AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT. The Adams-Novikov spectral sequence for the Brown-Peterson spectrum

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1 AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT MICHA L ADAMASZEK The Adams-Novikov spectral sequence for the Brown-Peterson spectrum E s,t = Ext s,t BP BP (BP, BP ) = π S s t(s 0 ) (p) has been one of the most successful tools in the understanding of stable homotopy groups of spheres. However, already the calculation of the algebraic E -page presents some prohibitive difficulties and led to the development of computational tools such as the chromatic spectral sequence of []. The purpose of this exposition is to introduce the reader to the Adams-Novikov Ext without using any of the heavy machinery. In particular, we will calculate the -line of the Adams- Novikov spectral sequence, that is Ext, (BP ). This was originally the result of Novikov [3], who also related Ext with the image of the J-homomorphism and Miller-Ravenel-Wilson [], who used the chromatic spectral sequence to obtain a lot more general results. We will then move on to other examples, including the image of Ext in the classical Adams spectral sequence and a short detour of the higher Greek letter elements.. Hopf algebroids and BP We assume the reader is familiar with the general theory of Hopf algebroids and their homological algebra and with the construction of the Adams-Novikov spectral sequence. The reference for these is [5, A., A.,.]. We will briefly recall the relevant algebraic notions. A Hopf algebroid over a ground ring K is a pair (A, Γ) of commutative K-algebras equipped with the left and right units η L, η R : A Γ, counit Γ A, conjugation Γ Γ and a coproduct : Γ Γ A Γ all of which are K-algebra homomorphisms satisfying a number of compatibility axioms. In the definition of we treat Γ as a left A-module via η L and as a right A-module via η R. An example of a Hopf algebroid is any Hopf algebra, in particular (F p, A p) where A p denotes the dual mod p Steenrod algebra. A left Γ-comodule is a left A-module M equipped with a map M Γ A M, which again satisfies the usual compatibility axioms. Right comodules are defined analogously. Note that A itself is a left Γ-comodule via the map a a, which can be identified with the right unit η R : A Γ = Γ A A. Similarly, A is a right Γ-comodule via the map a a which can be identified with η L. For a right Γ-comodule M and a left Γ-comodule N we define the cotensor product M N as M N = ker(ψ M id N id M ψ N : M A N M A Γ A N). In particular, if M = A with its right module structure then A N is the submodule of primitive elements of N (i.e. n N such that ψ N (n) = n). For every M the functor Date: October 8, 00.

2 MICHA L ADAMASZEK Cotor i Γ (M, N) is the i-th right derived functor of N M ΓN. When M = A we abbreviate notation to Cotor i (N). In due course this will always be denoted by Ext i (N), due to the isomorphism between Hom Γ (A, ) and A Γ, as explained in [5, p.30]. For the calculation of Ext(N) it is customary to treat the left Γ-comodule N as a right Γ-comodule via a device which employs the conjugation map of the algebroid. The standard left and right Γ-comodule structures on A correspond to each other under this operation. Now it follows (see [5, A..]) that Ext i (N) is the i-th cohomology group of the cobar complex {C s (N)} s 0, where C s (N) = N A Γ As : () N d N A Γ d N A Γ A Γ We write a typical element of C s (N) as nγ γ s and we skip n if n =. The differential is given by the formula s d(nγ γ s ) = ψ N (n)γ γ s + ( ) i nγ (γ i ) γ s + ( ) s+ nγ γ s i= where ψ N is the right Γ-comodule map for N. It is convenient to note that the differentials are determined by () d(n) = ψ N (n) n, d(γ) = γ + γ (γ) and extended to other elements of the cobar complex by requiring that d be a graded derivation with respect to the tensor product grading as follows: s d(nγ γ s ) = d(n)γ γ s + ( ) i nγ d(γ i ) γ s. When N = A then the complex () for computing Ext(A) can be identified with (3) A d Γ d Γ A Γ with d(a) = η R (a) a and d(γ) = γ + γ (γ). Eventually, if N is a comodule algebra (like A), then Ext(N) is equipped with a product [5, A...5]. In case when N = A this product is represented in the cobar construction by concatenation (4) γ γ s γ γ r = γ γ s γ γ r. Next we shall review the structure of the Hopf algebroid (BP, BP BP ) associated with the Brown-Peterson spectrum BP for a fixed prime p. These formulae and Quillen s construction of the spectrum BP can be found in [5, Ch.4] and the relation with formal group laws is described in [5, Appendix A]. A good introduction to the subject is [6]. We have the isomorphisms of graded rings i= BP = Z (p) [v, v,...], v i = (p i ), BP BP = BP [t, t,...], t i = (p i ). The structural maps of the Hopf algebroid are given in terms of the rationalization BP Q = Q[l, l,...] where l i = (p i ). In terms of the formal group laws the l i are the coefficients

3 AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT 3 of the logarithm for the universal p-typical formal group law. The generators of BP and BP Q are related by Hazewinkel relations: (5) pl n = l i v pi n i, l 0 = v 0 =. 0 i<n The left unit is just the inclusion η L : BP BP BP and the right unit is given on the generators of BP Q by (6) η R (l n ) = l i t pi n i 0 i n and the diagonal of BP BP is determined by: (7) l i (t j ) pi = i,j i,j,k l i t p i j t pj+k k. These equations determine the Hopf algebroid structure uniquely. Consider the invariant ideals I n = (p, v..., v n ) BP and I = (p, v, v,...) BP. The following relations, which hold in (BP, BP BP ), can be easily derived from (5)-(7), see eg. [4, B.5.5] or [5] for a proof. (8) (9) (0) () () (3) η R (v ) = v + pt η R (v n ) v n (mod I n ) η R (v n+j ) v n+j + v n t pn j (t ) = t + t v pj n t j (mod I n, t,..., t j ), j (t ) = t + t v t t + t t for p = η R (v ) = v 5v t 3v t + t 4t 3 for p = We conclude by making a notational convention: in the sequel, unless indicated otherwise, the pair (A, Γ) always denotes (BP, BP BP ).. Calculating Ext (A) This calculation follows precisely the strategy outlined in [6]. First, since I n A are invariant ideals, we can consider the Γ-comodules A/I n for n 0. We start by computing Ext 0 for these comodules. Lemma. (Landweber, []). Ext 0 (A) = Z (p), Ext 0 (A/I n ) = F p [v n ] Proof. By definition, Ext 0 (A/I n ) is the kernel of (η R id) : A/I n Γ/I n, so it contains all powers of v n by (9). Now suppose x is an element in A/I n such that η R (x) x 0 (mod I n ). Let v n+j be the highest of the v which appear in x and write x as x = v l n+jf l v n+j f + f 0, where each f i is a polynomial in v n,..., v n+j. Let J n,j be the ideal (p, v,..., v n, t,..., t j ) Γ. By (0) we have the following relations modulo J n,j : η R (v n+j ) v n+j + v n t pn j vn pj t j (mod J n,j ), η R (v k ) v k (mod J n,j ) for k = n,..., n + j.

4 4 MICHA L ADAMASZEK The second one implies that η R (f i ) f i (mod J n,j ), so modulo this ideal we have a congruence 0 η R (x) x l i=0 ((v n+j + v n t pn j v pj n t j ) i f i v i n+jf i ) (mod J n,j ), but the sum equals vnt l lpn j f l + (lower powers of t j ), so it follows that f l 0 (mod I n ). A repeated application of this argument proves that x is a monomial in v n. We can now describe the generators of Ext (A). First of all, since Ext i (A Q) = 0 for i > 0 (for a proof see [5, Thm. 5..]), and since (A, Γ) is Z (p) -local, all Ext i (A) are p-torsion groups for all i > 0. The short exact sequence of Γ-comodules 0 A p A A/(p) 0 yields a long sequence of Ext groups (4) 0 Ext 0 (A) By Lemma. for n = it becomes p Ext 0 (A) Ext 0 (A/(p)) δ Ext (A) p Ext (A). (5) 0 Z (p) p Z (p) F p [v ] δ Ext (A) p Ext (A). We define elements α t Ext,t(p ) (A) by α t = δ(v t ). By exactness of (5) the elements α t are nonzero, have order p and each group Ext,t(p ) (A) is a cyclic p-group, while all other groups in Ext, (A) are trivial. It remains to determine the orders of the groups, i.e. the divisibility of α t by p. We can perform this calculation in the cobar construction and write explicit representatives for α t. We have the diagram in which each column is the cobar resolution for the respective comodule: (6) 0 A p A A/(p) 0 0 Γ η R id p Γ Γ/(p) 0 Applying the definition of the connecting homomorphism δ and (8) we obtain (7) α t = δ(v t ) = p (η R(v t ) v t ) = p ((v + pt ) t v t ) From now on we fix a factorization t = sp i where s is not divisible by p. If p is odd then we have and when p = we have α t = sp i v spi t + (terms divisible by p i+ ) α t = s i v si t + s i (s i )v si t + (terms divisible by i+ ).

5 AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT 5 In either case α t is certainly divisible by p i and we define α t/j = α t /p j for j =,..., i +, so that α t/j is an element of order p j in Ext (A). If p is odd then α t/i+ = sv spi t + (terms divisible by p). It follows from the portion of the long exact sequence Ext (A) p Ext (A) Ext (A/(p)) that an element of Ext (A) is divisible by p if and only if its image in Ext (A/(p)) is zero. The image of α t/i+ in Ext (A/(p)) is represented by sv spi t and it is nonzero by Lemma.3.i. Therefore α t/i+ is the generator of Ext,(p )spi (A) Z/(p i+ ). If p = the situation is more complicated. If t is odd (i.e. i = 0) we have (8) α t/ = v s t + (terms divisible by ). As before, the mod reduction of this element is v s t which is nonzero in Ext (A/()) by Lemma.3.i, hence Ext,s (A) Z/(). If t = (i.e. s =, i = and we are in Ext,4 (A)) then we have exactly α / = v t + t. Note that in dimension 4 the image d(a/()) Γ/() is spanned by η R (v ) v = 4(v t + t ) 0, so α / necessarily gives a nonzero element in Ext (A/()). It means that Ext,4 (A) Z/(4). If t 4 is even (i.e. i ) then α t/i+ = v t t + v t t + (terms divisible by ) The image of α t/i+ in Ext (A/()) is v t t +v t using (9), (0) we have t, but this time it is trivial. Indeed, η R (v ) v (mod ), η R (v ) v + v t + v t (mod ), and we easily see that α t/i+ η R (v v t 3 ) v v t 3 = d(v v t 3 ) (mod ). It means that the element α t/i+ + (η R (v v t 3 ) v v t 3 ) in divisible by in Γ, so we can define α t/i+ = ( αt/i+ + (η R (v v t 3 ) v v t 3 ) ) ((v = i+ + t ) t v t) + (η R(v v t 3 ) v v t 3 ). In order to prove that this element is not further divisible by we compute its image in Ext (A/()), which is equivalent to computing the numerator of (8) mod 4. We don t need an exact formula; it suffices to recall from (3) that η R (v ) v 5v t 3v t + t (mod 4) and it quickly follows that the image of α t/i+ in Ext (A/()) has the form α t/i+ t v t 3 + (terms divisible by t ) (mod )

6 6 MICHA L ADAMASZEK and a cocycle of this form is nonzero in Ext (A/()) by Lemma.3.ii. All of this can be summarized in the following theorem [5, 5..6]. Theorem.. Let t = sp i. The generator of Ext,(p )t (A) is α t/i+ = p i+ (η R(v t ) v t ) = p i+ ((v + pt ) t v t ) unless p = and t 4 is even, when the generator of Ext,t (A) is α t/i+ given by (8). It follows that Ext,(p )t (A) = Z/(p i+ ) unless p = and t 4 is even, when Ext,t (A) = Z/( i+ ). It remains to prove the following lemma, which identifies some nonzero elements in Ext (A/(p)). The proof is similar to that of Lemma.. Lemma.3. For any prime p: i) The cocycle v k t Γ/(p) represents a nonzero element in Ext (A/(p)). ii) If α Γ/(p) is a cocycle such that α t v k (mod (p, t )) then α represents a nonzero element in Ext (A/(p)). Proof. We first prove (i). Suppose that x A/(p) is an element such that Then, in particular η R (x) x = v k t in Γ/(p). η R (x) x 0 (mod (p, t )). Now we use an argument identical to that of Lemma.. Let v +j be the highest v occurring in x and note, after (0): η R (v +j ) = v +j + v t p j vpj t j (mod (p, t,..., t j )) η R (v k ) = v k (mod (p, t,..., t j )) for k j. If j we argue as in Lemma. that the coefficient at the highest power of v +j in x must be 0. Therefore x is a polynomial in v, v only. Write x as x = v l f l v f + f 0 where each f i is a multiple of an appropriate power of v. Computing modulo p we now have l vt k η R (x) x ((v + v t p vp t ) i f i vf i i ) i=0 which is v l tlp f l + (lower powers of t ), so we must have f l 0 (mod p) for l. An inductive repetition proves that x = f 0 but then η R (x) x 0 (mod p). This means we have a contradiction. Now we move to (ii). The argument is similar, but one step longer. Suppose η R (x) x = α in Γ/(p). Then η R (x) x 0 (mod (p, t, t )) and the same method proves that x is a polynomial in v, v, v 3. The presence of v 3 is eliminated by computing modulo (p, t ) and

7 AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT 7 eliminating excessive powers of t which originate from η R (v 3 ) v 3 +v t p vp t (mod (p, t )). It remains to consider the case when x is a polynomial in v, v. Note that η R (v ) v (mod (p, t )), η R (v ) v (mod (p, t )), hence η R (x) x 0 (mod (p, t )) for any such element x, which contradicts η R (x) x t v k (mod (p, t )). 3. Hopf invariant one In this, and the following sections, we perform some simple explicit calculations with the elements we have just defined. We begin with the relation between the Adams-Novikov spectral sequence for BP and the classical Adams mod p spectral sequence for H/(p). The Brown-Peterson spectrum BP comes with a map Θ : BP H/(p) to the mod p Eilenberg-MacLane spectrum H/(p), which induces a map from the Adams-Novikov spectral sequence to the classical Adams spectral sequence E s,t = Ext A p (F p, F p ) = π S s t(s 0 ) (p). where A p denotes the mod p dual Steenrod algebra. For our purposes it will suffice to describe the map of Hopf algebroids (A, Γ) (F p, A p) induced by Θ. Let us remind the structure of A p as a Hopf algebra. When p = we have and for odd p: A = F [ξ, ξ,...], ξ i = i A p = F p [ξ, ξ,...] Λ[τ 0, τ,...], ξ i = (p i ), τ i = p i. In each case the diagonal is given on the polynomial part by (9) (ξ k ) = ξ pi k i ξ i, 0 i k which, in the case p = implies also (0) (ξ k ) = 0 i k (ξ k i )pi ξ i, It follows from the defining relations (5)-(7) that the diagonal of (A, Γ) satisfies () (t k ) t i t pi k i (mod I) 0 i k (see eg. [4, B.5.5]). For odd p (9) and () imply that the assignment v i 0, t i ξ i extends to a map of Hopf algebroids from (A, Γ) to the opposite Hopf algebra (F p, A p), or, in other words, the assignment v i 0, t i c(ξ i ) where c is the conjugation of A p, extends to a map of Hopf algebroids (A, Γ) (F p, A p). In a similar fashion, (0) and () imply that for p = the assignment v i 0, t i c(ξ i ) extends to a map of Hopf algebroids (A, Γ) (F, A ). In each case this is the map induced by the map of spectra BP H/(p).

8 8 MICHA L ADAMASZEK It follows that computing the image of the map () Ext Γ (A, A) Ext A p (F p, F p ) is equivalent to computing the reduction mod I in Ext Γ (A, A) and substituting t i c(ξ i ) for odd p or t i c(ξi ) for p =. Note, in particular, that the Adams -line Ext, A (F p, F p p ) is generated by the elements h i = [ξ pi ] (and an additional a 0 = [τ 0 ] for odd p). Moreover, we have c(ξ ) = ξ and c(ξ ) = ξ in A p. We will show that for p = the only elements in Ext, with nonzero image in the Adams spectral sequence are α Ext,, α / Ext,4 and α 4/4 Ext,8 with images h, h, h 3. It is the content of [5, Thm.5..8]. This follows from a direct reduction mod I (which we denote I ) using the formulas of Theorem.. α =t ξ = h, α / = 4 ((v + t ) v ) I t ξ 4 = h α 4/4 = ( 8 ((v + t ) 4 v 4 ) (η R (v v ) v v )) I t 4 ξ 8 = h 3. All the remaining generators of Ext (A) are mapped to zero. Indeed, we only need to check this for α i /i+ where i 3. In that case α i /i+ is given by (8) with t = i. By (3) we have η R (v ) t (mod 4, v, v,...), hence α i /i+ I i i t + i 3 t t I 0. If p is odd and t = sp i > the image of the generator α t/i+ Ext,(p )spi Adams spectral sequence is zero, because When t = the image of α is α t/i+ = p i+ ((v + pt ) t v t ) I p spi i t spi I 0. in the mod p α = p ((v + pt ) v ) I t ξ = h The β-family in Ext The elements β t in Ext are defined as images of v t under the composition (3) F p [v ] = Ext 0 (A/(p, v )) δ Ext (A/(p)) δ 0 Ext (A) where δ n is the connecting homomorphism corresponding to the short exact sequence of comodules (4) 0 A/I n v n A/I n A/I n+ 0. In particular, at the level of cobar constructions, we have δ (v t ) = v (η R (v t ) v t ) Γ/(p).

9 AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT 9 We also denote this element by β t Ext (A/(p)) and we let β t/i Ext (A/(p)) to be defined by the condition β t/i v i = β t, whenever it exists. Recalling from (0) that η R (v ) v + v t p vp t (mod p), we obtain the formula for a representative in Γ/(p): β t = δ (v t ) = v ((v + v t p vp t ) t v t ) which is particularly useful when t = p i. We then pass to Ext (A) defining elements β t/i Ext (A) as images of β t/i Ext (A/()) under δ 0 and then β t/i,j Ext (A) by the condition β t/i,j p j = β t/i. In short: (5) β t/i,j = p j δ0 ( v i δ (v t )) whenever the right-hand side makes sense. As usually we also abbreviate β t/i = β t/i,. We will now find the representatives for some of the elements of the β-family (namely β j / j, β j / j and β 4/, ), together with their images in the Adams spectral sequence. This will prove part of [5, Thm.5.4.6]. In particular, with β 4/, we will experience similar divisibility issues we had in the -line. Before we start, let us make the following observation which will simplify some of the calculations. Lemma 4.. If j and a p j, the cocycles v atpj and v a+pj t represent the same element of Ext (A/(p)). Proof. Suppose n 0 and m is a power of p. Using we obtain η R (v ) v (mod p), η R (v ) v + v t p vp t (mod p), η R (v n v m ) v n v m v n (v + v t p vp t ) m v n v m v n+m t pm v n+pm t m (mod p) which means that v n+m t pm and v n+pm t m represent the same element of Ext (A/(p)). The result follows by repeated application of this fact. We proceed with the description of a few selected elements of the β-family. β j / j for p =. The image of δ (v j ) is represented in Γ/() by β j v ((v + v t + v t ) j v j ) v j t j+ + v j+ t j. From this we get an element β j / j Ext (A/()) represented in Γ/() by β j / j = v j β j = t j+ + v j t j. By Lemma 4., the same cohomology class in Ext (A/()) is represented by β j / j = tj+ + v j+ t.

10 0 MICHA L ADAMASZEK It follows that the element β j / j Ext (A) is represented by β j / j = δ0 (t j+ + v j+ t ) = ( d(t j+ ) + d(v j+ ) t + v j+ d(t ) ). From () we have d(t ) = 0 and d(t k ) = tk + tk ( t + t ) k, so β j / j = ( ( ) j+ t i i t j+ i + ((v + t ) j+ v j+ ) t ). 0<i< j+ The only binomial coefficient in the j+ -th row of the Pascal triangle which is divisible only by but not by 4 is the middle one. It follows that β j / j = tj t j + v j+ t t + (terms divisible by ). It means that β j / j I t j tj, and this element maps to ξj+ ξ j+ = h j+ in the Adams spectral sequence. β p j /p j for odd p. For comparison, we see that an analogous calculation shows that β p j /p j Ext (A/(p)) is represented by β p j /p j = tpj+ + v pj+ t. It follows from our calculation of Ext (A) that v pj+ t is the mod p reduction of α p j+ /j+ Ext (A). The portion of the long exact sequence (4) Ext (A) Ext (A/(p)) δ 0 Ext (A) now implies that δ 0 (v pj+ t ) = 0, so in Ext (A) we have β p j /p j = δ0 (t pj+ ) = ( ( ) p j+ ) t i p i t pj+ i. 0<i<p j+ β j / j for p =. Following the previous calculation we get in Ext (A/()) β j / j = v j β j = v t j+ + v j t j. By Lemma 4. this element is also represented by v t j+ + v j+ t, hence in Ext (A) we have β j / j = d(v t j+ + v j+ t ) = ( ) d(v ) t j+ + v d(t j+ ) + d(v j+ ) t = ( ) t t j+ + v d(t j+ ) + ((v + t ) j+ v j+ ) t I t t j+ which maps to ξ ξj+ = h h j+ in the Adams spectral sequence.

11 AN ELEMENTARY GUIDE TO THE ADAMS-NOVIKOV EXT β 4/, for p =. First, we find β 4 Ext (A/()): β 4 = δ (v 4 ) = v ( (v + v t v t ) 4 v 4 ) v 3 t 8 + v 7 t 4 (mod ) so we can consider the element Its image in Ext (A) is β 4/ = v β 4 = v t 8 + v 6 t 4. β 4/ = (d(v ) t 8 + vd(t 8 ) + d(v) t vd(t 6 4 )) = ( ( ) 8 (4v t + 4t ) t 8 + v t i i t 8 i + 0<i<8 + ((v + t ) 6 v 6 ) t 4 + v 6 0<i<4 = v t 4 t 4 + v 6 t t + (terms divisible by ) ( ) 4 i t i t 4 i In order to see that β 4/, exists, we must know that β 4/ is divisible by or, equivalently, its reduction to Ext (A/()) is zero. This time it is not so easy to guess the expression as a coboundary, but it turns out that (6) v t 4 t 4 + v 6 t t d(v t 4 + v 4 t ) (mod ). The proof is straightforward using the mod reductions of (8)-(3): Therefore we can define d(v ) v t 4 + v 4 t, d(t 4 ) 0, d(v 4 ) 0, d(t ) v t t + t t 4. β 4/, = (β 4/ + d(v t 4 + v 4 t )) and after a rather tedious, but straightforward calculation we get β 4/, t t 8 + v t t (mod, v ) so β 4,/ I t t8 and its image in the Adams spectral sequence is ξ4 ξ6 = h h An example in higher Ext groups As a final example we prove [5, Prop.5..]. First, recall that as a generalization of the α- and β-families, we define the n-th Greek letter element α (n) t as the image of vn t under the composition of connecting homomorphisms of (4) ) (7) F p [v n ] = Ext 0 (A/I n ) δn Ext (A/I n ) δn δ 0 Ext n (A) so that, for example, α t = α () t, β t = α () t, etc. The proposition we want to prove is (8) α (n+) = α (n) p α for n. (It is equivalent with the formulation of [5, Prop.5..] using graded commutativity of product in Ext.)

12 MICHA L ADAMASZEK First, observe that every connecting homomorphism δ k : Ext (A/I k+ ) Ext + (A/I k ) of (7) satisfies δ k (x t ) = δ k (x) t which follows from δ k (x t ) = v k d(x t ) = v k (d(x) t ± x d(t )) = ( v k d(x)) t = δ k (x) t because d(t ) = 0. Now we can calculate the element α (n+). We start with v n+ F p [v n+ ] = Ext 0 (A/I n+ ) and its image in Ext (A/I n ): δ n (v n+ ) = (η R (v n+ ) v n+ ) v n v n (v n t pn vp nt ) = t pn vp n t (mod I n ) The image of this element in Ext (A/I n ) is in turn given by δ n δ n (v n+ ) = ( d(t p n v ) d(vp n ) t vn p d(t ) ) n n ) v n = d(vp t = δ n (v p n ) t because d(t pn ) = ( p n) 0<i<p n i t i t pn i 0 (mod p) and (p) I n. We now apply all the remaining connecting homomorphisms δ to obtain δ 0 δ δ n δ n (v n+ ) = δ 0 δ n ( δ n (vn p ) t ) which is exactly the cochain level version of (8). = δ 0 δ n δ n (v p n ) t Acknowledgements. The author thanks John Jones and Rupert Swarbrick for helpful discussions and encouragement and the Centre for Discrete Mathematics and its Applications (DIMAP), EPSRC award EP/D0639/, for the support during the preparation of this work. References [] P.S.Landweber, Annihilator ideals and primitive elements in complex cobordism, Ill.J.Math. 7 (973), 7-84 [] H.R.Miller, D.C.Ravenel, W.S.Wilson, Periodic phenomena in the Adams-Novikov spectral sequence, Annals of Math. 06 (977), [3] S.P.Novikov, The methods of algebraic topology from the viewpoint of cobordism theories, Math. U.S.S.R.-Izvestiia (967), [4] D.C.Ravenel, Nilpotence and Periodicity in Stable Homotopy Theory, Ann. of Math. Studies 8, Princeton, New Jersey, 99 [5] D.C.Ravenel, Complex cobordism and stable homotopy groups of spheres (Green book), AMS, 004 [6] D.C.Ravenel, A novice s guide to the Adams-Novikov spectral sequence, Geometric Applications of Homotopy Theory, LNM 658 (979), Warwick Mathematics Institute and DIMAP, University of Warwick, Coventry, CV4 7AL, UK address: aszek@mimuw.edu.pl