Automorphisms of finitary incidence rings

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Дата:	2010
Автор:	Khripchenko, N.
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Опубліковано:	Інститут прикладної математики і механіки НАН України 2010
Назва видання:	Algebra and Discrete Mathematics
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Цитувати:	Automorphisms of finitary incidence rings / N. Khripchenko // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 2. — С. 76–95. — Бібліогр.: 9 назв. — англ.

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spelling	irk-123456789-1546042019-06-16T01:31:50Z Automorphisms of finitary incidence rings Khripchenko, N. 2010 Article Automorphisms of finitary incidence rings / N. Khripchenko // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 2. — С. 76–95. — Бібліогр.: 9 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:18E05, 18B35, 16S50, 16S60,16G20, 08A35. http://dspace.nbuv.gov.ua/handle/123456789/154604 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
format	Article
author	Khripchenko, N.
spellingShingle	Khripchenko, N. Automorphisms of finitary incidence rings Algebra and Discrete Mathematics
author_facet	Khripchenko, N.
author_sort	Khripchenko, N.
title	Automorphisms of finitary incidence rings
title_short	Automorphisms of finitary incidence rings
title_full	Automorphisms of finitary incidence rings
title_fullStr	Automorphisms of finitary incidence rings
title_full_unstemmed	Automorphisms of finitary incidence rings
title_sort	automorphisms of finitary incidence rings
publisher	Інститут прикладної математики і механіки НАН України
publishDate	2010
url	http://dspace.nbuv.gov.ua/handle/123456789/154604
citation_txt	Automorphisms of finitary incidence rings / N. Khripchenko // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 2. — С. 76–95. — Бібліогр.: 9 назв. — англ.
series	Algebra and Discrete Mathematics
work_keys_str_mv	AT khripchenkon automorphismsoffinitaryincidencerings
first_indexed	2025-07-14T06:39:14Z
last_indexed	2025-07-14T06:39:14Z
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fulltext	A D M D R A F T Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 9 (2010). Number 2. pp. 76 – 95 c© Journal “Algebra and Discrete Mathematics” Automorphisms of finitary incidence rings Nikolay Khripchenko Communicated by B. V. Novikov Abstract. Let P be a quasiordered set, R an associative unital ring, C(P,R) a partially ordered category associated with the pair (P,R) [6], FI(P,R) a finitary incidence ring of C(P,R) [6]. We prove that the group OutFI of outer automorphisms of FI(P,R) is isomorphic to the group OutC of outer automorphisms of C(P,R) under the assumption that R is indecomposable. In particular, if R is local, the equivalence classes of P are finite and P = ⋃ i∈I Pi is the decomposition of P into the disjoint union of the connected com- ponents, then OutFI ∼= (H1(P ,C(R)∗)⋊ ∏ i∈I OutR)⋊OutP . Here H1(P ,C(R)∗) is the first cohomology group of the order complex of the induced poset P with the values in the multiplicative group of central invertible elements of R. As a consequences, Theorem 2 [9], Theorem 5 [2] and Theorem 1.2 [8] are obtained. Introduction Recall that an incidence algebra I(P,R) of a locally finite poset P over a ring R is the set of formal sums of the form α = ∑ x≤y α(x, y)[x, y], where α(x, y) ∈ R, [x, y] = {z ∈ P \| x ≤ z ≤ y} is a segment of the partial order. The study of the automorphism group of an incidence algebra was 2000 Mathematics Subject Classification: 18E05, 18B35, 16S50, 16S60, 16G20, 08A35. Key words and phrases: finitary incidence algebra, partially ordered category, quasiordered set, automorphism. A D M D R A F T N. Khripchenko 77 started by Stanley [9]. He showed that the group of outer automorphisms of an incidence algebra of a finite poset P over a field R is isomorphic to the semidirect product (R∗)n ⋊OutP where R∗ is the group of invertible elements of the field R, OutP is the group of outer automorphisms of the poset P and n is such that (R∗)n ∼= H1(P,R∗). This result was first generalized by Baclawski [2] (P is a locally finite quasiordered set, R is a field), then by Scharlau [8] (P is a finite quasiordered set with 0 or 1, R is a division ring, finite-dimensional over its center) and by Coelho [4] (P is a finite quasiordered set, R is a simple algebra, finite-dimensional over its center, or an indecomposable semiprime ring whose center is a unique factorization domain). After the notion of the finitary incidence algebra, which generalizes the notion of the incidence algebra to the cases of the arbitrary partially ordered [7] and quasiordered [6] sets, had been introduced, the task to describe the automorphism group of this type of algebras has arisen. Let P (4) be a quasiordered set, R an associative unital ring. As in [6], C(P,R) denotes the preadditive category associated with the pair (P,R), namely: 1. Ob C(P,R) = P = P/∼ with the induced order ≤. 2. For any x̄, ȳ ∈ P , x̄ ≤ ȳ the set of morphisms Mor(x̄, ȳ) =Mx̄×ȳ(R) (if x̄ � ȳ, then Mor(x̄, ȳ) = 0x̄ȳ). Here Mx̄×ȳ(R) is the additive group of matrices over R, whose rows and columns are indexed by the elements of the classes x̄ and ȳ, respectively, and each row has only a finite number of nonzero elements. For any two such matrices αx̄z̄ ∈ Mor(x̄, z̄), αz̄ȳ ∈ Mor(z̄, ȳ) the product αx̄z̄αz̄ȳ ∈ Mor(x̄, ȳ) is defined and gives the composition of the morphisms αx̄z̄ and αz̄ȳ in C(P,R). The category C(P,R) is a particular case of the so-called partially ordered category (pocategory), which was considered in [6]. For such categories the notion of the finitary incidence ring was introduced [6]. We shall formulate its definition for C(P,R). Consider the set of formal sums of the form α = ∑ x̄≤ȳ αx̄ȳ[x̄, ȳ], (1) where [x̄, ȳ] is a segment of the partial order, αx̄ȳ ∈ Mor(x̄, ȳ). The sum (1) is called a finitary series if for any [x̄, ȳ] there exists only a finite number of [ū, v̄] ⊂ [x̄, ȳ], ū < v̄ such that αūv̄ 6= 0ūv̄. The set of the finitary series forms a ring under the convolution [6, Theorem 1]. It is denoted by FI(P,R) (in fact FI(P,R) is an algebra over the center of R, but for the most part we are going to use only its ring properties). FI(P,R) has the unity element δ, where δx̄x̄ is the identity matrix of size \|x̄\|× \|x̄\|, δx̄ȳ = 0x̄ȳ A D M D R A F T 78 Automorphisms of finitary incidence rings for x̄ < ȳ. The finitary series can also be considered as the functions on the set of the segments of P with the values in R, namely: α(x, y) means the element of the matrix αx̄ȳ, which is situated in the intersection of the x-th row and y-th column. In this article we study the automorphism group AutFI of the ring FI(P,R) under the assumption that R is indecomposable. In the first section it is proved that the group OutFI = AutFI/InnFI of outer automorphisms of the finitary ring is isomorphic to the group OutC of outer automorphisms of the category C(P,R). After that in the second section we prove that under some additional assumptions on R the group OutC is isomorphic to the semidirect product Out0C⋊OutP , where Out0C belongs to the exact sequence 1 → H1(P ,C(R)∗) → Out0C → ∏ x̄∈P OutMx̄×x̄(R) (here C(R)∗ is the multiplicative group of the central invertible elements of the ring R, OutMx̄×x̄(R) is the group of outer automorphisms of the ring Mx̄×x̄(R)). In particular, if R is a local ring, P is a class finite quasiordered set and P = ⋃ i∈I Pi is the decomposition of P into the disjoint union of the connected components, then OutFI ∼= (H1(P ,C(R)∗)⋊ ∏ i∈I OutR)⋊OutP , as proved in the third section. Finally in the last section we investigate the group K-OutFI = K-AutFI/InnFI, where K-AutFI means the subgroup of AutFI consisting of those automorphisms, which agree with the structure of algebra over K = C(R). As the consequences, we obtain the results of Stanley, Scharlau and Baclawski about the automorphism group of incidence algebra. 1. The connection with the automorphisms of C(P,R) In what follows if no additional information is given P (4) is meant to be an arbitrary quasiordered set, R an indecomposable associative unital ring. The restriction of an element α ∈ FI(P,R) to the equivalence class x̄ ∈ P is by definition the series αx̄ = αx̄x̄[x̄, x̄]. The diagonal of α is αD = ∑ x̄∈P αx̄x̄[x̄, x̄]. Accordingly α is said to be diagonal iff αD = α. Note that (αβ)D = αDβD, (αβ)x̄ = αx̄βx̄ and αx̄βγȳ = αx̄x̄βx̄ȳγȳȳ[x̄, ȳ]. (2) A D M D R A F T N. Khripchenko 79 As a consequence, αx̄βγx̄ = αx̄βx̄γx̄, αx̄βȳ = 0 for x ≁ y. In particular, {δx̄}x̄∈P is a set of orthogonal idempotents and δx̄αδȳ = αx̄ȳ[x̄, ȳ], δx̄αδx̄ = αx̄. (3) First of all we shall be interested in the action of the automorphisms of the finitary ring on δx̄. Lemma 1. Let Φ ∈ AutFI. Then the image Φ(δx̄) is the conjugate of δϕ(x̄) for some order preserving bijection ϕ : P → P . Proof. By [6, Theorem 3] it is sufficient to prove that there is an order preserving bijection ϕ : P → P , such that Φ(δx̄)D = δϕ(x̄). (4) Consider an idempotent δx ∈ FI(P,R), which is defined for any x ∈ P as follows: δx(u, v) = { 1, if u = v = x, 0, otherwise. Obviously, δx = (δx)x̄, (δxαδy)(x, y) = α(x, y). By the indecomposability of R all δx are primitive. Indeed, if δx = α + β, where α and β are the orthogonal idempotents, then α = δxα = αδx = δxαδx, i. e. α(u, v) = α(x, x)δx(u, v). Since α(x, x) is an idempotent in R, α(x, x) equals 0 or 1 because R is indecomposable. Then either α coincides with δx, or it is equal to zero. Take an equivalence class x̄ ∈ P and choose an arbitrary element x′ ∈ x̄. The image Φ(δx′) is the primitive idempotent and by [6, Theorem 3] it is the conjugate of Φ(δx′)D. Since the restrictions of Φ(δx′) to the different classes are the orthogonal idempotents, the primitivity of Φ(δx′)D implies that there exists ȳ ∈ P , such that Φ(δx′)D coincides with Φ(δx′)ȳ. Note that δx′ and δx′′ are the conjugates iff x′ ∼ x′′. Hence the class ȳ does not depend on the choice of the representative x′ ∈ x̄. Thus Φ induces the mapping ϕ : P → P , such that Φ(δx′) is the conjugate of Φ(δx′)ϕ(x̄). Similarly we can consider Φ−1 and build ψ : P → P . Show that they are mutually inverse. Let ϕ(x̄) 6= v̄. Then for each x′ ∈ x̄: δvΦ(δx′)ϕ(x̄) = 0, i. e. δvβΦ(δx′)β −1 = 0 for some invertible β. Therefore, Φ−1(δv)Φ −1(β)δx′ = 0. This means that (Φ−1(δv)Φ −1(β))x̄ = 0, thus ψ(v̄) 6= x̄. The implication ψ(v̄) 6= x̄⇒ ϕ(x̄) 6= v̄ is proved similarly. So, ψ = ϕ−1. ConsiderΦ(δx̄) and prove that its diagonal coincides with the restriction on ϕ(x̄). Suppose that there are v′, v′′ ∈ v̄ 6= ϕ(x̄), such that δv′Φ(δx̄)δv′′ 6= 0. Then Φ−1(δv′)δx̄Φ −1(δv′′) 6= 0. But Φ−1(δv′) and Φ−1(δv′′) are the A D M D R A F T 80 Automorphisms of finitary incidence rings conjugates of Φ−1(δv′)ϕ−1(v̄) and Φ−1(δv′′)ϕ−1(v̄) respectively. This means that there are invertible β and γ, such that (see (2)) Φ−1(δv′)ϕ−1(v̄)β −1δx̄γΦ −1(δv′′)ϕ−1(v̄) = (Φ−1(δv′)β −1δx̄γΦ −1(δv′′))ϕ−1(v̄) is different from zero. Therefore (δx̄)ϕ−1(v̄) 6= 0, i. e. ϕ−1(v̄) = x̄, which con- tradicts the supposition. Hence Φ(δx̄)D = Φ(δx̄)ϕ(x̄). Similarly Φ−1(δϕ(x̄))D = Φ−1(δϕ(x̄))x̄. Using (3) we obtain that Φ−1(δϕ(x̄)) is the con- jugate of δx̄Φ −1(δϕ(x̄))δx̄. Therefore δϕ(x̄) is the conjugate ofΦ(δx̄)δϕ(x̄)Φ(δx̄). Since Φ(δx̄) is an idempotent and the diagonal of δx̄ is stable under the conjugation, we conclude that Φ(δx̄)D = δϕ(x̄). Prove that ϕ preserves the partial order. Let x̄ ≤ ȳ. Consider α = αx̄ȳ[x̄, ȳ] for some nonzero αx̄ȳ ∈ Mor(x̄, ȳ). Then α = δx̄αδȳ by (3). So Φ(α) = Φ(δx̄)Φ(α)Φ(δȳ) = βδϕ(x̄)β −1Φ(α)γδϕ(ȳ)γ −1 for some invertible β, γ ∈ FI(P,R). Hence β−1Φ(α)γ = δϕ(x̄)β −1Φ(α)γδϕ(ȳ) = (β−1Φ(α)γ)ϕ(x̄)ϕ(ȳ)[ϕ(x̄), ϕ(ȳ)]. Since α 6= 0, we have β−1Φ(α)γ 6= 0, and therefore (β−1Φ(α)γ)ϕ(x̄)ϕ(ȳ) 6= 0 by the previous equality. Thus ϕ(x̄) ≤ ϕ(ȳ). Remark 1. The lemma implies that the correspondence Φ 7→ ϕ agrees with the composition of the mappings. In particular, Φ−1 7→ ϕ−1, and ϕ−1 preserves the partial order. Let X ⊂ P . Denote by δX the diagonal finitary series ∑ x̄∈X δx̄x̄[x̄, x̄]. We shall need the following technical lemma. Lemma 2. Let Φ ∈ AutFI, ϕ : P → P be the bijection defined by (4), x 4 y, Z ⊂ P . Then 1. Φ(δx̄)ϕ(x̄)ϕ(ȳ) = 0ϕ(x̄)ϕ(ȳ) ⇔ Φ−1(δϕ(ȳ))x̄ȳ = 0x̄ȳ. 2. Φ(δZ)ϕ(x̄)ϕ(ȳ) = Φ(δZ′)ϕ(x̄)ϕ(ȳ), where Z ′ consists of those z̄ ∈ Z, for which Φ(δz̄)ϕ(x̄)ϕ(z̄) 6= 0ϕ(x̄)ϕ(z̄) and Φ(δz̄)ϕ(z̄)ϕ(ȳ) 6= 0ϕ(z̄)ϕ(ȳ). Proof. Prove the first statement. Write ū = ϕ(x̄), v̄ = ϕ(ȳ) for short. Let Φ(δx̄)ūv̄ = 0ūv̄. By (3) this is equivalent to the equality δūΦ(δx̄)δv̄ = 0 (5) in the ring FI(P,R). Apply Φ−1 to this equality. By the Remark 1 there are invertible β, γ ∈ FI(P,R), such that Φ−1(δū) = βδx̄β −1, Φ−1(δv̄) = γδȳγ −1. (6) A D M D R A F T N. Khripchenko 81 Then it follows from (5) that δx̄β −1δx̄γδȳ = 0, which is equivalent to (β−1)x̄x̄γx̄ȳ = 0x̄ȳ (see (3)). According to [6, Theorem 2], (β−1)x̄x̄ and (γ−1)ȳȳ are the invertible elements of the rings Mx̄×x̄(R) and Mȳ×ȳ(R) respectively, hence γx̄ȳ(γ −1)ȳȳ = 0x̄ȳ. This means that (γδȳγ −1)x̄ȳ = 0x̄ȳ, i. e. Φ−1(δv̄)x̄ȳ = 0x̄ȳ by (6). Let us turn to the proof of the second statement. Instead of Φ(δZ) we consider δūΦ(δZ)δv̄ (by (3) this series has the same value at the segment [ū, v̄] as the initial one). Using (6) we see that its preimage under Φ is equal to βδx̄β −1δZγδȳγ −1. It is sufficient to prove that in this product Z can be replaced by Z ′. According to (3) and the definition of the convolution, the product δx̄β −1δZγδȳ depends only on those z̄ ∈ Z, for which (β−1)x̄z̄ 6= 0x̄z̄ and γz̄ȳ 6= 0z̄ȳ. By the finitarity of β−1 and γ there is a finite number of such z̄. Note that the first inequality is equivalent to (βδx̄β −1)x̄z̄ 6= 0x̄z̄, i. e. Φ−1(δū)x̄z̄ 6= 0x̄z̄. Similarly the second one means that Φ−1(δv̄)z̄ȳ 6= 0z̄ȳ. Applying the first statement of the lemma to Φ−1, we obtain the required inequalities. For an arbitrary invertible β ∈ FI(P,R) denote by τβ ∈ InnFI the conjugation by the element β. If Φ ∈ AutFI, then, as it is mentioned above, for each x̄ ∈ P there is β, such that (τβΦ)(δx̄) = δϕ(x̄). It turns out that such a β can be chosen independently of the class x̄. Lemma 3. Let Φ ∈ AutFI, ϕ : P → P be the bijection defined by (4). Then there is τβ ∈ InnFI, such that (τβΦ)(δx̄) = δϕ(x̄) (7) for all x̄. Proof. Define β by the formal equality β = ∑ ū≤v̄ Φ(δϕ−1(ū))ūv̄[ū, v̄]. (8) Obviously, δϕ(x̄)β = δϕ(x̄)Φ(δx̄) for each x̄ ∈ P . Consider the product βΦ(δx̄). According to (8) and the definition of the convolution: (βΦ(δx̄))ūv̄ = ∑ ū≤w̄≤v̄ Φ(δϕ−1(ū))ūw̄Φ(δx̄)w̄v̄ = (Φ(δϕ−1(ū))Φ(δx̄))ūv̄. Since {δx̄}x̄∈P is a family of orthogonal idempotents in FI(P,R) and Φ is an isomorphism, we obtain that (βΦ(δx̄))ūv̄ = Φ(δx̄)ϕ(x̄)v̄ if ū = ϕ(x̄) and 0 otherwise. Thus, βΦ(δx̄) = δϕ(x̄)Φ(δx̄), i. e. βΦ(δx̄) = δϕ(x̄)β (9) A D M D R A F T 82 Automorphisms of finitary incidence rings for an arbitrary x̄ ∈ P . Note that βx̄ = Φ(δϕ−1(x̄))x̄ = δx̄ by (4). To prove the lemma it is sufficient to establish the finitarity of β. Indeed, then by [6, Theorem 2] β will be invertible and therefore βΦ(δx̄)β −1 = δx̄ from (9). Suppose that the set [ūs, v̄s]s∈S , ūs < v̄s of all different nontrivial subsegments of some fixed segment [ū, v̄] ⊂ P , for which βūsv̄s 6= 0ūsv̄s , is infinite. By the definition of β this means that Φ(δϕ−1(ūs))ūsv̄s 6= 0ūsv̄s . (10) According to the Lemma 2 Φ−1(δv̄s)ϕ−1(ūs)ϕ−1(v̄s) 6= 0ϕ−1(ūs)ϕ−1(v̄s). (11) It follows from (10) that for each ū0 ∈ P there is only a finite number of ūs, which coincide with ū0. Indeed, if ūs = ū0 ∈ [u, v] for some set of indexes S0 ⊂ S, then Φ(δϕ−1(ū0))ū0v̄s 6= 0ū0v̄s for this set of indexes by (10). Since Φ(δϕ−1(ū0)) is a finitary series and [ū0, v̄s] are the different nontrivial subsegments of the segment [ū, v̄], S0 must be finite. Similarly only a finite number of v̄s can coincide with some v̄0 ∈ P by (11) and the Remark 1. Consider an arbitrary segment [ū1, v̄1] from {[ūs, v̄s]}. According to our remark, there is only a finite number of segments in {[ūs, v̄s]}, one of whose end points coincides with one of the end points of [ū1, v̄1], i. e. {ūs, v̄s} ∩ {ū1, v̄1} 6= ∅. Throw away all such segments except [ū1, v̄1]. Then among the remaining segments choose [ū2, v̄2] 6= [ū1, v̄1]. Repeat the procedure for this segment, i. e. throw away all [ūs, v̄s] 6= [ū2, v̄2], for which {ūs, v̄s} ∩ {ū2, v̄2} 6= ∅ (there is a finite number of such segments). Note that [ū1, v̄1] will remain because {ū1, v̄1} ∩ {ū2, v̄2} = ∅ by the result of the previous step. Again, chose some [ū3, v̄3] 6= [ū1, v̄1], [ū2, v̄2] and so on. By iterating this process, we finally obtain the infinite set {[ūi, v̄i]} ∞ i=1 of segments, for which (10) and (11) are fulfilled, and, moreover, for each i there is a unique segment with the left end point ūi and a unique segment with the right end point v̄i (and there are no segments with the right end point ūi or with the left end point v̄i). Take X = {ϕ−1(ūi)} and consider the finitary series δX . According to the second statement of the Lemma 2, the value of Φ(δX)ūiv̄i must coincide with Φ(δX′)ūiv̄i , where X ′ consists of those ūj , for which Φ(δϕ−1(ūj))ūiūj 6= 0ūiūj and Φ(δϕ−1(ūj))ūj v̄i 6= 0ūj v̄i . In our case the only possibility for j is to be equal to i. Thus, Φ(δX)ūiv̄i = Φ(δϕ−1(ūi))ūiv̄i 6= 0ūiv̄i for all i. This contradicts the finitarity of Φ(δX). Remark 2. The series β from the previous lemma is determined up to the multiplication by the diagonal series. A D M D R A F T N. Khripchenko 83 Proof. Obviously, we need to prove that if τγ(δx̄) = δx̄ for all x̄, then γ is diagonal. Indeed, γδx̄ = δx̄γ means that γx̄ȳ = 0x̄ȳ, γz̄x̄ = 0z̄x̄ for all ȳ, z̄ 6= x̄. Since this is true for all x̄, γ is diagonal. Denote by AutC the automorphism group of the category C(P,R). An automorphism ϕ ∈ AutC is called inner if there is a diagonal invertible series β ∈ FI(P,R), such that for each αx̄ȳ ∈ Mor(x̄, ȳ) we have ϕ(αx̄ȳ) = βx̄αx̄ȳβ −1 ȳ . The set of inner automorphisms forms a normal subgroup of AutC, which is denoted by InnC. Accordingly, OutC = AutC/InnC denotes the group of outer automorphisms of the category C(P,R). The following theorem is the main result of this section. Theorem 1. The group OutFI is isomorphic to OutC. Proof. We shall build an epimorphism f : AutFI → OutC and prove that its kernel coincides with InnFI. Let Φ ∈ AutFI. There is a bijection ϕ : Ob C(P,R) → Ob C(P,R) given by (4). Define the corresponding mapping of the morphisms ϕ : Mor(x̄, ȳ) → Mor(ϕ(x̄), ϕ(ȳ)) (we denote it by the same letter). According to the Lemma 3 there is τβ ∈ InnFI, such that (7) is satisfied. Consider αx̄ȳ ∈ Mor(x̄, ȳ) and identify it with the series ε(αx̄ȳ), where ε is the embedding of the semigroup Mor C(P,R) in the multiplicative semigroup FI(P,R), namely: ε(αx̄ȳ) = αx̄ȳ[x̄, ȳ]. Then by (3) we have ε(αx̄ȳ) = δx̄ε(αx̄ȳ)δȳ. Therefore,Φε(αx̄ȳ) = Φ(δx̄)Φε(αx̄ȳ)Φ(δȳ). Using (7) we obtain βΦε(αx̄ȳ)β −1 = δϕ(x̄)βΦε(αx̄ȳ)β −1δϕ(ȳ). In other words, τβΦε(αx̄ȳ) = ε((τβΦε(αx̄ȳ))ϕ(x̄)ϕ(ȳ)). Thus, ϕ = ε−1τβΦε (12) defines the required mapping. Obviously, it is an isomorphism of the abelian groups and ϕ(δx̄x̄) = δϕ(x̄)ϕ(x̄). Moreover, ϕ agrees with the composition, because ε does. So, there is a mapping f : AutFI → OutC, namely f(Φ) = ϕ · InnC. (13) According to the Remark 2, the definition of f is correct. Prove that f is a homomorphism. Consider another automorphism Ψ ∈ AutFI, f(Ψ) = ψ·InnC. As it was mentioned above,Φ(δx̄) = τβ−1(δϕ(x̄)). Applying Lemma 3 to Ψ, we obtain ΨΦ(δx) = τΨ(β−1)Ψ(δϕ(x̄)) = τΨ(β−1)γ−1(δψ◦ϕ(x̄)) for some invertible γ ∈ FI(P,R). Therefore, τγΨ(β)ΨΦ(δx) = δψ◦ϕ(x̄). Thus, f(ΨΦ) = χ · InnC, where χ acts on objects as ψ ◦ ϕ and on mor- phisms as ε−1τγΨ(β)ΨΦε = (ε−1τγΨε)(ε −1τβΦε) (see (12)); hence f is a homomorphism. A D M D R A F T 84 Automorphisms of finitary incidence rings Conversely, let ϕ ∈ AutC, α ∈ FI(P,R). Define ϕ̂(α) as follows: ϕ̂(α)x̄ȳ = ϕ(αϕ−1(x̄)ϕ−1(ȳ)). Obviously, ϕ̂ is linear. Furthermore, since ϕ and ϕ−1, being the functions on P , preserve the partial order, ϕ̂(αβ)x̄ȳ = ∑ x̄≤z̄≤ȳ ϕ(αϕ−1(x̄)ϕ−1(z̄))ϕ(βϕ−1(z̄)ϕ−1(ȳ)) = (ϕ̂(α)ϕ̂(β))x̄ȳ. Therefore, ϕ̂ ∈ AutFI. Obviously, ϕ̂(δx̄) = δϕ(x̄) and hence f(ϕ̂) = ϕ·InnC. By (12) and (13) Kerf consists of the automorphisms Φ, for which the image ε−1τβΦε(αx̄ȳ) coincides with γx̄αx̄ȳγ −1 ȳ for all x̄ ≤ ȳ, αx̄ȳ ∈ Mor(x̄, ȳ) and for some diagonal invertible γ ∈ FI(P,R). This is equivalent to τγ−1βΦ(αx̄ȳ[x̄, ȳ]) = αx̄ȳ[x̄, ȳ]. In particular, τγ−1βΦ(δx̄) = δx̄. Denote Φ1 = τγ−1βΦ for short. Then, using (3), for an arbitrary α ∈ FI(P,R) we have: Φ1(α)x̄ȳ[x̄, ȳ] = δx̄Φ1(α)δȳ = Φ1(δx̄αδȳ) = Φ1(αx̄ȳ[x̄, ȳ]) = αx̄ȳ[x̄, ȳ]. Thus, τγ−1βΦ = idFI(P,R), i. e. Φ = τβ−1γ . 2. The group OutC Theorem 1 shows that the study of the group of outer automorphisms of the finitary ring is reduced to the study of the group of outer automorphisms of the category C(P,R). Denote by Aut0C the subgroup of AutC, consisting of the automor- phisms of C(P,R), which act identically on the objects. Let Out0C denote the image of Aut0C in OutC. Theorem 2. The following sequence of groups is exact: 1 → Out0C → OutC → AutP , where AutP is the automorphism group of the poset P . Proof. Let ϕ ∈ AutC. Then obviously ϕOb ∈ AutP , where ϕOb is the restriction of ϕ to the set Ob C = P . Note that if ϕ ∈ InnC, then ϕOb = id. Hence f : OutC → AutP is defined, namely: f(ϕ · InnC) = ϕOb. (14) Obviously, f is a homomorphism and its kernel consists of the cosets ϕ · InnC, for which ϕ(x̄) = x̄, i. e. Kerf = Out0C. A D M D R A F T N. Khripchenko 85 We are interested in the image of OutC in AutP . For this reason suppose that the ring R has the following property: MX×X(R) ∼=MY×Y (R) ⇒ \|X\| = \|Y \|. (15) In particular, commutative rings satisfy (15) for finite X and Y (see [3, Corollary 5.13]); we shall give another class of such rings below. Let AutP denote the automorphism group of the quasiordered set P . The image of an arbitrary class x̄ ⊂ P under ϕ ∈ AutP is again a class ϕ(x), such that \|ϕ(x̄)\| = \|x̄\|. An automorphism ϕ is called inner if ϕ(x̄) = x̄. The subgroup of inner automorphisms is denoted by InnP , then the group of outer automorphisms is OutP = AutP/InnP . Lemma 4. Under the condition (15) the image of the group OutC in AutP is isomorphic to the group OutP . Proof. Taking into account the remark before the lemma, it is easy to show that the group OutP is isomorphic to the subgroup G of AutP , consisting of the automorphisms ψ, such that \|ψ(x̄)\| = \|x̄\| for all x̄ ∈ P . Therefore, we need to prove that f(OutC) = G, where f is the homomorphism defined by (14). Let ϕ ∈ AutC. Since ϕ is an automorphism, Mx̄×x̄(R) is isomorphic to Mϕ(x̄)×ϕ(x̄)(R). Therefore, by (15) \|ϕ(x̄)\| = \|x̄\| and hence ϕOb ∈ G. Conversely, take ψ ∈ G and extend it arbitrarily to the automorphism of P . Define ψ̂(αx̄ȳ) ∈ Mor(ψ(x̄), ψ(ȳ)) as follows: ψ̂(αx̄ȳ)(ψ(x ′), ψ(y′)) = αx̄ȳ(x ′, y′), (16) where x′ ∈ x̄, y′ ∈ ȳ, αx̄ȳ(x ′, y′) is the element of the matrix αx̄ȳ, corre- sponding to the pair (x′, y′). The definition is correct, because ψ maps bijectively x̄ onto ψ(x̄) and ȳ onto ψ(ȳ). Moreover, ψ̂ is an isomorphism of the abelian groups Mor(x̄, ȳ) and Mor(ψ(x̄), ψ(ȳ)) with ψ̂(idx̄) = idψ(x̄). Furthermore, since ψ is an automorphism of P , ψ̂(αx̄ȳαȳz̄)(ψ(x ′), ψ(z′)) = (ψ̂(αx̄ȳ)ψ̂(αȳz̄))(ψ(x ′), ψ(z′)). Thus, ψ̂ ∈ AutC. Finally, note that f(ψ̂ · InnC) = ψ. Theorem 3. Let the ring R satisfy (15). Then the group OutC is isomor- phic to the semidirect product Out0C ⋊OutP . Proof. Identify OutP with the subgroup G of AutP . By the Theorem 2 and the Lemma 4 it is sufficient to build the monomorphism g : G→ OutC, such that fg = idG. Fix the numeration of the elements in each x̄ ⊂ P . A D M D R A F T 86 Automorphisms of finitary incidence rings Let ω(x) denote the number of the element x in the equivalence class x̄. We shall say that ϕ ∈ AutP agrees with ω if ω(ϕ(x)) = ω(x) for all x ∈ P . Note that in each coset of the subgroup InnP there is a unique automorphism, which agrees with ω, because an inner automorphism, which agrees with ω, is the identity. Let ψ ∈ G. Extend ψ to the automorphism ψω of the set P , which agrees with ω. By our remark this can be done uniquely. Then the mapping g(ψ) = ψ̂ω · InnC, where ψ̂ω is given by (16), is defined correctly. Obviously, (̂ψη)ω = ψ̂ωη̂ω. Thus, g is a homomorphism. Suppose that ψ̂ω ∈ InnC. Then, in particular, ψ̂ω(αx̄x̄) ∈ Mor(x̄, x̄), i. e. ψ(x̄) = x̄. Hence, ψ = idP and therefore g is a monomorphism. Finally (ψ̂ω)Ob = ψ by (16). This means that f(g(ψ)) = ψ. Show that the condition (15) is essential. Example 1. Let R be a ring, such that R2 R ∼= R3 R (see [1]). Take P with P = {x̄, ȳ, 1}, where x̄ = {x1, x2}, ȳ = {y1, y2, y3}, 1 is an one-element class; x̄ and ȳ are incomparable, x̄, ȳ < 1. Then OutC 6= Out0C ⋊OutP . Indeed, it is easy to see that OutP = 1. Therefore, we need to prove that OutC 6= Out0C, i. e. to find an automorphism ϕ of the category C(P,R), such that ϕOb 6= id. Note that Mor(x̄, x̄) = M2(R), Mor(ȳ, ȳ) = M3(R), Mor(1, 1) = R, Mor(x̄, 1) = R2 R, Mor(ȳ, 1) = R3 R, Mor(x̄, ȳ) = 0 (here Mn(R) denotes the ring of n × n matrices over R). It is convenient to represent the elements of R2 R and R3 R by the columns. Then M2(R) ∼= End(R2 R), M3(R) ∼= End(R3 R), where a matrix acts on a column by the left multiplication (since the modules are right). Let f : R2 R → R3 R be an isomorphism. For an arbitrary A ∈ M2(R) define g(A) ∈ M3(R) by its action on a column (r1, r2, r3) T ∈ R3 R: g(A)(r1, r2, r3) T = fAf−1(r1, r2, r3) T . Obviously, g is an isomorphism of the rings M2(R) and M3(R). Note that g(A)f(r1, r2) T = fA(r1, r2) T for an arbitrary (r1, r2) T ∈ R2 R. Define the mapping of the morphisms ϕ as follows: ϕ\|Mor(x̄,1) = f : Mor(x̄, 1) → Mor(ȳ, 1), ϕ\|Mor(x̄,x̄) = g : Mor(x̄, x̄) → Mor(ȳ, ȳ), ϕ\|Mor(1,1) = id. By the construction ϕ ∈ AutC and ϕOb(x̄) = ȳ. 3. The group Out0C In this section we are going to investigate the group Out0C. Obviously, the restriction of any automorphism ϕ ∈ Aut0C to the ring Mor(x̄, x̄) is an automorphism of this ring. Denote by Aut1C the subgroup consisting of those automorphisms ϕ from Aut0C, for which ϕ\|Mor(x̄,x̄) = id (17) A D M D R A F T N. Khripchenko 87 for all x̄ ∈ P . Let Out1C be an image of this subgroup in OutC. We shall first describe Out1C. Recall that the order complex K(X) of a poset X is the simplicial complex, whose n-dimensional faces are the chains of length n in X. Let Cn(X,A), Zn(X,A), Bn(X,A) and Hn(X,A) denote the groups of n- dimensional cochains, cocycles, coboundaries and cohomologies of the complex K(X) with the values in an abelian group A. Lemma 5. The group Out1C is isomorphic to H1(P ,C(R)∗), where C(R)∗ is the multiplicative group of the central invertible elements of the ring R. Proof. Prove that Aut1C ∼= Z1(P ,C(R)∗) and Aut1C ∩ InnC goes to B1(P ,C(R)∗) under this isomorphism. Let ϕ ∈ Aut1C, x̄, ȳ ∈ P , x̄ ≤ ȳ, x′ ∼ x, y′ ∼ y. Consider δx′y′ ∈ Mor(x̄, ȳ), defined as follows: δx′y′(u, v) = { 1, if u = x′, v = y′, 0, otherwise. (18) Note that δx′xαx̄ȳδyy′ = αx̄ȳ(x, y)δx′y′ (19) for each αx̄ȳ ∈ Mor(x̄, ȳ). In particular, δx′xδxyδyy′ = δx′y′ . Apply ϕ to this equality. Since δx′x ∈ Mor(x̄, x̄) and δyy′ ∈ Mor(ȳ, ȳ), using (17) we obtain δx′xϕ(δxy)δyy′ = ϕ(δx′y′). Therefore by (19) we have ϕ(δx′y′) = σ(x̄, ȳ)δx′y′ (20) for some σ(x̄, ȳ) ∈ R and for all x′ ∈ x̄, y′ ∈ ȳ. Prove that σ(x̄, ȳ) belongs to the center of R. Indeed, for an arbitrary r ∈ R according to (18) and (19) we have ϕ(rδxy) = ϕ(rδxxδxy) = ϕ(rδxx)ϕ(δxy) = rδxxσ(x̄, ȳ)δxy = rσ(x̄, ȳ)δxy. (21) Similarly ϕ(rδxy) = ϕ(δxyrδyy) = σ(x̄, ȳ)rδxy. Therefore rσ(x̄, ȳ) = σ(x̄, ȳ)r. Since r is arbitrary, σ(x̄, ȳ) ∈ C(R). Prove that σ(x̄, ȳ) is in- vertible. By (19) Rδxy = δxxMor(x̄, ȳ)δyy. Hence ϕ(Rδxy) = Rδxy. This means that there is r ∈ R, such that ϕ(rδxy) = δxy. Then it follows from (21) that rσ(x̄, ȳ) = 1. Since σ(x̄, ȳ) ∈ C(R), r = σ(x̄, ȳ)−1. So σ ∈ C1(P ,C(R)∗). Prove that σ is actually a cocycle. Indeed, it is easy to see that δxyδyz = δxz for arbitrary x 4 y 4 z. Hence by (20) σ(x̄, ȳ)σ(ȳ, z̄) = σ(x̄, z̄). Now determine how ϕ acts on an arbitrary αx̄ȳ ∈ Mor(x̄, ȳ). According to (19), ϕ(αx̄ȳ)(x, y)δxy = δxxϕ(αx̄ȳ)δyy. By (17) the last product is A D M D R A F T 88 Automorphisms of finitary incidence rings equal to ϕ(δxxαx̄ȳδyy). But δxxαx̄ȳδyy = αx̄ȳ(x, y)δxy and hence by (21) ϕ(δxxαx̄ȳδyy) = αx̄ȳ(x, y)σ(x̄, ȳ)δxy. Finally ϕ(αx̄ȳ) = σ(x̄, ȳ)αx̄ȳ. (22) Conversely, each σ ∈ Z1(P ,C(R)∗) defines an automorphism ϕ ∈ Aut1C with the help of (22). Obviously, the correspondence ϕ↔ σ is bijective and agrees with the multiplication in Aut1C and Z1(P ,C(R)∗). Now let ϕ ∈ Aut1C ∩ InnC and β ∈ FI(P,R) be the corresponding diagonal invertible series. Take arbitrary x̄ ∈ P , x′, x′′ ∼ x. By (17) and the definition of the conjugation βx̄x̄δx′x′ = δx′x′βx̄x̄. If x′ 6= x′′, then the value of the left-hand side of this equality at the segment [x′, x′′] obviously equals zero, while the value of the right-hand side equals βx̄x̄(x ′, x′′). Since x′, x′′ are the arbitrary elements of the class x̄, βx̄x̄ is a diagonal matrix for each x̄. Furthermore, βx̄x̄δx′x′′ = δx′x′′βx̄x̄ implies βx̄x̄(x ′, x′) = βx̄x̄(x ′′, x′′). Thus, βx̄x̄ = λ(x̄)δx̄x̄ for some function λ : P → R∗. Then βx̄x̄αx̄ȳβ −1 ȳȳ = λ(x̄)αx̄ȳλ(ȳ) −1. Taking x = y and αx̄ȳ = rδx̄x̄ by (17) we obtain λ(x̄)r = rλ(x̄). Therefore, λ(x̄) ∈ C(R)∗. Thus, a cocycle, corresponding to ϕ, satisfies σ(x̄, ȳ) = λ(x̄)λ(ȳ)−1, i. e. it is a coboundary. Conversely, let σ(x̄, ȳ) = λ(x̄)λ(ȳ)−1 for some λ ∈ C0(P ,C(R)∗). Define β = ∑ x̄∈P λ(x̄)δx̄x̄[x̄, x̄]. Then, obviously, the conjugation by β coincides with the action of σ. Denote by OutMx̄×x̄(R) the group of outer automorphisms of the ring Mx̄×x̄(R). Theorem 4. The following sequence of groups is exact: 1 → H1(P ,C(R)∗) → Out0C → ∏ x̄∈P OutMx̄×x̄(R). (23) Proof. Define f : Aut0C → ∏ x̄∈P AutMx̄×x̄(R) as follows: f(ϕ) = {ϕ\|Mor(x̄,x̄)}x̄∈P for an arbitrary ϕ ∈ Aut0C. Obviously, f is a homomorphism, under which InnC goes to ∏ x̄∈P InnMx̄×x̄(R). Hence a mapping f̄ : Out0C → ∏ x̄∈P OutMx̄×x̄(R) is defined, namely f̄(ϕ · InnC) = f(ϕ) · ∏ x̄∈P InnMx̄×x̄(R). Moreover, the kernel of f̄ consists of those ϕ·InnC, for which ϕ\|Mor(x̄,x̄) = id for all x̄ ∈ P . Therefore, Kerf coincides with the group Out1C, which is isomorphic to H1(P ,C(R)∗) by the previous lemma. A D M D R A F T N. Khripchenko 89 Corollary 1. Let P be an arbitrary quasiordered set, R an indecomposable unital ring, such that for any sets X and Y an isomorphism MX×X(R) ∼= MY×Y (R) implies \|X\| = \|Y \|. Then OutFI ∼= Out0C ⋊ OutP , where Out0C belongs to the exact sequence (23). The description of the image of Out0C in ∏ x̄∈P OutMx̄×x̄(R) seems to be difficult in general situation, so we shall restrict ourselves to one special case. Recall that the ring R is called local if R/RadR is a division ring. In particular, R is indecomposable. Prove that R satisfies (15) in the case of finite X and Y . Denote by Mn(R) the ring of n× n matrices over R. Suppose that Mn(R) ∼=Mm(R). Consider the matrix units δij ∈Mn(R), i, j = 1, . . . , n. By definition {δii} n i=1 is the decomposition of the unit of Mn(R). Obviously, δiiMn(R)δjj = Rδij . (24) In particular, δiiMn(R)δii ∼= R is a local ring. Therefore, δii is completely primitive [5, p. 59, Definition 2] for all i. If ϕ : Mn(R) → Mm(R) is an isomorphism, then {ϕ(δii)} n i=1 is the decomposition of the unit of Mm(R), consisting of the completely primitive idempotents. But Mm(R) already has such decomposition of the unit of cardinality m. Then by [5, p. 59, Theorem 2] n = m. Let ψ ∈ AutR, α ∈Mn(R). Define (ψ̂(α))ij = ψ(αij). (25) Obviously, ψ̂ ∈ AutMn(R). It turns out that in the case when R is local, each automorphism of the ring Mn(R) can be represented as ψ̂ up to conjugacy. Lemma 6. Let R be a local ring, ϕ ∈ AutMn(R). Then there is a unique up to conjugacy in R automorphism ψ ∈ AutR and an invertible matrix β ∈Mn(R), such that ϕ = τβψ̂, where τβ is the conjugation by β and ψ̂ is defined by (25). Proof. Let {δij} n i,j=1 be matrix units of the ring Mn(R). According to the reasoning before the theorem, {δii} n i=1 and {ϕ(δii)} n i=1 are two decomposi- tions of the unit consisting of the completely primitive idempotents. By [5, p. 59, Theorem 2] there is an invertible matrix β1 ∈ Mn(R), such that τβ1ϕ(δii) = δii. Therefore, by (24) there are ϕi ∈ AutR, such that τβ1ϕ(rδii) = ϕi(r)δii (26) for an arbitrary r ∈ R. Since it is easy to show that β1 is determined up to the diagonal multiplier, each ϕi is determined up to the inner A D M D R A F T 90 Automorphisms of finitary incidence rings automorphism of the ring R. According to (24) denote by σij the element of the ring R, such that τβ1ϕ(δij) = σijδij . (27) In particular, σii = 1. Furthermore δijδjk = δik implies σijσjk = σik. Therefore, σij is invertible and σij = σi1σ −1 j1 . (28) Now take an arbitrary r ∈ R and consider rδi1. By the definition of the matrix units rδiiδi1 = δi1rδ11 and hence ϕi(r)σi1 = σi1ϕ1(r). Therefore, ϕi = τσi1ϕ1. (29) Consider a diagonal matrix β2 = n∑ i=1 σi1δii. The equalities (26), (27), (28) and (29) imply τβ1ϕ(rδij) = (τβ1ϕ(rδii))(τβ1ϕ(δij)) = ϕi(r)δiiσijδij = σi1ϕ1(r)σ −1 i1 σi1σ −1 j1 δij = σi1ϕ1(r)σ −1 j1 δij = τβ2(ϕ1(r)δij). Hence τβϕ(rδij) = ϕ1(r)δij , where β = β−1 2 β1. Take an arbitrary matrix α ∈Mn(R). Since δiiαδjj = αijδij , we have δii(τβϕ(α))δjj = τβϕ(δiiαδjj) = τβϕ(αijδij) = ϕ1(αij)δij and hence (τβϕ(α))ij = ϕ1(αij). Thus, ϕ = τβ−1ϕ̂1. Corollary 2. Let R be a local ring. Then OutMn(R) ∼= OutR. Proof. Let ϕ ∈ AutMn(R). According to the previous lemma, ϕ = τβψ̂ and ψ is defined up to conjugacy in R. Put f(ϕ · InnMn(R)) = ψ · InnR. Obviously, f is defined correctly and it is a homomorphism of the groups OutMn(R) and OutR. Prove that f is a monomorphism. Indeed, if ψ is an inner automorphism of R, then ψ̂ is a conjugation in Mn(R) by a scalar matrix and ϕ ∈ InnMn(R). The surjectivity of f is obvious, because f(ψ̂ · InnMn(R)) = ψ · InnR. Theorem 5. Let R be a local ring, P a quasiordered set whose classes are finite, P = ⋃ i∈I Pi the decomposition of P into the disjoint union of the connected components. Then the group Out0C is isomorphic to the semidirect product H1(P ,C(R)∗)⋊ ∏ i∈I OutR. A D M D R A F T N. Khripchenko 91 Proof. Let ϕ ∈ Aut0C. Applying Lemma 6 to ϕ\|Mor(x̄,x̄) for each x̄ ∈ P , we obtain the representation of ϕ\|Mor(x̄,x̄) as τβx̄x̄ϕ̂x̄, where βx̄x̄ ∈ Mx̄×x̄(R) is an invertible matrix, ϕx̄ ∈ AutR and ϕ̂x̄ is given by (25). Let β = ∑ x̄∈P βx̄x̄[x̄, x̄]. Then β ∈ FI(C) is a diagonal invertible series, such that (τβ−1ϕ)\|Mor(x̄,x̄) = ϕ̂x̄ for all x̄ ∈ P (here τβ−1 means the inner automorphism of C(P,R) corresponding to β−1). Thus, we can assume up to the conjugation by β that ϕ(αx̄x̄)(x ′, x′′) = ϕx̄(αx̄x̄(x ′, x′′)) (30) for any x′, x′′ ∈ x̄, αx̄x̄ ∈ Mor(x̄, x̄). Take x ≺ y and show that ϕx̄ differs from ϕȳ by an inner automorphism of the ring R. Since by (30) ϕ(δx′x′′) = δx′x′′ and ϕ(δy′y′′) = δy′y′′ for any x′, x′′ ∈ x̄, y′, y′′ ∈ ȳ, we obtain as in the proof of the Lemma 5 that ϕ(δx′y′′) = σ(x̄, ȳ)δx′y′′ for some constant σ(x̄, ȳ) ∈ R, which depends only on the classes x̄ and ȳ. Therefore, ϕx̄(r)σ(x̄, ȳ)δxy = ϕ(rδxy) = σ(x̄, ȳ)ϕȳ(r)δxy for any r ∈ R. These equalities guarantee the invertibility of σ(x̄, ȳ), because it follows from (30) that ϕ(Rδxy) = Rδxy (it is sufficient to take r, such that ϕ(rδxy) = δxy). Thus, ϕx̄ = τσ(x̄,ȳ)ϕȳ. Choose xi ∈ Pi for each i ∈ I and put g(ϕ·InnC) = {ϕx̄i ·InnR}i∈I . Our reasoning shows that g is defined correctly and it is a homomorphism of the groups Out0C and ∏ i∈I OutR with the kernel Out1C which is isomorphic to H1(P ,C(R)∗). It remains only to build and embedding h : ∏ i∈I OutR→ Out0C, such that gh = id∏ i∈I OutR. Let ϕi ∈ AutR, i ∈ I. For arbitrary x̄ ≤ ȳ, x, y ∈ Pi and αx̄ȳ ∈ Mor(x̄, ȳ) define ϕ̂(αx̄ȳ) ∈ Mor(x̄, ȳ) as follows: ϕ̂(αx̄ȳ)(x ′, y′) = ϕi(αx̄ȳ(x ′, y′)), (31) where x′ ∈ x̄, y′ ∈ ȳ. It is easy to see that ϕ̂ ∈ Aut0C, moreover, if all ϕi are inner, then ϕ̂ is also inner. Therefore, h({ϕi ·InnR}i∈I) = ϕ̂·InnC is defined. Obviously, h is a homomorphism. Suppose that ϕ̂ ∈ InnC, i. e. ϕ̂(αx̄ȳ) = γx̄x̄αx̄ȳγ −1 ȳȳ for any x̄ ≤ ȳ. Since ϕ̂(δx′x′′) = δx′x′′ for all x′, x′′ ∈ x̄, γx̄x̄ is a scalar matrix, similarly so is γȳȳ. Furthermore, since ϕ̂(δx′y′) is by definition equal to δx′y′ , γx̄x̄(x ′, x′) = γȳȳ(y ′, y′). Therefore, γx̄x̄ = siδx̄x̄, γȳȳ = siδȳȳ for some si ∈ R∗. Thus, ϕ̂(αx̄ȳ)(x ′, y′) = τsi(αx̄ȳ(x ′, y′)), where si depends only on the connected component Pi, which contains x and y. This means that h is a monomorphism. Obviously, g(ϕ̂ · InnC) = {ϕi · InnR}i∈I . Corollary 3. Let R be a local ring, P a quasiordered set whose classes are finite, P = ⋃ i∈I Pi the decomposition of P into the disjoint union A D M D R A F T 92 Automorphisms of finitary incidence rings of the connected components. Then the group OutFI is isomorphic to (H1(P ,C(R)∗)⋊ ∏ i∈I OutR)⋊OutP . Recall that the restriction (15) on R was imposed in order to assert that the isomorphism Mϕ(x̄)×ϕ(x̄)(R) ∼=Mx̄×x̄(R), where ϕ ∈ AutC, implies the equality \|ϕ(x̄)\| = \|x̄\|. Suppose that P is partially ordered. Then x ∼ y iff x = y, i. e. all the equivalence classes under ∼ are one-element, P = P and OutP = AutP . Therefore, we don’t need to require (15). Furthermore, since Mx̄×x̄(R) = R for all x̄, ϕ\|Mor(x̄,x̄) ∈ AutR and the Theorem 5 can be proved without using the Lemma 6. Thus, in the case of the partial order we can refuse the locality of R. Remark 3. Let P be a partially ordered set, R an indecomposable ring, P = ⋃ i∈I Pi the decomposition of P into the disjoint union of the connected components. Then the group OutFI is isomorphic to (H1(P,C(R)∗) ⋊∏ i∈I OutR)⋊AutP . 4. C(R)-automorphisms of the ring FI(P,R) Let A be a unital algebra over a commutative ring K, AutA denote the group of its ring automorphisms, InnA be a subgroup of inner automor- phisms, OutA = AutA/InnA. We say that an automorphism ϕ ∈ AutA is a K-automorphism, if it agrees with the structure of K-algebra. K- automorphisms form a subgroup of AutA, which we denote by K-AutA. Note that an automorphism ϕ belongs to K-AutA iff ϕ(k · 1) = k · 1 for all k ∈ K. Since K is commutative, InnA ⊂ K-AutA and hence the group K-OutA = K-AutA/InnA is defined. Now let P be a quasiordered set, R an arbitrary associative unital ring. Put K = C(R). Then both R and FI(P,R) are K-algebras. Therefore, the groups K-AutR, K-OutR, K-AutFI := K-AutFI(P,R), K-OutFI := K-OutFI(P,R) are defined. By the Theorem 1 we can identify K-OutFI with the subgroup of OutC, which we shall denote by K-OutC. It is easy to see that K-OutC consists of the cosets ϕ ·InnC, where ϕ(kδx̄x̄) = kδϕ(x̄)ϕ(x̄) for all x̄ ∈ P and k ∈ K (recall that δx̄x̄ denotes the identity matrix in the ring Mx̄×x̄(R) = Mor(x̄, x̄)). In this section we describe K-OutFI in the case when the classes of P are finite and R is local. Lemma 7. Let R be a local ring, P a quasiordered set whose classes are finite, P = ⋃ i∈I Pi the decomposition of P into the disjoin union of the connected components, f : (H1(P ,K∗)⋊ ∏ i∈I OutR)⋊OutP → OutC the isomorphism from the previous section. Then A D M D R A F T N. Khripchenko 93 1. f(H1(P ,K∗)) ⊂ K-OutC, 2. f( ∏ i∈I OutR) ∩K-OutC = f( ∏ i∈I K-OutR), 3. f(OutP ) ⊂ K-OutC. Proof. Let σ ∈ Z1(P ,K∗). Then f(σ ·B1(P ,K∗)) = σ̂ ·InnC, where for an arbitrary αx̄ȳ ∈ Mor(x̄, ȳ) the value of σ̂(αx̄ȳ) is defined by the right-hand side of (22). Since σ(x̄, x̄) = 1 for any x̄ ∈ P , σ̂(kδx̄x̄) = kδx̄x̄ for all k ∈ K. Thus, f(H1(P ,K∗)) ⊂ K-OutC. Now let {ϕi · InnR}i∈I ∈ ∏ i∈I OutR. Then f({ϕi · InnR}i∈I) = ϕ̂ · InnC, where ϕ̂ is given by means of (31). Therefore, ϕ̂(kδx̄x̄) = ϕi(k)δx̄x̄ for any x ∈ Pi, k ∈ K. Hence ϕ̂ · InnC ∈ K-OutC iff ϕi · InnR ∈ K-OutR for all i ∈ I. In other words, f( ∏ i∈I OutR) ∩K-OutC = f( ∏ i∈I K-OutR). Consider ψ ∈ AutP . An image f(ψ·InnP ) is a coset ψ̂·InnC, where ψ̂ is defined by the equation (16). Take k ∈ K and note that ψ̂(kδx̄x̄)(x ′, x′′) = kδ(ψ−1(x′), ψ−1(x′′)), where x′, x′′ ∈ x̄. If x′ = x′′, then ψ̂(kδx̄x̄)(x ′, x′′) = k, otherwise, ψ̂(kδx̄x̄)(x ′, x′′) = 0. Therefore, ψ̂(kδx̄x̄) = kδ ψ̂(x̄)ψ̂(x̄) and f(OutP ) ⊂ K-OutC. Theorem 6. Let R be a local ring, P a quasiordered set whose classes are finite, P = ⋃ i∈I Pi the decomposition of P into the disjoin union of the connected components. Then the group K-OutFI is isomorphic to the semidirect product (H1(P ,K∗)⋊ ∏ i∈I K-OutR)⋊OutP . Proof. Identify K-OutFI with K-OutC ⊂ OutC. Let f : (H1(P ,K∗) ⋊∏ i∈I OutR)⋊OutP → OutC be the isomorphism from the previous section. Recall that the image of H1(P ,K∗) ⋊ ∏ i∈I OutR under f coincides with Out0C. Denote the subgroup Out0C∩K-OutC byK-Out0C. We shall prove that K-OutC = K-Out0C ⋊ f(OutP ) and K-Out0C = f(H1(P ,K∗)) ⋊ f( ∏ i∈I K-OutR)). Consider an arbitrary χ ∈ OutC. Then [χ] = [σ̂ϕ̂ψ̂], where σ̂, ϕ̂, ψ̂ ∈ AutC are the isomorphisms from the previous lemma, namely [σ̂] ∈ f(H1(P ,K∗)), [ϕ̂] ∈ f( ∏ i∈I OutR), [ψ̂] ∈ f(OutP ) (here and below the square brackets mean that we consider the coset of the subgroup InnC). By the previous lemma [σ̂], [ψ̂] ∈ K-OutC. Therefore, χ(kδx̄x̄) = ϕ̂(kδx̄x̄). Hence [χ] ∈ K-OutC iff [ϕ̂] ∈ K-OutC. According to the second state- ment of the previous lemma this is equivalent to [ϕ̂] ∈ f( ∏ i∈I K-OutR). A D M D R A F T 94 Automorphisms of finitary incidence rings Thus, K-OutC = f(H1(P ,K∗))f( ∏ i∈I K-OutR)f(OutP ). Similar reason- ing shows that K-Out0C = f(H1(P ,K∗))f( ∏ i∈I K-OutR). Furthermore, note that the intersection of f(H1(P ,K∗)) and f( ∏ i∈I K-OutR) is trivial, because f(H1(P ,K∗)) ∩ f( ∏ i∈I OutR) = {1}, and similarly K-Out0C ∩ f(OutP ) = {1}. Hence it is sufficient to prove that f(H1(P ,K∗)) is nor- mal in K-Out0C and K-Out0C is normal in K-OutC. The first assertion is obvious: since f(H1(P ,K∗)) is normal in Out0C, it will be normal in its subgroup K-Out0C. For the proof of the second assertion consider [ϕ] ∈ K-Out0C and conjugate it by [ψ] ∈ K-OutC. The result of the con- jugation belongs to Out0C ∩K-OutC, because Out0C is normal in OutC. But by definition Out0C ∩ K-OutC = K-Out0C and hence K-Out0C is normal in K-OutC. If P is partially ordered, then, as in the Remark 3, it is sufficient to require that R is indecomposable. Remark 4. Let P be a partially ordered set, R an indecomposable ring, P = ⋃ i∈I Pi the decomposition of P into the disjoint union of the connected components. Then the group K-OutFI is isomorphic to (H1(P,K∗) ⋊∏ i∈I K-OutR)⋊AutP . If R is a simple algebra, finite-dimensional over its center, then by Skolem-Noether theorem K-OutR = 1 and hence K-OutFI is isomorphic to H1(P ,K∗)⋊OutP . Thus we obtain a generalization of [9, Theorem 2] and [2, Theorem 5]. In the case when P has 0 or 1, H1(P ,K∗) = 1 and K-OutFI ∼= OutP . This generalizes [8, Theorem 1.2]. References [1] G.Abrams, J. Haefner and Á. del Ŕıo, Corrections and addenda to ’The isomor- phism problem for incidence rings’, Pacific J. Math., 207(2002), no 2, 497–506. [2] K. Baclawski, Automorphisms and derivations of incidence algebras, Proc. Amer. Math. Soc., 38(1972), 351–356. [3] W. C. Brown, Matrices over Commutative Rings, Monographs and Textbooks in Pure and Applied Mathematics, Vol. 169, Marcel Dekker, New York, 1993. [4] S. P. Coelho, Automorphism groups of certain structural matrix rings, Communi- cations in Algebra, 22(1994), no.14, 5567–5586. [5] N. Jacobson, Structure of Rings, Amer. Math. Soc. Colloq. Publ. 37 , Providence, 2nd ed., 1964. [6] N. S.Khripchenko, Finitary incidence algebras of quasiorders, Matematychni Studii (submitted). A D M D R A F T N. Khripchenko 95 [7] N. S.Khripchenko and B.V.Novikov, Finitary incidence algebras, Communica- tions in Algebra, 37(2009), no. 5, 1670–1676. [8] W. Scharlau, Automorphisms and involution of incidence algebras in “Representa- tions of Algebra, Ottawa, 1974”, pp. 340–350, Lecture Notes in Math. No. 488, Springer-Verlag, Berlin/Heidelberg/New York, 1975. [9] R. P. Stanley, Structure of incidence algebras and their automorphism groups, Bull. Amer. Math. Soc. 76(1970), 1236–1239. Contact information Nikolay Khripchenko Department of Mechanics and Mathematics, Kharkov V. N. Karazin National University, 4 Svobody sq, 61077, Kharkov, Ukraine E-Mail: NSKhripchenko@mail.ru Received by the editors: 24.05.2010 and in final form ????. Nikolay Khripchenko

Automorphisms of finitary incidence rings

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