The Ring of Polynomial Functors of Prime Degree

Let $\hat\Z_p$ be the ring of $p$-adic integers. We prove in the present paper that the category of polynomial functors from finitely generated free abelian groups to $\hat \Z_p$-modules of degree at most $p$ is equivalent to the category of finitely generated modules over a particularly well understood ring, called Green order. That this is the case was conjectured by Yuri Drozd.


Introduction
Polynomial functors attained a lot of interest in recent years by at least two major discoveries. First, in [19] Henn, Lannes and Schwartz showed that the category of analytic functors from the category of finite dimensional vector spaces to to the category of vector spaces over the same field of characteristic p is equivalent to the category of unstable modules over the mod p Steenrod algebra modulo nilpotent objects. Second, Franjou, Friedlander, Scorichenko and Suslin in [15], Friedlander and Suslin in [16], Touzé [35] as well as Touzé and van der Kallen [36] use strict polynomial functor to prove the finite generation of cohomology of group schemes and to compute Ext-groups of modules over general linear groups. More recently Djament and Vespa studied stable homology of orthogonal, symplectic and unitary groups using some category of polynomial functors [37,38,39,5,6,7]. For definitions and more ample remarks of these concepts we refer to section 1.
The category A R of polynomial functors Z − f ree −→ R − mod for a commutative ring R is a classical object in algebraic topology (cf Eilenberg, MacLane [10]). Let A n R be the full subcategory of at most degree n polynomial functors in A R . Quadratic functors were characterized by Baues [2] as modules over a particular algebra. Baues, Dreckmann, Franjou and Pirashvili show in [3] that A n R is a module category of finitely generated R-algebra Γ n R as well. This description was used by Drozd to show in [8] that A 2 Z 2 and in [9] that A 3 Z 3 are two very explicitly given classical orders overẐ 2 andẐ 3 respectively, whose representation theory is completely understood. In particular each of them admit only a finite number of indecomposable lattices. Here, and in the sequel, we denote byẐ p the ring of p-adic integers, and by F q the field with q elements. These orders were introduced by Roggenkamp in [34].
Recall that an R-order over an integral domain R is an R-algebra Λ, finitely generated projective as R-modules and so that K ⊗ R Λ is a semisimple K-algebra, for K being the field of fractions of R. Drozd conjectures at the end of [9] that A p Zp should be equivalent to the module category of a particular Green order Λ p overẐ p for all primes p. Drozd proves the case p = 2 and p = 3 by explicitly associating the generators of the ring given by Baues or Baues, Dreckmann, Franjou and Pirashvili respectively to matrices in the corresponding matrix rings and computes the kernel and the image of the so-defined mapping. The relations in [3] are sufficiently involved so that going beyond p = 3 by this method seems to be not realistic.
In this paper we prove Drozd's conjecture. Our method is conceptual. We develop a recollement diagram of A n Fp by A n−1 Fp and the module category of the group ring F p S n , analogous to the one described by Schwartz [40,Section 5.5] for functors F p − mod −→ F p − mod. This recollement diagram for A n Fp may be of independent interest since it is completely general. It actually appears already in work of Pirashvili [27], as it was indicated by the referee. Comparison of these two diagrams gives many informations. A second ingredient then is the study of various Ext-groups between simple functors, using work of Franjou, Friedlander, Scorichenko and Suslin [15]. The third main ingredient is the explicit projective functor mapping to the reduction modulo p functor. It should be noted that we do not actually use the ring defined by Baues, Dreckmann, Franjou and Pirashvili in [3]. We just use that there is an algebra which is finitely generated, so that the Krull-Schmidt property, lifting of idempotents and similar properties are valid for A n Zp . For this reason we do not give a Morita bimodule between the Baues, Dreckmann, Franjou and Pirashvili-ring and the order we get. As an application of our result, we count the number of 'torsion free' indecomposable polynomial functors in A p Zp . Our paper is organized as follows. In Section 1 we give the essential definitions and relate the different concepts. In Section 2 we recall some of the most important discoveries used in the sequel. Section 3 describes the classical recollement diagrams as well as the new one we have to use for A Fp , and we derive first consequences. The first main result is proved in Section 5. We give the structure of A p Fp there. Finally, in Section 6 we determine A p Recently we received numerous encouragements to publish our manuscript. Steffen König 1 wrote to me that he will not find the time to finish the paper, and he gave me the autorisation to publish the paper alone. I wish to thank Steffen König for having shared his insight with me, and for allowing me to publish the paper.
I wish to thank the referee for many helpful comments and in particular for indicating the treatment of Section 3.2.
1. Generalities on polynomial functors 1.1. Definitions. Let A be a category with direct sums and B be a category with direct sums and kernels. Then, following Eilenberg and MacLane [10] define the cross effect F (1) of a functor F : A −→ B to be the bifunctor A × A −→ B defined on objects by F (1) (V |W ) := ker(F (V ⊕ W ) −→ F (V ) ⊕ F (W )), and on morphism by the naturality of the construction. For n ≥ 1, the n-th cross effect F (n) is the cross effect of the n − 1-st cross effect of F , seen as functor in the first variable. Hence, for objects V, W, V 1 , . . . , V n−1 in A. Suppose in the sequel F (0) = 0. A functor F is said to be polynomial of degree at most n if F (n) = 0 for an n ∈ N (see Pirashvili [28]). Given a commutative ring R, let R − mod be the category of finitely generated R-modules, let R − M od be the category of all R-modules, and let R − f ree be the category of finitely generated free R-modules. Further, call F n R the category of polynomial functors of degree at most n from R − mod to R − mod, and F R := lim − → n F n R . Moreover, let A n R be the category 1 email to the author from April 12, 2013 of polynomial functors of degree at most n from Z − f ree to R − mod, and A R := lim − → n A n R . All these categories are abelian. Observe that additive functors are exactly the degree 1 polynomial functors. The only degree 0 polynomial functor is the trivial functor due to our hypothesis that F (0) = 0.
Friedlander and Suslin define in [16, Definition 2.1] the category of strict polynomial functors P k over a field k. A strict polynomial functor F is defined by associating to each finite dimensional k-vector space a k-vector space F (V ) and to associate for any two finite dimensional k-vector spaces V and W an element in which in addition satisfy the usual compatibility relations for compositions and the identity. Each of these elements can be interpreted as mapping Hom k (V, W ) −→ Hom k (F (V ), F (W )) by interpreting the formal polynomial as polynomial mapping, so that any strict polynomial functor induces a polynomial functor k − mod −→ k − M od. Hence, we have a forgetful functor P k −→ F k . It is shown in [16] that the category of exact degree n strict polynomial functors P n k from k − mod to k − M od for a field k is equivalent to the category of modules over the Schur algebra S k (n, n). Moreover, P k = n P n k . Finally, the forgetful functor P k −→ F k sends a strict polynomial functor of degree at most d to a polynomial functor of degree at most d (cf [29,Remark 4.1]).
A main theme in Section 1 in particular is the question of base change. In Section 5 we shall define for any commutative rings R and S a particular degree d polynomial functor which assigns to every free S-module V the quotient of the group ring R[V ] of the additive group V over R by the d + 1-th power of the augmentation ideal I. Look at the case S = Z.
A key idea, due to Pirashvili, is to replace the morphisms Hom Z (Z n , Z m ) in the category of free abelian groups by R[Hom Z (Z n , Z m )]/(I d+1 ). This construction does not allow base change. Strict polynomial functors basically replace Hom Z (Z n , Z m ) by S d (Hom Z (Z n , Z m )), which does admit base change. Comparison between these two constructions is the main theme in Section 1.
If the degree of F is 1, there is nothing to show since then the functor is linear. Let n > 1. Now, we know that Hence, the restriction f ′ i of each of the f i to Z n−1 and the restriction f ′′ i to the last component In the first two cases, the dimension of the source space is less than n, while the degree of F is unchanged, whereas in the third case the dimension of the source space is n, but the degree of the functor is d − 1. So, in any of these cases by the induction hypothesis we can express F ( k i=1 λ i f i ) a polynomial of degree n in the variables λ 1 , λ 2 , . . . , λ k .
We are left with the case n = 1. The very same reduction applied to the image and induction on m + d implies that one can suppose that m = 1. But then, Eilenberg and MacLane [10, (8.3)] show that for λ ∈ Z one has where F (1) is the identity. Now, where by the induction hypothesis, since the degree of F (1) is less than the degree of F , the right hand side F (1) + F (1) We shall now adapt an argument of Kuhn [22,Lemma 4.8] to this slightly more general situation. We claim that a function f : Z −→ R is a polynomial if and only if some derivative We assume for the moment that R contains Q. Suppose f is a polynomial. Then, it is clear that f (deg(f )+1) = 0. Suppose to the contrary that f (r) = 0. The polynomials . By induction on n, the values g (1) (n) = g(n) − g(n − 1) determine the values g(n) up to the value of g(0). Therefore, up to this constant value, f (s−1) = h (s−1) . Now, suppose R an integral domain of characteristic 0. Then, since R ⊆ f rac(R), f can be considered as being in values f rac(R) which contains Q. This proves the claim.
Now, define f (λ) := F (λ·) and apply the claim to conclude that F (λ·) is polynomial of degree d.
Let K be a field. A priori the category A K is different from the category F K . Nevertheless, in some cases we get one inclusion. Lemma 1.2. Let K be either a prime field of finite characteristic or let K be a field of characteristic 0. Let (•(K⊗ Z −)) * : F K −→ A K be the functor defined by (•(K⊗ Z −)) * (F ) := F • (K ⊗ Z −). Then, (•(K ⊗ Z −)) * induces a fully faithful embedding F K ֒→ A K .
Proof: The functor •(− ⊗ Z K) : F K −→ A K induces for any two functors F and G in F K a mapping ϕ : We shall need to show that this mapping is an isomorphism.
Injectivity: Let η 1 and η 2 be two objects in Hom F K (F, G). Suppose ϕ(η 1 ) = ϕ(η 2 ). Observe that for any V ∈ Z − f ree we have and likewise for η 2 , satisfying that for any V and W and any ρ ∈ Hom Z (V, W ) one has Since •(− ⊗ Z K) is dense, η 1 and η 2 coincide on every object of K − mod, and so η 1 = η 2 . As a consequence, without any further hypothesis, .
) be a natural transformation. We need to show that there is a natural transformation η ′ ∈ Hom F K (F, G) so that ϕ(η ′ ) = η.
In the case K being a prime field of finite characteristic define for any V ∈ K − mod the mapping η ′ (V ) := η(P V ), where P V is a fixed chosen projective cover of V , so that K ⊗ Z P V = V .
In the case K being of characteristic 0, fix for any K-vector space V a free abelian subgroup P V so that K ⊗ Z P V = V . Define η ′ (V ) := η(P V ).
We need to show that η ′ is a natural transformation. Let ϕ ∈ Hom K (V, W ). Consider first the case of K being a prime field of finite characteristic. Since P V and P W are projective covers of V and W as abelian groups, there is aφ ∈ Hom Z (P V , P W ) so that K ⊗ Zφ = ϕ under the identification K ⊗ Z P V = V and K ⊗ Z P W = W . Since η is a natural transformation,Ĝ(φ) • η(P V ) = η(P W ) •F (φ). But, by definition,Ĝ(φ) = G(ϕ) andF (φ) = F (ϕ), as well as η(P W ) = η ′ (W ) and η(P V ) = η ′ (V ). So, η ′ is a natural transformation.
Suppose now that K is a field of characteristic 0. Since F K = P K in this case, we know that F (and G resp.) are polynomial laws transforming any linear mapping V −→ W into a linear mapping F (V ) −→ F (W ) (and G(V ) −→ G(W ) resp.) which depends polynomially in the coefficients of any matrix representation with respect to any fixed bases. We know that for any Z-linear mappingφ : P V −→ P W that the equation Since this equation holds evaluated in infinitely coefficients, since Z and K are both infinite, this above equation ( ‡) holds as polynomial equation.
Therefore, the equation hold as well for ϕ, since there the only difference is that the polynomials are evaluated not only on integer coefficients, but also on coefficients in K.
Since the equation holds as polynomials, this equation holds true also evaluated on K. Therefore This proves that η ′ is a natural transformation.
We are now concerned with the question when a polynomial functor Z−f ree −→ R−mod can be extended to a polynomial functor R − mod −→ R − mod by composing with the 'extending scalars' functor Z − f ree In other words we study the question when (•(K ⊗ Z −)) * is an equivalence A R ≃ F R . In order to prove this, by Lemma 1.2, one needs to show that (•(K ⊗ Z −)) * is dense as well.
We have to deal with mainly two cases: the case of R being a field of characteristic 0 and the case of R being a field of characteristic p. We shall see that fields with characteristic 0 behave more like characteristic ∞. The remarks at the beginning of this section on base change properties are particularly visible in the proof of the following lemma. Lemma 1.3. Let R be a field of characteristic 0 and let F : Z − f ree −→ R − mod be a polynomial functor of degree d. Then, F extends to a polynomial functorF : We have to show that this embedding is dense. Let F be a degree d polynomial functor in A d K . By Lemma 1.1 we know that for any n and m the functor F induces a degree d polynomial mapping with coefficients in K in the coordinate functions of matrices in Hom Z (Z n , Z m ). Moreover, F is a functor, that is F (αβ) = F (α)F (β) and F (id A ) = id A for any free abelian group A and any two composable morphisms of abelian groups α and β.
Let α : Z n −→ Z m and β : Z k −→ Z n . The equation F (αβ) = F (α)F (β) translates into an equation between the evaluation of the corresponding polynomials in each degree. Since Z and K are of characteristic 0, the polynomial equation holds if evaluated on infinitely many values, and so the polynomial equations actually holds as polynomials. Friedlander and Suslin remark in [16, remark after Definition 2.1] that this is actually equivalent to saying that F actually is a strict polynomial functorĜ ∈ P K of degree d. Now, for K infinite, P ≤d K ≃ F d K . ( [16]). So, actuallyĜ, and in turn F is a degree d polynomial functor in F d K . This proves the lemma. Then, for any homomorphism α : Z n −→ Z m one gets F (p · α) = 0.
Proof. Let M = Z n and N = Z m . We write p · α in the diagram where δ is the diagonal mapping and σ is the summation mapping, on M or on N respectively. Denote by F (i) the i th cross effect of the functor F . Since F is polynomial of degree at most p, for p abelian groups A 1 , . . . , A p , and this decomposition is functorial with respect to these groups. Moreover, since deg(F ) = p, and since F (0) = 0, one gets F (p−1) = 0. Since and π i is the projection of F (M ⊕ M ⊕ · · · ⊕ M ) to this direct factor. But now, for i < p, Corollary 1.5. Let F be a field of characteristic p and let F : Z − f ree −→ F − mod be polynomial functor of degree at most p − 1 which preserves the initial object, i.e. F (0) = 0. Then, for any two homomorphisms α : Z n −→ Z m and β : Proof. This is a consequence of the previous lemma and [10, p. 76, formula 8.5] and [10, by Lemma 1.4. Lemma 1.6. Let F be the prime field of characteristic p and let F : Z − f ree −→ F − mod be a polynomial functor of degree less or equal to p − 1. Then, F factors through the functor For any V ∈ F − mod choose P V a projective cover as abelian group. Then, P V is in Z-free. By the universal property of projective covers one has for any α ∈ Hom F (V, W ) a (non-unique)α ∈ Hom Z (P V , P W ) so that . We need to show that this gives a functorF : then choosing liftsα : P U −→ P V and β : P V −→ P W , one gets αβ −αβ lifts the 0-mapping. So, αβ −αβ ∈ p · Hom Z (P U , P W ) and again by Corollary 1.5 one has F ( αβ) = F (αβ).

A review on polynomial functors and functor cohomology
2.1. Polynomial functors are modules. Let R be a commutative ring. We know that by a result of Baues, Dreckmann, Franjou and Pirashvili [3] a polynomial functor of degree at most n from free abelian groups to R-modules is defined by giving R-modules F m (Z|Z| . . . |Z) for all m ≤ n and mappings for k ≤ m ≤ n − 1 satisfying the relations : Define the algebra Γ n R over R by a quiver with n vertices F m (Z|Z| . . . |Z) for any m with 1 ≤ m ≤ n and arrows h m subject to the relations ( * ). Observe that the relations above do not form a set of admissible relations. The third relation though should be read as the defining equation for the symbols t n i , and this way the relations ( * ) is a set of admissible relations. The result [3] of Baues et alii implies that Γ n R − mod is equivalent to the category of polynomial functors of degree at most n.
Since the ring homomorphismẐ p −→ F p induces an embedding F p − mod −→Ẑ p − mod, we get an induced embedding Γ n Fp − mod −→ Γ n Zp − mod which is also induced by the surjective ring homomorphism Γ n Zp −→ Γ n Fp .
Remark 2.1. We should mention that the description of [3] was recently generalised by Hartl, Pirashvili and Vespa [18] to show an isomorphism of the category of functors Z − f ree −→ Z − M od with the category of what they call pseudo-Mackey functors.
2.2. Some facts on functor cohomology. We shall give some facts that we will need from Franjou, Friedlander, Scorichenko and Suslin [15]. Basically, these results reduce the computation of extension groups between polynomial functors to questions between extension groups between strict polynomial functors. Let k be a commutative ring. Let F(k) be the category of functors k − mod −→ k − mod and let P(k) the category of strict polynomial functors between k-modules. Let F n = F n (k) be the category of degree n polynomial functors from k−mod to k−mod. If k = F q for q = p s , the field with q elements, we write F n (k) = F n (q). In this case, for any strict polynomial functor P in P(k) let P (m) be the functor twisted by the Frobenius endomorphism defined by Theorem 1. [15] Given any two homogeneous strict polynomial functors P and Q between F q -vector spaces for q = p s . If the degrees of P and Q are different and strictly smaller than q, then Ext * F (k) (P, Q) = 0. Moreover, if the degree of P and Q coincide, then lim Frobenius twisting decreases the 'degree of homological triviality' as is shown in a result of H.H. Andersen. For two homogeneous strict polynomial functors P and Q between F q -vector spaces of the same degree, for q = p s and for m ∈ N ∪ {0} we get The first of the two statements in Theorem 1 actually is due to Kuhn: This induces a decomposition of the category of functors between F q -vector spaces F(q) = q−1 i=0 F(q) i . Finally, a result due to Kuhn will be essential in the sequel. [23,24]) The injective envelope I Fp of the trivial module in the category of analytic functors F ω (F p ) from finite dimensional F p -vector spaces to F p -vector spaces is uniserial and the only composition factors in F p (F p ) are the two composition factors of soc 2 (I Fp )), where, as usual, soc 2 denotes the second layer in the socle series [ is the exterior algebra in one variable with generator in degree 2p − 1 and e h are generators in degree 2p h . Moreover, Remark that F p ⊗ Z id = id as functors on the category F p − mod. As a consequence, [12, 7.3] the following four term sequence is a non zero where S p is the degree p homogeneous part of the coinvariants under the S p action on the tensor algebra, and where S p is the degree p homogeneous part of the invariants of the tensor algebra.

On recollement diagrams
We remind the reader to the notion of a recollement diagram. A recollement diagram is given by three categories A, B and C with functors (1) (l, e) and (e, r) are adjoint pairs.
(3) i is a full embedding and e( (4) the adjointness morphisms e • r −→ id C and id C −→ e • l are isomorphisms. We denote a recollement diagram as above by (A, B, C, (e, l, r), (i, q, p)).
We shall give a result which is a special case of a recent result of Chrysostomos Psaroudakis and Jorge Vitoria [32]. We shall give our original proof below, since in our special case the proof is much easier than the proof for the general statement from [32].
Proof. Since e has a left and a right adjoint, e is exact. Therefore, e = Hom B (P, −) where P is a projective object. Since l and r are left and right adjoints to e, we get that l = P ⊗ End B (P ) −, that r = Hom End B (P ) (Hom B (P, B), −) and that C ≃ End B (P ). Since B is a Krull-Schmidt category, then up to Morita equivalence, one can choose P = Be ′ for an idempotent e ′2 = e ′ ∈ B and we get C is Morita equivalent to e ′ Be ′ .
The third condition in the definition of a recollement diagram implies that A can be identified with those B-modules M for which e ′ M = 0. Hence, This proves the proposition.
Another important observation is that, by the adjointness properties, l maps projective object in C to projective objects in B, and that r maps injective objects in C to injective objects in B.
3.1. Analyzing Schwartz' recollement for polynomial functors. Let q = p s for a prime p. Then, using the notation of Section 2.2, following Kuhn [ where S k is the symmetric group on k elements and where n(λ) := λ 0 + · · · + λ s−1 . Moreover, the functor F n (q) −→ n(λ)=n F q S λ − mod is representable by id λ and for a partition λ = (λ 0 ≥ · · · ≥ λ s−1 ), we set id λ := ⊕ s−1 j=1 id ⊗λ j . Remark 3.2. Hence, in case s = 1 and n < p, the recollement becomes since F p S n is semisimple, since F p is a splitting field, and therefore its module category is equivalent to a direct product of copies F p − mod.
In case s = 1 and n = p, the recollement becomes We have an immediate consequence of the above result. Lemma 3.3. For any simple polynomial functor F there is a strict polynomial functorF such that the forgetful functor, which assigns to every strict polynomial functor its polynomial functor by evaluating the polynomial as mapping, mapsF to F .
Proof. This is done by induction on the degree. The simple objects in F d (q) are in bijection with the union of the simple objects in F d−1 (q) and the simple objects in F q S d − mod. Now, any simple F q S d -module is image of a simple module of the Schur algebra S Fq (d, d) under the Schur functor as is a classical fact. Since the category of degree d strict polynomial functors is equivalent to the category of modules over the Schur algebra S(d, d), the simple objects in F d (q) of degree d are images of a strict polynomial functor. By induction, the simple objects of degree less than d are images of strict polynomial functors.

Recollement for
For the category A Fp we get a similar recollement diagram.
Remark 3.4. The treatment we give for Section 3.2 was suggested by the referee, and follows paths which are not easily documented. They appeared in the special case of degree 2 in introductory remarks in [11,12]. An alternative proof can be obtained, and actually this was our initial approach, using Piriou's thesis [30,31] along its first chapter. I am very grateful to the referee for this useful hint.
The recollement diagram was proved in a more general situation in the meantime by Djament and Vespa [ We shall mainly work with properties of adjoint functors. Lemma 3.5 below seems to be a well-known result. It appears maybe for the first time in the proof of [12, Lemma 0.4] without further reference. In order to keep the presentation as self-contained as possible, we shall provide a short proof. We shall need a very simple observation.
Lemma 3.6. Let A and B be two polynomial functors of degree at most n and let be an exact sequence of functors. Then, the degree of C is at most n as well. Moreover, taking cross effects is exact.
Proof: We get a commutative diagram and the snake lemma implies that is exact. Induction on the degree gives the result.
We shall now prove a general statement on functor categories which is an adaption from Franjou [11, Section 1].
Lemma 3.7. Let R be a commutative ring and let G is an object in A n R . Then we get Hom A R ((R ⊗ id ⊗n , G) = G (n−1) .

Proof. Let
n − Z − f ree = (Z − f ree) × · · · × (Z − f ree) n copies be the category with objects F 1 × · · · × F n for F i being an object in Z − f ree for each i ∈ {1, . . . , n} and morphisms being for all objects F i , G j of Z − f ree. Composition of morphisms is given by composition of mappings in Z − f ree. We shall define (F 1 , . . . , F n ) := F 1 ⊕ · · · ⊕ F n and ∆ n (F ) := (F, . . . , F ). Then Π n is left and right adjoint to ∆ n as is easily seen, almost by definition. Now, we consider the category n − A R of functors n − Z − f ree −→ R − M od with morphisms being natural transformations. We get functors By Lemma 3.5, we see that the functor Π n is left and right adjoint to ∆ n . Put and obtain id ⊗n = ⊠ n • ∆ n and R ⊗ id ⊗n = (R ⊗ ⊠ n ) • ∆ n . But then By definition of the cross effect of a functor we get Moreover, using that ι n I is left adjoint to π n I , in case m = n we get ..,im} , G (m−1) ) = 0 since at least one of the factors in the tensor product is 0. Therefore Since G is of degree at most n, the functor G (n−1) is additive in each variable. Hence its value is given by the R-module G (n−1) (Z| . . . |Z). However, (R ⊗ id ⊠ n )(Z| . . . |Z) = R and Hom R (R, G (n−1) (Z| . . . |Z)) = G (n−1) (Z| . . . |Z). Hence a natural transformation in Hom n−A R ((R ⊗ ⊠ n ), G (n−1) ) induces an element in G (n−1) (Z| . . . |Z). On the other hand, any element in G (n−1) (Z| . . . |Z) induces a natural transformation in Hom n−A R ((R ⊗ ⊠ n ), G (n−1) ). Therefore Hom A R (R ⊗ id ⊗n , G) = G (n−1) as claimed and we obtain the statement. Proposition 3.8.
Proof. We shall show the following auxiliary lemma needed for the proof of the proposition.
Lemma 3.9. F p ⊗ id ⊗n is projective and injective in A n Fp . Proof. Taking cross effects is exact. By Lemma 3.7 we see that F p ⊗ id ⊗n is projective. Using that the duality D on functors F : Z − f ree −→ F p − M od given by and observing that F p ⊗ id ⊗n is self-dual, we obtain that F p ⊗ id ⊗n is injective as well.
We need to prove that . But this is clear by Lemma 3.7 Since F p ⊗ id ⊗n is projective there is a right adjoint and a left adjoint to Hom A Fp (F p ⊗ id ⊗n , −), namely the functor M → (F p ⊗ id ⊗n ⊗ M ) Sn is the right adjoint, and the functor M → (F p ⊗ id ⊗n ⊗ M ) Sn is the left adjoint.
Moreover, the unit and the counit of the adjunctions induce the identity on F p S n -mod. This can be done literally as in Piriou [ This shows Proposition 3.8.
As a consequence we show the following lemma.
Therefore the number of simple objects in A n Fp and in F n Fp coincides. The statement is clear for n ≤ p−1 by Lemma 1.6. Let n ≥ p and let S be a simple object in F n Fp . We may suppose, using the induction hypothesis that e F n (S) = 0. Then, suppose X is a simple and proper subobject of i n S. Since e A n is exact, e A n (X) is a subobject of e A n (i n S) = e F n (S). Since e F n is exact, e F n (S) is simple. So, e A n (X) is either 0 or isomorphic to e F n (S). Since X is a proper subobject of i n S we see that e A n (X) = 0, and hence X ∈ A n−1 . Since the simple objects of F n−1 and of A n−1 coincide by the induction hypothesis, X is a proper non zero subobject of S. This gives the contradiction.
We have seen in Lemma 1.6 and Remark 3.2 that

Short Review of Brauer tree algebras
Proposition 3.8 shows that the representation theory of the symmetric group is closely related to A n (p) we shall need some information from group representations. We give Benson [4, Section 4.18 and Section 6.5] as a general reference.
Let k be a field of characteristic p > 0 and let G be a finite group so that k is a splitting field for G, i.e. the endomorphism ring of each simple kG-module is k. Then it is well-known that there are only a finite number of isomorphism classes of indecomposable kG-modules if and only if the Sylow p subgroups of G are all cyclic. In particular the symmetric group S p is a group with every Sylow p subgroup being cyclic.
An indecomposable ring-direct factor of kG is called a p-block of G, and the representation theory of a block B is completely understood in case the Sylow p subgroups of G are cyclic. The algebras occurring in this case are so-called Brauer tree algebras.
We explain briefly the theory of Brauer tree algebras and refer to Benson [4, Section 4.18 and Section 6.5] or to Auslander-Reiten-Smalø [1, Section X.3] for more details.
A Brauer tree is a finite connected tree Γ, i.e. a connected graph without cycles and multiple edges, together with additional structure as we shall describe now: Let Γ 0 be the vertices of Γ and let Γ 1 be the edges of Γ. For each e ∈ Γ 1 let v 1 (e) ∈ Γ 0 and v 2 (e) ∈ Γ 0 be the vertices adjacent to e.
• Then for each vertex v ∈ Γ 0 let E v be the edges adjacent to v, and suppose given a transitive permutation σ v ∈ S Ev . • Moreover, we choose one vertex v 0 ∈ Γ v and an integer µ > 0, called the exceptional multiplicity. A Brauer tree algebra B is then a finite dimensional symmetric k-algebra so that • the isomorphism classes of simple B-modules are parameterised by Γ 1 , denoting by S e a simple B-module whose isomorphism class corresponds to the edge e ∈ Γ 1 , • the projective cover P e of S e has the property that rad(P e )/soc(P e ) = U v 1 (e) ⊕U v 2 (e) , where U v 1 (e) and U v 2 (e) are both uniserial modules.
is not the exceptional vertex, then the composition length of U v j (e) is equal to |E v j (e) | − 1 -if v j (e) is the exceptional vertex, then the composition length of U v j (e) is equal to µ · |E v j (e) | − 1 • for each i, for which the i-th radical quotient is not 0, and One observes that in case µ = 1, then there is no difference between the exceptional vertex and a non exceptional vertex. Hence, in this case we do not need to fix an exceptional vertex and we say that the Brauer tree has no exceptional vertex if the exceptional multiplicity µ is 1.
Observe further that if a vertex v is a leaf, i.e. E v = {e}, then the projective cover of S e is uniserial and U v = 0.
A particular case is when each vertex has at most 2 edges adjacent to it, or in other words if |E v | ≤ 2 for all v ∈ Γ 0 . In this case we call the Brauer tree a stem. If moreover the Brauer tree has no exceptional vertex then the tree can be visualised as The projective indecomposable modules P are then all of Loewy length 3, i.e. P has simple top, simple socle and rad(P )/soc(P ) is semisimple of composition length at most 2.
This situation occurs for F p S p which has several blocks for p ≥ 5, one of which is a Brauer tree algebra associated to a stem without exceptional vertex and p − 1 edges, and all the other blocks are simple algebras.

The structure of polynomial functors modulo p
The situation of polynomial functors of degree p is different from those of degree n < p. We are going to describe in this section their structure completely. From now on we assume that p ≥ 5 since the representation theory of F 2 S 2 and of F 3 S 3 is slightly different from the case p ≥ 5.
Remark 5.1. Let us recall the relation between the Schur algebra S Fp (p, p) and the group algebra F p S p . We shall use the structure of the Schur algebra and the corresponding Hecke algebra as it is proved in [20,Corollary 1.3]. The group algebra of the symmetric group is a special case of the Hecke algebra. For general informations on Schur algebras and Brauer tree algebras see [17] and [21]. The algebra S Fp (p, p) admits p simple modules S 1 , . . . , S p−1 , S p whereas the group algebra F p S p admits p − 1 simple modules S ′ 1 , . . . , S ′ p−1 . The projective indecomposable S Fp (p, p)-modules have composition series whereas the projective indecomposable F p S p -modules have composition series Lemma 5.2. Let L be the simple polynomial functor in F p Fp so that L is of degree p and so that L corresponds to the trivial representation of F p S p . Then, Ext 1 F Fp (id, L) = 0 = Ext 1 F Fp (L, id). Moreover, if Ext 1 F Fp (id, S) = 0 or 0 = Ext 1 F Fp (S, id) for a simple degree p-functor S, then L ≃ S.
Proof. Let S 1 be the simple polynomial functor of degree 1. The identity functor id is trivially of degree 1 and simple, which implies S 1 = id. Then, Ext 1 F p (L, id) is not necessarily zero for L being an irreducible polynomial functor of degree p. Now, since we are working over F p , we get that id (1) ≃ id as polynomial functor, but not as strict polynomial functor. As strict polynomial functor, I (1) is of degree p.
Theorem 2 implies that there is only one simple functor L of degree p with Ext 1 F p (L, id) = 0. Proposition 2.2 in connection with Theorem 1, imply that L is the simple functor corresponding to the trivial F p S p -module, since this is the module which has an extension with the unique simple module of the Schur algebra S Fp (p, p) which is not a simple F p S p -module (cf Remark 5.1).
This implies that . Since the category of degree p strict polynomial functors is equivalent to the category of modules over the Schur algebra S Fp (p, p), one sees that (1) ) ≃ Ext 1 S Fp (p,p) (V, I 0 ) for I 0 being the simple S Fp (p, p)-module corresponding to the p singular partition of p and V being the simple S Fp (p, p)-module corresponding to L. Finally, it is a classical fact (cf e.g. [20]) that Ext 1 S Fp (p,p) (I 0 , V ) = 0 or Ext 1 S Fp (p,p) (V, I 0 ) = 0 if and only if V corresponds to the trivial module of the symmetric group, and in this case, the dimension of Ext 1 S Fp (p,p) (I 0 , V ) and of Ext 1 S Fp (p,p) (V, I 0 ) is 1. This proves the statement. Using the embedding F p Fp ֒→ A p Fp from Lemma 1.2, the functor L of F p Fp induces a functor L(F p ⊗−) in A p Fp . In order to avoid additional notational burden we shall denote this functor L(F p ⊗ −) by L as well.
We get as a corollary the following statement.
Proof. We know that Ext 1 is simple by Lemma 3.10. This sequence is non split since the functor pre-composing with F p ⊗ Z id is a fully faithful embedding. Hence Ext 1 id, L) = 0. Therefore, L is a direct factor of the top of the radical of the projective cover of Actually, the argument of Lemma 5.2 gives another slightly different statement.
Lemma 5.4. Let X and Y be two simple functors of degree at most p. Then, Proof. By Lemma 3.10 we know that the simple functors X and Y can be considered to lie in F p Fp . Proposition 2.2 in connection with Theorem 1 imply this result. Denoting by ρ p (i) the number of p-regular partitions of i, the algebra Γ p Fp is Morita equivalent to a direct product of p−1 i=1 ρ p (i) − 1 copies of F p and of an indecomposable ring Γ p Fp,0 . By the recollement diagram preceding Lemma 3.3 this ring Γ p Fp,0 has a projective module P = Γ p Fp,0 · e so that the endomorphism ring of P is Morita equivalent to the Brauer tree algebra corresponding to F p S p . There is a projective indecomposable Γ p Fp,0 -module P 0 so that P 0 ⊕ P is a progenerator of Γ p Fp,0 and the endomorphism ring of P 0 ⊕ P is basic and Morita equivalent to Γ p Fp,0 . Moreover, Proof. For any degree n polynomial functor F one gets Hom A Fp (proj m n , F ) ≃ F (Z m ). This is true if one does not factorizes the power of the augmentation ideal, and since all functors are of degree at most n, each natural transformation from proj m ∞ to F is zero on I n+1 (see [12, Section 1]). So, Hom A (proj m n , −) is exact, as evaluation on exact sequences of functors is exact. Hence, proj m n is a projective object in A n Fp and since Hom A (proj m n , id) = id(Z m ) = Z m = 0, the projective cover of the reduction modulo p functor is a direct summand in proj m n . Remark 5.7. The situation is different for F Fp . Indeed, the functor F p [Hom Fp (F m p , −)]/I n+1 is projective in F n Fp again since . But, automatically I p = 0 for m = 1 and the evaluation at F p in this case, so that the endomorphism ring of F p [Hom Fp (F p , −)]/I n+1 is an ℓ-dimensional vector space where ℓ = min(p, n).
Recall the embedding F n Fp −→ A n Fp given by pre- As we will see, the projective indecomposable cover Fp is a direct factor of the functor F p [Hom Z (Z, −)]/I p+1 and this is the only indecomposable functor which is not in the image of the embedding F p Fp −→ A p Fp .
Lemma 5.8. The projective functor proj 1 n has one composition factor of degree d for each 0 ≤ d ≤ n for all n ≤ p − 1. In particular proj 1 n contains a simple constant functor as a direct summand.
Proof. First, proj 1 n (0) = F p , and so the simple functor of degree 0 is a direct factor of proj 1 n . Furthermore, proj 1 n −→ → proj 1 n−1 for trivial reasons. Moreover, proj 1 n is of degree n and not of degree n − 1. We compute that End A n Fp (proj 1 n ) = proj 1 n (Z) = F p [Z]/I n+1 is an n + 1dimensional vector space. Moreover, by Lemma 1.6 and Remark 3.2 the projective module Fp is a semisimple category. Observe that proj 1 n has exactly one composition factor more than proj 1 n−1 . This composition factor is of degree n since is the projective object corresponding to degree n polynomial functors in the recollement diagram.
Recall that L denotes the simple functor in A p Fp mapping to the trivial F p S p -module in the recollement diagram.
Fp is uniserial with top and socle being F p ⊗ Z id and with rad(P Fp⊗ Z id )/soc(P Fp⊗ Z id ) ≃ L. Moreover, Proof. We shall divide the proof into various claims.
On the other hand, we know from the recollement diagram that (F p ⊗ Z id) ⊗p is the projective cover of the simple modules coming from F p S p , that is those of degree p. Now, each projective indecomposable module of F p S p has the property that the top of this module is isomorphic to the socle of this module and that the top and the socle of this projective indecomposable module are different. Hence, suppose a simple polynomial functor of degree p would be in the top of proj 1 p , then let Q be its projective cover in A p Fp . Further, Q is a direct summand of (F p ⊗ Z id) ⊗p . Since the top of (F p ⊗ Z id) ⊗p is isomorphic to the socle of (F p ⊗ Z id) ⊗p , the above homomorphism space would be at least 2-dimensional, corresponding to the mapping of proj 1 p on the top and on the socle of (F p ⊗ Z id) ⊗p . Proof. We know by Lemma 1.6 that A p−1 Fp is semisimple. Moreover, the category of constant functors is a direct factor in the category of polynomial functors. Using Lemma 3.6 this shows the statement.
We denote by P V the projective cover of the functor V in A p Fp .
Claim 5.12. proj 1 p ≃ S 0 ⊕ S 2 ⊕ S 3 ⊕ · · · ⊕ S p−1 ⊕ M for simple projective functors S i of degree i and the projective cover M of the functor F p ⊗ Z id.
We know already that proj 1 p−1 is a quotient of proj 1 p and that this is a semisimple functor with p − 1 direct factors. So, every direct summand of the semisimple functor proj 1 p−1 is a direct factor of the head of proj 1 p . Denote by S 0 , S 1 , S 2 · · · S p−1 the simple direct factors of proj 1 p−1 and let S i be of degree i. Then, since the degree 0 functors split off in any case, P S 0 = S 0 . Moreover, S 0 ⊕ P S 0 ⊕ P S 1 ⊕ · · · ⊕ P S p−1 is a direct factor of proj 1 p . We need to study the functors P S i . If all the composition factors of P S i for an i ≤ p − 1 are of degree p − 1 at most, then by Lemma 3.6 we get that P S i is of degree at most p − 1 as well. Since the category A p−1 Fp is semisimple, we get P S i = S i . Suppose that a degree p simple polynomial functor S is a composition factor of P S i 0 . Then, since A p−1 Fp is semisimple, again by Lemma 3.6, Ext (proj 1 p , F p ⊗ Z −) is a one-dimensional F p -vector space, this happens for precisely one i 0 ∈ {1, 2, . . . , p − 1}. We already know by Claim 5.3 that i 0 = 1 and S is the simple module corresponding to the trivial representation of F p S p . This proves the claim.
Claim 5.13. For the projective cover P L of L we get rad(P L )/rad 2 (P L ) ≃ (F p ⊗ id) ⊕ L 2 , where L 2 is simple of degree p, L 2 ≃ L and rad 2 (P L ) ≃ L as well as rad 3 (P L ) = 0.
Remark 5.14. We do not claim here that soc(P L ) is simple. However, the radical layer structure of P L can be described by and where it is not clear if soc(P L ) is simple and isomorphic to L or if the socle is isomorphic to L ⊕ L 2 or to L ⊕ F p ⊗ id. But Hom A p Fp (F p ⊗ id ⊗p , P L ) is the projective cover of the trivial F p S p -module. This projective cover is uniserial with composition series Since L 2 is simple of degree p, its image in F p S p is given by the known module structure of F p S p . In particular, L ≃ L 2 . This shows that the only uniserial module of length 3 which is a quotient of P L , if there is any, can have composition series In particular, soc(P L ) ≃ L ⊕ L 2 .
Proof of Claim 5.13. By Claim 5.3 we know that (F p ⊗ id) is composition factor of top(rad(P L )) and since Hom A p Fp (proj 1 p , F p ⊗ Z −) is a one-dimensional F p -vector space we know that it has multiplicity 1. Since the image of P L in F p S p − mod is uniserial with top and socle L and simple rad(P L )/soc(P L ) ≃ L 2 , we have the above structure.
Claim 5.15. The projective cover M = P Fp⊗id of F p ⊗ Z id has rad(M )/rad 2 (M ) ≃ L.
Proof. By Claim 5.3 we know that L is a direct factor of top(rad(M )). Since A p−1 Fp is semisimple, using Lemma 3.6 we see that no simple functor of degree p − 1 at most can be a direct factor of top(rad(M )). Suppose top(rad(M )) has a second simple direct factor T of degree p. Then, Ext 1 Simple functors are self-dual (cf as above [23] for functors in F Fp and by Lemma 3.10 for simple functors in A Fp ). So, Ext 1 But, we have seen in Corollary 5.3 that there is one simple functor of degree p with a non trivial extension group with F p ⊗ Z id, namely L. Moreover, since proj 1 p contains the projective cover of the simple degree 1-functor as a direct factor (see Claim 5.12), and since by Lemma 5.11 this is the only degree where non trivial first extension groups can occur, we see that T ≃ L.
Suppose L ⊕ L|rad(M ). Then, Ext 1 is two-dimensional at least, and again by the self-duality of the simple functors, Ext 1 Therefore, F p ⊗ Z id occurs twice in top(rad(P L )). Since M is a direct factor of proj 1 p , the space Hom A p Fp (proj 1 p , (F p ⊗ Z id) ⊗ p ) would be two-dimensional at least. This contradiction shows that rad(M ) has simple top L. Hence, top(rad(M )) ≃ L.
Remark 5.16. The radical layer structure of M is therefore given by (F p ⊗ id) L rad 2 (M ) and by Claim 5.13 we get rad 3 (P L ) = 0, and therefore we obtain rad 4 (M ) = 0.
Claim 5.17. For the projective cover M of F p ⊗ Z id we get that top(rad 2 (M )) ≃ F p ⊗ Z id.
Proof. Degree p functors can only have extensions with degree p-functors or degree 1functors by Lemma 5.11. Moreover, the structure of P L implies that we get that top(rad 2 (M )) is a direct summand of top(rad(P L )), whence is isomorphic to either 0, or to L 2 (which is defined in Claim 5.13), or to F p ⊗ Z id, or to L 2 ⊕ (F p ⊗ Z id).
Suppose L 2 ⊕ (F p ⊗ Z id) ≃ top(rad 2 (M )). We shall use the fact that by Claim 5.13 we know the structure of P L .
We get two possibilities for the projective resolution of F p ⊗ Z id. Either In the second case, Ext 2 A p Fp ((F p ⊗ Z id), L) = 0. By the self-duality of the simple functors,

Ext 2
A p Fp (L, (F p ⊗ Z id)) = 0. Our information is sufficient for being able to write down the first terms of the projective resolution of L, for some projective P L 3 , for some simple object L 3 of degree p, given by the known projective resolution of the trivial F p S p -module. Since p = 2, we get L 3 ≃ L. In an case Hom(P L ⊕ P L 3 , (F p ⊗ Z id)) = 0, and therefore Ext 2 is exact, the projective dimension of (F p ⊗ Z id) is 1. But, we know by Theorem 3 and the example following it, that Ext 2 is at least 2.
If rad 3 (M ) = 0, then we get a non split exact sequence Our knowledge of the various projective covers of simples is sufficient to write down the first terms of the projective resolution of L 2 . We get for projective objects P L 4 and P L 5 corresponding to degree p simple functors L 4 and L 5 , given by the known projective resolution of the trivial F p S p -module. Moreover, L 3 and L 5 are both different from L 2 , since p ≥ 5. This implies Ext 3 Hence, L 2 ≃ top(rad 2 (M )).
We still have the possibility that rad 2 (M ) = 0. But again, this would imply that This proves the claim. But this contradicts the structure of P L as described in Claim 5.13 and in particular Remark 5.14.
This proves the claim.
Examining what we showed implies that proj 1 p is as stated in Proposition 5.9.
We now come to our first main result in describing the structure of Γ Fp,0 . For the relevant definitions on Brauer tree algebras we refer to Section 4. Theorem 4. Γ Fp,0 is a Brauer tree algebra over F p without exceptional vertex and associated to a stem with p edges.
Proof. The case p ≤ 3 is a consequence of Drozd's results. By Proposition 5.9 we know that the projective cover of the functor F p ⊗ Z − is uniserial with top and socle F p ⊗ Z − and with second layer V , where V is the simple functor corresponding to the trivial F p S p -module.
We know furthermore that except the projective cover of F p ⊗ Z − only the projective indecomposable functor P L has a composition factor F p ⊗ Z − and that this composition factor is a direct summand of top(rad(P L )).
Since we know that the principal block of F p S p is a Brauer tree algebra without exceptional vertex associated to a stem with p vertices, this means that the we only need to show that F p ⊗ Z − is not in the socle of P L , since the only basic algebra with the composition series as a Brauer tree algebra associated to a stem is actually a Brauer tree algebra associated to a stem.
For this we use the duality D on the category of polynomial functors. The projective indecomposable functor P L is a direct factor of F p ⊗ Z id ⊗ Z · · · ⊗ Z id p factors , since this is the projective cover of all the degree p simple functors.
It is clear that F p ⊗ Z id ⊗ Z · · · ⊗ Z id p factors is self dual. Since all the simple functors are self-dual, This proves the Theorem.
6. Lifting to characteristic 0 6.1. Lifting Brauer tree algebras to orders. Proposition 6.1. Let R be a complete discrete valuation ring with residue field k and field of fractions K. Let B be a Brauer tree algebra over k associated to a Brauer tree which is a stem without exceptional vertex. Let Λ be an R-order. Then, for any proper two-sided ideal I = 0 of B we get that Proof. We shall first suppose that Λ is indecomposable and that I ≤ rad(B). Let S 1 , S 2 , . . . , S n be representatives of the simple B-modules. The projective cover P i of S i has then a composition series where soc(P i ) ≃ S i and for all i ∈ {2, 3 . . . , n − 1}, rad(P 1 )/soc(P 1 ) ≃ S 2 and rad(P n )/soc(P n ) ≃ S n−1 . Denote B := B/I. Since I ≤ rad(B), we get B has the same number of simple modules, and moreover, the simple B-modules and the simple B-modules coincide by the epimorphism B −→ B. Therefore the projective indecomposable B-modules are P i := B ⊗ B P i for i ∈ {1, 2, . . . , n}. Moreover, P i is the projective cover of S i as B-module.
Extending k if necessary, we may assume that the field of fractions K of R is a splitting field for Λ, since extending K does not decrease the rank of the Grothendieck group, using the Noether-Deuring theorem. Since k is a splitting field for B and for B/I, and since k ⊗ R Λ ≃ B, the Cartan matrix of B is symmetric (cf e.g. [ we see that the composition length of P i differs from the composition length of P i by at most 1. Indeed, if this would not be the case, then the composition matrix of B would be decomposable into at least two blocks and B would be decomposable as algebra. But, since Λ is indecomposable, so is k ⊗ R Λ ≃ B. So, I ≤ soc(B).
Since R is complete, we may assume that B and Λ are both basic algebras. Let Q i be the projective cover of P i as Λ-module. Hence, k ⊗ R Q i ≃ P i . Since we know that KQ i and KQ j do not have a character in common if |i − j| > 1 and do have one character in common if |i − j| = 1. Since dim k Hom B (P i , P i ) ∈ {1, 2}, the character of KQ i for all i ∈ {1, . . . , n} is a sum of at most two irreducible characters, and in case of two characters these are non isomorphic. Now, since Otherwise, the character of KQ i 0 would be a constituent of KQ i 0 +1 , of KQ i 0 and of KQ i 0 −1 , which implies then Hom KΛ (KQ i 0 +1 , KQ i 0 −1 ) = 0. This would give a contradiction. This gives that I equals either S 1 or S n or S 1 ⊕ S n .
Suppose now that S 1 is a direct factor of I (as left module) and suppose rank Z (K 0 (K ⊗ R Λ)) > rank Z (K 0 (B)).
We shall prove that S 1 is not a direct factor of I. By symmetry, then neither S n is a direct factor of I, and therefore, I = 0. Under these hypotheses, dim(P 1 ) = 2 and as a consequence also dim R (Q 1 ) = 2. So, for the Wedderburn components corresponding to K ⊗ R Q 1 in K ⊗ R Λ we have two possibilities. Either K ⊗ R Q 1 is a sum of two one-dimensional characters or K ⊗ R Q 1 is isomorphic to one two-dimensional character. Since K⊗ R Λ admits at least n+1 irreducible characters, K⊗ R Q 1 must have two constituents. So, K ⊗ R Q 1 is a sum of two one-dimensional characters.
But now, let {e i |i ∈ {1, . . . , n}} be an orthogonal set of primitive idempotents with Λe i ≃ Q i . Then, since e 1 and e 2 must be non zero on the common Wedderburn component of KQ 1 and KQ 2 , we get e 2 e 1 = 0. This is a contradiction to the fact that e 1 and e 2 are orthogonal.
We have to deal with the case Λ being decomposable. The structure of B implies that in this case, B/I is a direct product of algebras we have dealt with in the earlier case, and copies of k. By induction on the number of simple modules of each indecomposable factor the result holds for each of the pieces as well. Summing up for all of these pieces, we get the desired result.
Finally, we have to deal with the case that I is not contained in rad(B). The same argument as for Λ decomposable applies here as well.
This proves the Proposition. admits at most p simple modules. This proves the proposition. Proof. This is a consequence of Proposition 6.2, Theorem 4 and Proposition 6.
is a Green order with p isomorphism classes of indecomposable projective modules.
Remark 6.4. Roggenkamp described the orders Λ which admit a set of lattices with periodic projective resolutions encoded by a Brauer tree ( [34], see also [21]). Roggenkamp called these orders Green-orders and he described their structure in great detail.
Proof of the theorem. -modules M i . These periodic resolutions are encoded by the same Brauer tree. It remains to show that the modules M i are lattices. Actually, this is automatic. Indeed, since the resolution is periodic, each module M i is also a kernel of a differential, after a complete period of the periodic projective resolution. Hence, Λ p Zp,0 is a Green order with Brauer tree being a stem with p edges and without exceptional vertex.
We have to show that the maximal overorder of the Green order Λ p Zp,0 is a direct product of matrix rings overẐ p and that the image of Λ p Zp,0 in each of the matrix rings is a hereditary order.
The first part is clear sinceQ p is a splitting field of Λ p Zp,0 , and Λ p Zp,0 can be embedded into a direct product of matrix rings over the ring of integers inQ p (see e.g. [33] ⊗Ẑ p F p is a Brauer tree algebra without exceptional vertex, x 1 = x 2 = · · · = x p−1 and as a consequence, if one of the matrix rings is hereditary, all of them are hereditary. The structure theory of Green orders (cf Roggenkamp [34]; see also [21,Section 4.4]) and of hereditary orders, (cf e.g. Reiner [33]) then gives the statement.

Define a functor
where we use again that the functorẐ p ⊗ Z ( p j=1 id) : Z−f ree −→Ẑ p −mod carries a natural Z p -linear S p action. Denote for notational simplicity E := Hom A p Zp Ẑ p ⊗ Z ( p j=1 id), − . By Lemma 3.7 this functor is just the p − 1-th cross effect, and by Lemma 3.6 this functor E is exact.
Since E is exact, and since by definition E is represented by P :=Ẑ p ⊗ Z ( p j=1 id), this object P is projective. Let e be an idempotent in Γ p We need to show that End AẐ p (P ) ≃Ẑ p S p , where the action is given by permutation of components in the tensor product. Once this is done, we know that End AẐ p (P ) ≃ e·Γ p Zp ·e ≃ Z p S p and we observe that for all idempotents in Γ p

Zp
we get that this product e · Γ p Zp · e is again a product of Green orders with the same order of congruences. SinceẐ p S p is a Green order with congruences modulo p only, we get that x = 1.

Identifying the lattices as functors
We shall identify the indecomposable functors of A p Zp which correspond to indecomposable ΓẐ p -lattices. We call such polynomial functors 'polynomial lattices'. The structure of ΓẐ p,0 implies that there are exactly 3p − 2 indecomposable ΓẐ p ,0 lattices. Indeed, the indecomposable lattices are the p projective indecomposable modules P 1 , P 2 , . . . , P p , the p − 1 kernels of any fixed non zero homomorphism P i −→ P i+1 for i = 1, 2, . . . , p−1, as well as the p−1 kernels of any fixed non zero homomorphism P i −→ P i−1 for i = 2, 3, . . . , p. Therefore, there are exactly 3p − 2 indecomposable 'lattices' of exactly degree p polynomial functors in A p Zp . Moreover, the proof of Theorem 5 shows that exactly the projective indecomposable ΓẐ pmodules of degree d ∈ {2, . . . , p − 1} will give rise to indecomposable lattices. Denote by ρ(k) the number of partitions of k into non zero integers, we get the following corollary to Theorem 5.