j i ; R ) and { j , · · · , j k } / ∈ K .Proof. This proof modifies the proofs in [3, 5]. Consider homotopy cofibration sequence( X, ∗ ) K ı → m Y i =1 X i → C, where C is the mapping cone of ı and is the inclusion. Suspend it and obtain a diagram ofhomotopy cofibration sequences(2) Σ( X , ∗ ) K Σ ı / / a (cid:15) (cid:15) Σ ( Q mi =1 X i ) Σ / / b (cid:15) (cid:15) Σ C c (cid:15) (cid:15) W J ∈ K Σ X ∧ J ¯ ı / / W J ∈ [ m ] Σ X ∧ J ¯ / / W J / ∈ K Σ X ∧ J where X ∧ J = X j ∧ · · · ∧ X j k for J = { j , · · · , j k } , ¯ ı is the inclusion, ¯ is the pinch map, a is a homotopy equivalence by [3, Theorem 2.21], b is a homotopy equivalence, and c is aninduced homotopy equivalence. Take cohomology and get the diagram0 / / L J / ∈ K ˜ H ∗ ( X ∧ J ) ¯ ∗ / / c ∗ (cid:15) (cid:15) L J ∈ [ m ] ˜ H ∗ ( X ∧ J ) ¯ ı ∗ / / b ∗ (cid:15) (cid:15) L J ∈ K ˜ H ∗ ( X ∧ J ) / / a ∗ (cid:15) (cid:15) / / ˜ H ∗ ( C ) ∗ / / ˜ H ∗ ( Q mi =1 X i ) ı ∗ / / ˜ H ∗ (( X, ∗ ) K ) / / H ( X ) ⊗ J = ˜ H ∗ ( X j ) ⊗ · · · ⊗ ˜ H ∗ ( X j k ) and ˜ H f ( X ) ⊗ J = ˜ H ∗ f ( X j ) ⊗ · · · ⊗ ˜ H ∗ f ( X j k ). Therows are split exact sequences. All vertical maps are isomorphisms. All maps are additivewhile ı ∗ is multiplicative. Apply Lemma 2.2 to the diagram and get0 / / L J / ∈ K ˜ H ∗ f ( X ∧ J ) ¯ ∗ f / / c ∗ f (cid:15) (cid:15) L J ∈ [ m ] ˜ H ∗ f ( X ∧ J ) ¯ ı ∗ f / / b ∗ f (cid:15) (cid:15) L J ∈ K ˜ H ∗ f ( X ∧ J ) / / a ∗ f (cid:15) (cid:15) / / ˜ H ∗ f ( C ) ∗ f / / ˜ H ∗ f ( Q mi =1 X i ) ı ∗ f / / ˜ H ∗ f (( X, ∗ ) K ) / / H ∗ f ( Q mi =1 X i ) ∼ = L mi =1 H ∗ f ( X i ) so there is a ring isomorphism H ∗ f (( X, ∗ ) K ) ∼ = m O i =1 H ∗ f ( X i ) ! / ker( ı ∗ f ) . We show that ker( ı ∗ f ) is generated by x j ⊗· · ·⊗ x j k for x j i ∈ ˜ H ∗ f ( X j i ; R ) and { j , · · · , j k } / ∈ K .For J = { j , · · · , j k } ⊂ [ m ], the restriction of a ∗ f to ˜ H ∗ f ( X ) ⊗ J is the composite˜ H ∗ f ( X ) ⊗ J ¯ µ f −→ ˜ H ∗ f ( X ∧ J ) q ∗ f −→ ˜ H ∗ f ( k Y i =1 X j i ) −→ H ∗ f ( m Y i =1 X i ) . So the image is µ f (cid:16) ˜ H ∗ f ( X ) ⊗ J (cid:17) by Lemma 3.2. Since ˜ ı ∗ f is the projection, µ f (cid:16) ˜ H ∗ f ( X ) ⊗ J (cid:17) isin ker( ı ∗ f ) if and only if J / ∈ K . So ker( ı ∗ f ) ∼ = I K . (cid:3) Proposition 3.3 can be refined as follows. If the quotient map H ∗ ( X i ) → H ∗ f ( X i ) has rightinverse for all i , then so does H ∗ (( X, ∗ ) K ) → H ∗ f (( X, ∗ ) K ). To formulate this, we introducenew definition. Definition 3.3.1.
A graded algebra A is free split if the quotient map π : A → A f hasa section as algebras. In other words, there is a ring morphism s : A f → A making the EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 11 following diagrams commute A f s / / ❆❆❆❆❆❆❆❆ ❆❆❆❆❆❆❆❆ A π (cid:15) (cid:15) A f A f ⊗ A fs ⊗ s (cid:15) (cid:15) m f / / A fs (cid:15) (cid:15) A ⊗ A m / / A , where m and m f are multiplications in A and A f . We call s a free splitting of A .In general, a free splitting of A is not unique. Any two free splittings s and s differ bya torsion element. Remark 3.4.
Not all cohomology rings of spaces are free split. Let C be the mapping coneof the composite P (2) ρ −→ S ı ,ı ] −→ S ∨ S , where P (2) is the mapping cone of degree map S → S , ρ is the quotient map and [ ı , ı ] is the Whitehead product. Then H ∗ ( C ) ∼ = Z [ a, b ] / ( a = b = 2 ab = 0) where | a | = | b | = 2 ,and it is not free split. Lemma 3.5.
Under the conditions of Proposition 3.3, if H ∗ ( X i ) is free split for all i ,then H ∗ (( X, ∗ ) K ) is free split.Proof. Use the notations in the proof of Proposition 3.3. For 1 ≤ i ≤ m , let s i : H ∗ f ( X i ) → H ∗ ( X i )be a free splitting and let s be the composite s : m O i =1 H ∗ f ( X i ) N mi =1 s i −→ m O i =1 H ∗ ( X i ) µ −→ H ∗ ( m Y i =1 X i ) . Then s is a free splitting of H ∗ ( Q mi =1 X i ). As ı ∗ f : H ∗ f ( Q mi =1 X i ) → H ∗ f (( X, ∗ ) K ) is surjective,define s ′ : H ∗ f (( X, ∗ ) K ) → H ∗ (( X, ∗ ) K ) by the diagram N mi =1 H ∗ f ( X i ) s / / ı ∗ f (cid:15) (cid:15) H ∗ ( Q mi =1 X i ) ı ∗ (cid:15) (cid:15) H ∗ f (( X, ∗ ) K ) s ′ / / H ∗ (( X, ∗ ) K )We need to show that s ′ is well defined. For x ∈ H ∗ f ( X, ∗ ) K , let y, y ′ ∈ N mi =1 H ∗ f ( X i )be two preimages of x . Then y − y ′ ∈ ker( ı ∗ f ) = I K . For J / ∈ K , s sends ˜ H ∗ f ( X ) ⊗ J to µ (cid:16) ˜ H ∗ ( X ) ⊗ J (cid:17) which is contained in ker( ı ∗ ). So ı ∗ ◦ s ( y − y ′ ) = 0 and s ′ is well defined.Since s , ı ∗ and ı ∗ f are multiplicative, so is s ′ . Furthermore, s ′ is right inverse to the quotientmap H ∗ (( X, ∗ ) K ) → H ∗ f (( X, ∗ ) K ). So s ′ is a free splitting. (cid:3) Realization of graded monomial ideal rings
Quotient rings of square-free ideals.
Let P = Z [ x , · · · , x m ] ⊗ Λ[ y , · · · , y n ] be agraded polynomial ring where x i ’s have arbitrary positive even degrees and y j ’s have arbitrarypositive odd degrees, and let I = ( M , · · · , M r ) be an ideal generated by monomials M j = x a j · · · x a mj m ⊗ y b j · · · y b nj n , where a ij ’s are non-negative integers and b ij ’s are either 0 or 1. Then A = P/I is a gradedmonomial ideal ring . We say that I is square-free if M j ’s are square-free monomials, that isall a ij ’s are either 0 or 1.In the following let • { i , · · · , i k } + { j , · · · , j l } = { i , · · · , i k , j + m, · · · , j l + m } for { i , · · · , i k } ⊂ [ m ]and { j , · · · , j l } ⊂ [ n ] and • X + Y = { X , · · · , X m , Y · · · , Y n } for X = { X i } mi =1 and Y = { Y j } nj =1 sequences ofspaces.Given a square-free ideal I , take K to be a [ m + n ]-simplex with faces { i , · · · , i k } + { j , · · · , j l } removed whenever x i · · · x i k ⊗ y j · · · y j l ∈ I . Then I is the generalized Stanley-Reisner idealof K . Lemma 4.1.
Let X = { K ( Z , | x i | ) } mi =1 and Y = { S | y j | } nj =1 and let K be the simplicial complexdefined as above. Then H ∗ f (( X + Y , ∗ ) K ) ∼ = A . Furthermore, H ∗ (( X + Y , ∗ ) K ) is free split.Proof. Since H ∗ f ( X i ) ∼ = Z [ x i ] and H ∗ ( Y j ) ∼ = Λ[ y j ], the first part follows from Proposition 3.3.For the second part, it suffices to show that H ∗ ( X i ) and H ∗ ( Y j ) are free split by Lemma 3.5.For 1 ≤ j ≤ n , H ∗ ( Y j ) is free and hence free split. For 1 ≤ i ≤ m , let x ′ i be a generatorof H | x i | ( X i ) ∼ = Z . Then inclusion ı : Z h x ′ i i → H ∗ ( X i ) extends to a ring morphism s : Z [ x ′ i ] ∼ = H ∗ f ( X i ) → H ∗ ( X i ) . Let π : H ∗ ( X i ) → H ∗ f ( X i ) be the quotient map. Since π ◦ ı sends x ′ i to itself, by universalproperty π ◦ s is the identity map. So s is a free splitting of H ∗ ( X i ). (cid:3) Polarization of graded monomial ideal rings.
Now drop the square-free assump-tion on I = ( x a j · · · x a mj m ⊗ y b j · · · y b nj n | ≤ j ≤ r ) and suppose some a ij ’s are greater than 1.Following ideas from [3] and [15], we use polarization to reduce the realization problem of A to the special case when I is square-free.For 1 ≤ i ≤ m , let a i = max { a i , · · · , a ir } be the largest index of x i among M j ’s, and letΩ = { ( i, j ) ∈ N × N | ≤ i ≤ m, ≤ j ≤ a i } where ( i, j ) ∈ Ω are ordered in left lexicographical order, that is ( i, j ) < ( i ′ , j ′ ) if i < i ′ , orif i = i ′ and j < j ′ . Let P ′ = Z [ x ′ ij | ( i, j ) ∈ Ω] ⊗ Λ[ y , · · · , y n ]= Z [ x ′ , · · · , x ′ a , x ′ , · · · , x ′ a , · · · , x ′ m , · · · , x ′ ma m ] ⊗ Λ[ y , · · · , y n ] , be a graded polynomial ring where | x ′ ij | = | x i | , let M ′ j = ( x ′ x ′ · · · x ′ a j )( x ′ x ′ · · · x ′ a j ) · · · ( x ′ m x ′ m · · · x ′ ma mj ) ⊗ ( y b j · · · y b nj n )and let I ′ = ( M ′ , · · · , M ′ r ). Then I ′ is square-free and A ′ = P ′ /I ′ is called the polarization of A .Conversely, we can reverse the polarization process and obtain A back from A ′ . Let¯Ω = { ( i, j ) ∈ N × N | ≤ i ≤ m, ≤ j ≤ a i } where ( i, j ) ∈ ¯Ω are ordered in left lexicographical order, and let W be a graded polynomialring W = Z [ w ij | ( i, j ) ∈ ¯Ω]= Z [ w , · · · , w a , w , · · · , w a , · · · , w m , · · · , w ma m ] EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 13 where | w ij | = | x i | . Define a ring morphism δ : W → P ′ by δ ( w ij ) = x ′ ij − x ′ i and make P ′ a W -module via δ . Then A ′ is a W -module and A ∼ = A ′ / ¯ W A ′ . Lemma 4.2.
Let A ′ be a square-free graded monomial ideal ring and let W and δ be definedas above. Then A ′ is a free W -module.Proof. Since A ′ / ¯ W A ′ is a free Z -module, by Lemma 2.8 it suffices to show that { w ij } ( i,j ) ∈ ¯Ω is a A ′ -regular sequence. Denote N = | ¯Ω | = P mi =1 a i − m . For 1 ≤ k ≤ N , let ( i k , j k ) ∈ ¯Ω bethe k th pair under lexicographical order and let I k = ( w , w , · · · , w i k j k ). We need to showthat multiplication w i k +1 j k +1 : A ′ /I k A ′ → A ′ /I k A ′ is injective.Observe that A ′ /I k A ′ = ˜ P / ˜ I , where˜ P = Z [ x ′ , x ′ , · · · , x ′ m , x ′ i k +1 j k +1 , x ′ i k +2 j k +2 , · · · x ′ i N j N ] ⊗ Λ[ y , · · · , y n ]and ˜ I = ( ˜ M , · · · , ˜ M r ) is generated by monomials ˜ M j obtained by identifying x ′ ij with x ′ i in M ′ j for ( i, j ) ≤ ( i k , j k ). Suppose there is a polynomial p ∈ ˜ P such that( x ′ i k +1 j k +1 − x ′ i k +1 ) · p ∈ ˜ I. We can use the combinatoric argument of [8, Page 31] to show p ∈ ˜ I . Here is an outlineof the argument. Write p = P α p α as a sum of monomials p α . For each monomial p α ,it can be shown that x ′ i k +1 j k +1 p α and x ′ i k +1 p α are in ˜ I . Counting the indices of variablesimplies p α ∈ ˜ I . So p is in ˜ I and multiplication w i k +1 j k +1 : A ′ /I k A ′ → A ′ /I k A ′ is injective.Therefore { w ij } ( i,j ) ∈ ¯Ω is a regular sequence and A ′ is a free W -module. (cid:3) Constructing a realization modulo torsion X A . Let A ′ = P ′ /I ′ be the polarizationof A and let K be a simplicial complex on ( P mi =1 a i + n ) vertices such that I ′ is the generalizedStanley-Reisner ideal of K . Construct a polyhedral product to realize A ′ . Take X = { X ij = K ( Z , | x i | ) | ( i, j ) ∈ Ω } = { K ( Z , | x | ) , · · · , K ( Z , | x | ) | {z } a , K ( Z , | x | ) , · · · , K ( Z , | x | ) | {z } a , · · · , K ( Z , | x m | ) , · · · , K ( Z , | x m | ) | {z } a m } and Y = { Y k = S | y k | | ≤ k ≤ n } = { S | y | , S | y | , · · · , S | y n | } . By Lemma 4.1 H ∗ f (( X + Y , ∗ ) K ) is isomorphic to A ′ .For 1 ≤ i ≤ m , define δ i : Q a i j =1 X ij → Q a i j =2 X ij by δ i ( u , · · · , u a i ) = ( u · u − , · · · , u a i · u − ),and define δ : ( X + Y , ∗ ) K → Q ( i,j ) ∈ ¯Ω X ij to be the composite δ : ( X + Y , ∗ ) K ֒ → Y ( i,j ) ∈ Ω X ij × n Y k =1 Y k proj −→ Y ( i,j ) ∈ Ω X ij Q mi =1 δ i −→ Y ( i,j ) ∈ ¯Ω X ij . As δ is a fibration, take X A to be its fiber. We claim H ∗ f ( X A ) ∼ = A .Apply Eilenberg-Moore spectral sequence to the fibration sequence(3) X A −→ ( X + Y , ∗ ) K δ −→ Y ( ij ) ∈ ¯Ω X ij , where H ∗ (( X + Y , ∗ ) K ) is an H ∗ ( Q ( ij ) ∈ ¯Ω X ij )-module via δ ∗ . Let { E p,qr } be the E r -page. Lemma 4.3.
For E ∞ -page, ( E ,q ∞ ) f ∼ = A q as modules and E − p,q ∞ = 0 for p = 0 . Proof.
The E -page is given by E − p, ∗ = Tor − p, ∗ H ∗ ( Q ( ij ) ∈ ¯Ω X ij ) ( H ∗ (( X + Y , ∗ ) K ) , Z ). By Lemma 2.5,there is a monomorphism π ′ : ( E − p, ∗ ) f −→ (cid:16) Tor − p, ∗ H ∗ f ( Q ( ij ) ∈ ¯Ω X ij ) ( H ∗ f (( X + Y , ∗ ) K ) , Z ) (cid:17) f , which is an isomorphism for p = 0. By Lemmas 3.2 and 3.3, H ∗ f (( X + Y , ∗ ) K ) ∼ = A ′ and H ∗ f ( Y ( ij ) ∈ ¯Ω X ij ) ∼ = Z [ w , · · · , w a , w , · · · , w a , · · · , w m , · · · , w ma m ] . Denote H ∗ f ( Q ( ij ) ∈ ¯Ω X ij ) by W . So A ′ is a W -module via δ ∗ . By Lemma 4.2 A ′ is a free W -module, so Tor − p,qW ( A ′ , Z ) ∼ = ( A q p = 00 otherwise . It follows that ( E − p,q ) f is A q for p = 0 and is zero otherwise.Suppose ( E − p,qr ) f is A q for p = 0 and is zero otherwise. Since ( E − p, ∗ r ) f is concentratedin the column p = 0, any differentials d r in and out of torsion-free elements is trivial.So we have ker( d r ) f = ( E − p,qr ) f and Im( d r ) f = 0. By Lemma 2.2, ( E − p,qr +1 ) f ∼ = ( E − p,qr ) f .Therefore ( E − p,q ∞ ) f is isomorphic to A q for p = 0 and is zero otherwise. (cid:3) Lemma 4.4.
There is a module isomorphism H qf ( X A ) ∼ = A q .Proof. Since Eilenberg-Moore spectral sequence strongly converges to H ∗ ( X A ), for any fixed q there is a decreasing filtration { F − p } of H q ( X A ) such that F −∞ = H q ( X A ) , F = 0 , E − p,p + q ∞ ∼ = F − p / F − p +1 . By Lemma 2.2 ( E − p,p + q ∞ ) f ∼ = (( F − p ) f / ( F − p +1 ) f ) f . By Lemma 4.3 ( E − p,p + q ∞ ) f is zero un-less p = 0, so H qf ( X A ) ∼ = ( E ,q ∞ ) f ∼ = A q as modules. (cid:3) Before going to the extension problem of the E ∞ -page, we look back at Lemma 4.3 andconsider the special case where all of the even degree generators of A are in degree 2. Thenwe do not need to mod out by torsion and get a generalization of the results of Bahri-Bendersky-Cohen-Gitler [4, Theorem 2.2] and Trevisan [15, Theorem 3.6]. Proposition 4.5.
Let A be a graded monomial ideal ring where its generators have eitherdegree 2 or arbitrary positive odd degrees. Then H ∗ ( X A ) ∼ = A .Proof. The E -page is given by E − p, ∗ = Tor − p, ∗ H ∗ ( Q ( ij ) ∈ ¯Ω X ij ) ( H ∗ (( X + Y , ∗ ) K ) , Z ). By hypothe-sis, X ij = CP ∞ for ( i, j ) ∈ Ω and H ∗ ( Q ( ij ) ∈ ¯Ω X ij ) and H ∗ (( X + Y , ∗ ) K are free. Followingthe argument in the proof of 4.3, ( E − p,q ) f is A q for p = 0 and is zero otherwise. Since the E -page is concentrated in the column p = 0, the spectral sequence collapses and H ∗ ( X A ) ∼ = A as algebras. (cid:3) The extension problem.
Lemma 4.4 shows that H ∗ f ( X A ) and A are free Z -modulesof same rank at each degree. We claim that they are isomorphic as algebras. The idea isto construct a space Z A similar to constructing X A so that H ∗ ( Z A ) is free and computable.Then we define a map g A : Z A → X A and compare H ∗ ( X A ) with H ∗ ( Z A ) via g ∗ A . EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 15
Construction of Z A . For 1 ≤ i ≤ m let | x i | = 2 c i , and let Z = { Z ij = ( CP ∞ ) c i | ( i, j ) ∈ Ω } = { ( CP ∞ ) c , · · · , ( CP ∞ ) c | {z } a , ( CP ∞ ) c , · · · , ( CP ∞ ) c | {z } a , · · · , ( CP ∞ ) c m , · · · , ( CP ∞ ) c m | {z } a m } and construct polyhedral product ( Z + Y , ∗ ) K . Fix a generator z of H ( CP ∞ ). For ( i, j ) ∈ Ωand 1 ≤ k ≤ c i , let π ijk : Z ij → CP ∞ be the projection onto the k th copy of CP ∞ andlet z ′ ijk = π ∗ ijk ( z ). By Theorem 3.1 we have H ∗ (( Z + Y , ∗ ) K ) ∼ = Q ′ /L ′ , where Q ′ = Z [ z ′ ijk | ( i, j ) ∈ Ω , ≤ k ≤ c i ] ⊗ Λ[ y , · · · , y n ] and L ′ is the ideal generated bymonomials z ′ i j k · · · z ′ i t j t k t ⊗ y l · · · y l τ for { j + P i − s =1 a s , · · · , j t + P i t − s =1 a s } + { l , · · · , l τ } / ∈ K . For 1 ≤ i ≤ m , define˜ δ i : a i Y j =1 Z ij → a i Y j =2 Z ij , ˜ δ i ( u , · · · , u a i ) = ( u · u − , · · · , u a i · u − ) , and define ˜ δ : ( Z + Y , ∗ ) K → Q ( i,j ) ∈ ¯Ω Z ij to be the composite˜ δ : ( Z + Y , ∗ ) K ֒ → Y ( i,j ) ∈ Ω Z ij × n Y k =1 Y k proj −→ Y ( i,j ) ∈ Ω Z ij Q mi =1 ˜ δ i −→ Y ( i,j ) ∈ ¯Ω Z ij . Lemma 4.6.
Let Z A be the fiber of δ ′ . Then H ∗ ( Z A ) ∼ = Q/L , where Q = Z [ z ik | ≤ i ≤ m, ≤ k ≤ c i ] ⊗ Λ[ y , · · · , y n ] and L is generated by monomials z i k · · · z i N k N ⊗ y b j · · · y b nj n satisfying ≤ j ≤ r , ≤ k l ≤ c i l and ( i , · · · , i N ) = (1 , · · · , | {z } a j , , · · · , | {z } a j , · · · , m, · · · , m | {z } a mj ) . Proof.
Apply Eilenberg-Moore spectral sequence to fibration sequence Z A −→ ( Z + Y , ∗ ) K ˜ δ −→ Y ( i,j ) ∈ ¯Ω Z ij . The E -page is given by ˜ E − p, ∗ = Tor − p, ∗ H ∗ ( Q ( i,j ) ∈ ¯Ω Z ij ) ( Z , H ∗ (( Z + Y , ∗ ) K )). By K¨unneth Theorem H ∗ ( Y ( i,j ) ∈ ¯Ω Z ij ) ∼ = Z [ v ijk | ( i, j ) ∈ ¯Ω , ≤ k ≤ c i ]where | v ijk | = 2. Denoted H ∗ ( Q ( i,j ) ∈ ¯Ω Z ij ) by V . By definition ˜ δ ∗ ( v ijk ) = z ′ ijk − z ′ i k . Thisgives an action of V on Q ′ . By Lemma 4.2 Q ′ /L ′ is a free V -module, soTor − p, ∗ V ( Q ′ /L ′ , Z ) = ( ( Q ′ /L ′ ) / ( z ′ ijk − z ′ ilk ) p = 00 otherwise.Modding out ( z ′ ijk − z ′ ilk ) identifies z ′ ijk with z ′ ilk in Q ′ /L ′ , so ( Q ′ /L ′ ) / ( z ′ ijk − z ′ ilk ) ∼ = Q/L .Since the E -page is concentrated in the column p = 0, H ∗ ( Z A ) ∼ = Q/L as algebras. (cid:3)
Construction of g A . Fix a generator z ∈ H ( CP ∞ ). For 1 ≤ j ≤ c i , let π j : ( CP ∞ ) c i → CP ∞ be the projection onto the j th copy of CP ∞ and let z j = π ∗ j ( z ). For 1 ≤ i ≤ m , take amap g i : ( CP ∞ ) c i → K ( Z , c i ) that represents the cocycle class z · · · z c i ∈ H c i (( CP ∞ ) c i ).For ( i, j ) ∈ Ω, let g ij : Z ij → X ij be g i , and for 1 ≤ k ≤ n , let h k : Y k → Y k be the identitymap. Then { g ij , h k | ( i, j ) ∈ Ω , ≤ k ≤ n } induces a map g K : ( Z + Y , ∗ ) K → ( X + Y , ∗ ) K by functoriality of polyhedral products. Lemma 4.7.
Let { x ′ ij , y k | ( i, j ) ∈ Ω , ≤ k ≤ n } be generators of H ∗ f (( X + Y , ∗ ) K ) ∼ = P ′ /I ′ and { z ′ ijl , y k | ( i, j ) ∈ Ω , ≤ l ≤ c i , ≤ k ≤ n } be generators of H ∗ (( Z + Y , ∗ ) K ) ∼ = Q ′ /L ′ .Then ( g ∗ K ) f ( x ′ ij ) = Q c i l =1 z ′ ijl and ( g ∗ K ) f ( y ′ k ) = y k .Proof. There is a commutative diagram( Z + Y , ∗ ) K / / g K (cid:15) (cid:15) Q ( i,j ) ∈ Ω Z ij × Q nk =1 Y kg (cid:15) (cid:15) ( X + Y , ∗ ) K ı / / Q ( i,j ) ∈ Ω X ij × Q nk =1 Y k where ı and are inclusions, g = Q ( i,j ) ∈ Ω g ij × Q nk =1 h k . Take cohomology, mod out torsionelements and identify the quotient algebras. Then we obtain the commutative diagram P ′ ı ∗ f / / g ∗ f (cid:15) (cid:15) P ′ /I ′ ( g ∗ K ) f (cid:15) (cid:15) Q ′ ∗ / / Q ′ /L ′ where ı ∗ f and ∗ are the quotient maps. Let ˜ x ij , ˜ y k ∈ P ′ and ˜ z ijl , ˜˜ y k ∈ Q ′ be generators suchthat ı ∗ f (˜ x ij ) = x ′ ij , ı ∗ f (˜ y k ) = ∗ f (˜˜ y k ) = y k and ∗ (˜ z ijl ) = z ′ ijl . By construction g ∗ f (˜ x ij ) = Q c i l =1 ˜ z ijl and g ∗ f (˜ y k ) = ˜˜ y k , so we have ( g ∗ K ) f ( x ′ ij ) = Q c i l =1 z ′ ijl and ( g ∗ K ) f ( y ′ k ) = y k . (cid:3) Lemma 4.8.
There is a map g A : Z A → X A satisfying commutative diagram Z Ag A (cid:15) (cid:15) / / ( Z + Y , ∗ ) Kg K (cid:15) (cid:15) X A / / ( X + Y , ∗ ) K Proof.
One may want to construct g A by showing the diagram( Z + Y , ∗ ) K ˜ δ / / g K (cid:15) (cid:15) Q ( i,j ) ∈ ¯Ω Z ij Q ( i,j ) ∈ ¯Ω g ij (cid:15) (cid:15) ( X + Y , ∗ ) K δ / / Q ( i,j ) ∈ ¯Ω X ij commutes. However, as ( Q ( i,j ) ∈ ¯Ω g ij ) ◦ δ ′ and δ ◦ g K induce different morphisms on cohomology,the diagram cannot commute. Instead, we show that the composite Z A −→ ( Z + Y , ∗ ) K g K −→ ( X + Y , ∗ ) K δ −→ Y ( i,j ) ∈ ¯Ω X ij EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 17 is trivial. If so, there will exist a map g A : Z A → X A as asserted since X A is the fiber of δ .By definition of δ there is a commutative diagram( Z + Y , ∗ ) K ˜ δ / / (cid:15) (cid:15) Q ( i,j ) ∈ ¯Ω Z ij Q ( i,j ) ∈ Ω Z ij × Q nk =1 Y k proj / / Q ( i,j ) ∈ Ω Z ij Q mi =1 ˜ δ i O O where is the inclusion. Denote ( Q mi =1 ˜ δ i ) ◦ proj by ˜ δ ′ and extend the diagram to Z A / / e (cid:15) (cid:15) ( Z + Y , ∗ ) K ˜ δ / / (cid:15) (cid:15) Q ( i,j ) ∈ ¯Ω Z ij Q mi =1 ( CP ∞ ) c i × Q nk =1 Y k △ ′ × h / / Q ( i,j ) ∈ Ω Z ij × Q nk =1 Y k ˜ δ ′ / / Q ( i,j ) ∈ ¯Ω Z ij where △ ′ : Q mi =1 ( CP ∞ ) c i → Q a i j =1 Z ij is the diagonal map, h : Q nk =1 Y k → Q nk =1 Y k is theidentity map, and e is an induced map. The top and the bottom row are fibration sequences.The left square fits into the following commutative diagram Z A / / e (cid:15) (cid:15) ( Z + Y , ∗ ) K g K / / (cid:15) (cid:15) ( X + Y , ∗ ) K δ / / ı (cid:15) (cid:15) Q ¯Ω X ij Q mi =1 ( CP ∞ ) c i × Q nj =1 Y j △ ′ × h / / Q g i × h * * ❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱❱ Q Ω Z ij × Q nj =1 Y j Q g ij × h / / Q Ω X ij × Q nk =1 Y k δ ′ / / Q ¯Ω X ij Q mi =1 K ( Z , | x i | ) × Q nj =1 Y j △× h ✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐✐ where ı is the inclusion, △ : Q mi =1 K ( Z , | x i | ) → Q a i j =1 X ij is the diagonal map, and δ ′ is thecomposite δ ′ : Y ( i,j ) ∈ Ω X ij × n Y k =1 Y k proj −→ Y ( i,j ) ∈ Ω X ij Q mi =1 δ i −→ Y ( i,j ) ∈ ¯Ω X ij . The middle square is due to the functoriality of polyhedral products, the right square is dueto the definition of δ and the bottom triangle is due to the naturality of diagonal maps.The composite of maps from Z A to Q ( i,j ) ∈ ¯Ω X ij round the bottom triangle is trivial, since m Y i =1 K ( Z , | x i | ) × n Y k =1 Y k △× h −→ Y ( i,j ) ∈ Ω X ij × n Y k =1 Y k δ ′ −→ Y ( i,j ) ∈ ¯Ω X ij is a fibration sequence. So the composite in the top row is trivial and this induces amap g A : Z A → X A as asserted. (cid:3) Proof of Main Theorem.
Lemma 4.9.
Let φ : X A → (( X + Y , ∗ ) K ) be the inclusion. Then the induced morphism φ ∗ f : H ∗ f (( X + Y , ∗ ) K ) −→ H ∗ f ( X A ) is surjective and ker ( φ ∗ f ) is generated by x ′ ij − x ′ i for ( i, j ) ∈ ¯Ω . Proof.
Fix a positive integer q and let ψ : Z A → ( Z + Y , ∗ ) K be the inclusion. Considercommutative diagram H q (( X + Y , ∗ ) K ) g ∗ K / / φ ∗ (cid:15) (cid:15) e w w ♦♦♦♦♦♦♦♦♦♦♦♦ H q (( Z + Y , ∗ ) K ) ψ ∗ (cid:15) (cid:15) E ,q ∞ h / / H q ( X A ) g ∗ A / / H q ( Z A )where e is the edge homomorphism, h is the inclusion, and the right square is inducedby Lemma 4.8. Mod out torsion elements and take a generator x ′ i j · · · x ′ i s j s ⊗ y l · · · y l t in H qf (( X + Y , ∗ ) K ). By Lemma 4.7 and the above diagram we have( g ∗ A ◦ h ◦ e ) f ( x ′ i j · · · x ′ i s j s ⊗ y l · · · y l t ) = ( ψ ∗ ◦ g ∗ K ) f ( x ′ i j · · · x ′ i s j s ⊗ y l · · · y l t )( g ∗ A ◦ h ) f ( x i · · · x i s ⊗ y l · · · y l t ) = s Y u =1 c iu Y k =1 z i u j u k ! ⊗ y l · · · y l t . Since x i · · · x i s ⊗ y l · · · y l t and (cid:0)Q su =1 Q c iu k =1 z i u j u k (cid:1) ⊗ y l · · · y l t are generators, ( g A ◦ h ) ∗ f is theinclusion of a direct summand into H qf ( Z A ). By Lemma 4.3 ( E ,q ∞ ) f and H qf ( X A ) are freemodules of same rank, so h f is an isomorphism. Since e f is a surjection, so is φ ∗ f .For the second part of the lemma, suppose there is a polynomial p ∈ ker( φ ∗ f ) not containedin ( x ′ ij − x ′ i ) ( i,j ) ∈ ¯Ω . Since φ ∗ f is a degree 0 morphism, we assume p = P α p α is a sum ofmonomials p α of some fixed degree q . Then p α ’s are linear dependent. So the rank of H qf ( X A )is less than the rank of A q , contradicting to Lemma 4.3. So ker( φ ∗ f ) = ( x ′ ij − x ′ i ) ( i,j ) ∈ ¯Ω . (cid:3) Restate the Main Theorem (Theorem 1.2) and prove it.
Theorem 4.10.
Let A be a graded monomial ideal ring. Then there exists a space X A suchthat H ∗ f ( X A ) is ring isomorphic to A . Moreover, H ∗ ( X A ) is free split.Proof. For the first part of the lemma, the ring isomorphism H ∗ ( X A ) ∼ = A follows fromLemma 4.9.In Lemma 4.1 we construct a free splitting s K : H ∗ f ( X + Y , ∗ ) K → H ∗ ( X + Y , ∗ ) K outof free splittings s ij : H ∗ f ( X ij ) → H ∗ ( X ij ) and the identity maps on H ∗ ( Y k ). Define amap s : H ∗ f ( X A ) → H ∗ ( X A ) by diagram H ∗ f (( X + Y , ∗ ) K ) s K / / φ ∗ f (cid:15) (cid:15) H ∗ (( X + Y , ∗ ) K ) φ ∗ (cid:15) (cid:15) H ∗ f ( X A ) s / / H ∗ ( X A )We need to show that s is well defined. By Lemma 4.9 φ ∗ f is a surjection and ker( φ ∗ f ) isgenerated by polynomials x ′ ij − x ′ i for ( i, j ) ∈ ¯Ω. It suffices to show φ ∗ ◦ s K ( x ′ ij − x ′ i ) = 0.Let x ij ∈ H c i ( X ij ) and ¯ x ij ∈ H c i f ( X ij ) be generators. Then s ij (¯ x ij ) = x ij . There is a string EALIZATION OF GRADED MONOMIAL IDEAL RINGS MODULO TORSION 19 of equations φ ∗ ◦ s K ( x ′ ij − x ′ i ) = φ ∗ ◦ µ ( s ij (¯ x ij ) − s i (¯ x i ))= φ ∗ ◦ µ ( x ij − x i )= φ ∗ ◦ δ ∗ ◦ µ (1 ⊗ · · · ⊗ x ij ⊗ · · · ⊗ s K , the third line is due to the naturality of µ ,and the last line is due to the fact that δ and φ are two consecutive maps in the fibrationsequence X A φ → ( X + Y , ∗ ) K δ → Q ( i,j ) ∈ ¯Ω X ij . So s is well defined.Obviously s is right inverse to the quotient map H ∗ ( X A ) → H ∗ f ( X A ). Since φ ∗ f , φ ∗ and s K are multiplicative, so is s . Therefore s is a free splitting. (cid:3) Example
Now we demonstrate how to construct X A for A = Z [ x ] ⊗ Λ[ y ] / ( x y ). First, polarize A by introducing two new variables x ′ and x ′ of degree 4 and let A ′ = Z [ x ′ , x ′ ] ⊗ Λ[ y ] / ( x ′ x ′ y ) . Let K be the boundary of a 2-simplex. Then ( x ′ x ′ y ) is the Stanley-Reisner ideal of K . Take X = { K ( Z , , K ( Z , } , Y = { S } and construct polyhedral product ( X + Y , ∗ ) K . By Proposition 3.3 H ∗ f (( X + Y , ∗ ) K ) ∼ = Z [ x ′ , x ′ ] ⊗ Λ[ y ] / ( x ′ x ′ y ) . Define δ : ( X + Y , ∗ ) K → K ( Z ,
4) by δ ( u , u , t ) = u · u − , and define X A to be the fiberof δ . By Theorem 4.10 H ∗ f ( X A ) ∼ = A .Next, we construct Z A and g A to illustrate the proof of the extension problem. In thiscase, take Z = { ( CP ∞ ) , ( CP ∞ ) } . Denote the first ( CP ∞ ) by Z and the second ( CP ∞ ) by Z . Then H ∗ ( Z ) = Z [ z ′ , z ′ ] and H ∗ ( Z ) = Z [ z ′ , z ′ ], where | z ij | = 2 for i, j ∈ { , } ,and H ∗ (( Z + Y , ∗ ) K ) ∼ = Z [ z ′ , z ′ , z ′ , z ′ ] ⊗ Λ[ y ] /L ′ where L ′ = ( z ′ z ′ y, z ′ z ′ y, z ′ z ′ y, z ′ z ′ y ). Define˜ δ : ( Z + Y , ∗ ) K → ( CP ∞ ) , ˜ δ ( v , v , t ) = v · v − , and define Z A to be the fiber of ˜ δ . Then H ∗ f ( Z A ) ∼ = Z [ z , z ] ⊗ Λ[ y ] /L , where | z | = | z | = 2and L = ( z y, z y, z z y ).For i = { , } , let g i : Z i → K ( Z ,
4) be a map representing z ′ i z ′ i ∈ H ( Z i ), andlet h : S → S be the identity map. Then g , g and h induce g K : ( Z + Y , ∗ ) K → ( X + Y , ∗ ) K such that g ∗ K ( x ′ i ) = z ′ i z ′ i and g ∗ K ( y ) = y . Lemma 4.8 gives a map g A : Z A → X A making thefollowing diagram commute Z A / / g A (cid:15) (cid:15) ( Z + Y , ∗ ) Kg K (cid:15) (cid:15) X A / / ( X + Y , ∗ ) K . acknowledgment This work was done during the first author’s PIMS postdoctoral Fellowship. Both authorsgratefully acknowledge the support of Fields Institute and NSERC.
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Department of Mathematics and Statistics, Univeristy of Regina, Regina, SK S4S 0A2,Canada
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