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Higher-order amalgamation of algebraic structures David Milovich http://dkmj.org Welkin Sciences Colorado Springs, CO UCCS Algebra Seminar Nov. 7, 2018 0 / 15 Overlapping structures Below are two overlapping group multiplication tables, G 1


  1. Higher-order amalgamation of algebraic structures David Milovich http://dkmj.org Welkin Sciences Colorado Springs, CO UCCS Algebra Seminar Nov. 7, 2018 0 / 15

  2. Overlapping structures Below are two overlapping group multiplication tables, G 1 ∼ = C 2 2 on the left and G 2 ∼ = S 3 on the right. y 2 xy 2 · z xz 1 x y xy z 1 x z xz 1 xz x xz z y 2 xy 2 1 z xz 1 x y xy xy 2 y 2 1 x xz z x xy y xy 2 y 2 y y x 1 xy y 2 xy 2 1 xy xy x y y 2 y 2 xy 2 xy 1 y x xy 2 xy 2 y 2 1 y x xy In general, an indexed set ( A i : i ∈ E ) of algebraic structures is overlapping if, for all distinct i , j ∈ E , the algebraic operations of A i and A j , including any distinguished elements, agree when restricted to A i ∩ A j . 1 / 15

  3. An amalgamation I say that G 1 and G 2 amalgamate as groups because there is a group K ∼ = C 2 × S 3 containing G 1 and G 2 as subgroups: y 2 z xy 2 z y 2 xy 2 yz xyz z xz 1 x y xy · y 2 xy 2 xy 2 z y 2 z yz x 1 xy y yz xz z xyz xy 2 y 2 y 2 z xy 2 z xyz 1 x y xy xyz z xz yz y 2 z xy 2 y 2 y 2 z xy 2 z 1 y x xy xyz z yz xz xy 2 z y 2 xy 2 xy 2 z y 2 z x xy 1 y yz xz xyz z y 2 xy 2 y 2 z xy 2 z z y xy 1 x z xz yz xyz xy 2 y 2 xy 2 z y 2 z xz xy y x 1 xz z xyz yz y 2 z xy 2 z y 2 xy 2 1 yz xyz z xz 1 x y xy xy 2 z y 2 z xy 2 y 2 x xyz yz xz z x 1 xy y y 2 z xy 2 z xy 2 y 2 y xz z xyz yz y x 1 xy xy 2 z y 2 z y 2 xy 2 xy z xz yz xyz xy 1 x y y 2 xy 2 z y 2 z y 2 xy 2 z yz xz xyz xy 1 y x xy 2 y 2 z xy 2 z xy 2 y 2 xz xyz z yz y x xy 1 2 / 15

  4. � � � Amalgamation formalized Given a category C consisting of a class of algebraic structures and all homomorphisms between them, and overlapping ( A i : i ∈ E ) in C , we say that A 1 , . . . , A n amalgamate in C if there are embeddings e i : A i → B in C that commute with the identity embeddings id: A i ∩ A j → A i , id( x ) = x . id A i ∩ A j A i e i id e j � B A j By embedding, I mean an injective homomorphism. (Embeddings are always monomorphisms (if you know what those are), and the converse is true in most categories of interest. The category of divisible abelian groups is an exception.) 3 / 15

  5. Schreier’s Theorem Call an indexed family of sets ( A i : i ∈ E ) a ∆-system or sunflower if there is root R such that A i ∩ A j = R for all distinct i , j . Theorem (Schreier, 1927) Every sunflower of groups ( G i : i ∈ E ) amalgamates in the category of groups. In particular, any two overlapping groups amalgamate as groups. About the proof. Choose isomorphisms ψ i : G i → H i such that H i ∩ H j = { 1 } . Let F be the free product consisting of all words a 1 a 2 · · · a n where adjacent letters a k , a k +1 are not in the same H i . Let K = F / N where N is the smallest normal subgroup of F containing all words of the form ψ i ( r ) ψ j ( r − 1 ) where r is in the root. Schreier proves that ψ i / N : G i → K is injective by means of a normal form lemma. 4 / 15

  6. Compactness I Typically, if there are no finitary obstructions to amalgamation, then there are no obstructions at all. Lemma Suppose that: ◮ A i ∩ A j ∈ C for all i , j ∈ E. ◮ All diagrams in C have colimits. ◮ ( A i : i ∈ F ) amalgamates in C for all finite F ⊂ E. Then ( A i : i ∈ E ) amalgamates in C . Proof sketch. For each finite F ⊂ E , let B F and φ i , F : A i → B F for i ∈ F be the colimit of the morphisms id: A i ∩ A j → A i for i , j ∈ F . The morphisms φ i , F must be injective. For G ⊂ F ⊂ E , there is a natural morphism χ F , G : B G → B F , and it must also be injective. Let C and ψ F : B F → C for finite F ⊂ E be the colimit of the morphisms χ F , G for G ⊂ F . Then ψ { i } : A i → C must be injective for each i ∈ E . 5 / 15

  7. Compactness II Lemma Suppose that: ◮ A i ∩ A j ∈ C for all i , j ∈ E. ◮ C is axiomatized by a set of first-order formulas. ◮ ( B i : i ∈ E ) amalgamates in C for all finitely generated substructures B i ⊂ A i with B i ∈ C . Then ( A i : i ∈ E ) amalgamates in C . Proof sketch. Let F denote the set of B = ( B i : i ∈ E ) where B i ∈ C , B i ⊂ A i , and B i is finitely generated. Partially F by B ≤ B ′ iff B i ⊂ B ′ i for all i . Let U be an ultrafilter on F such that { X | X ≥ B } ∈ U for all B . By hypothesis, there are embeddings φ B , i : B i → D B in C for each B ∈ F . Let D be the ultraproduct � B D B / U . Then x �→ ( φ B , i ( x ) : x ∈ B i ) / U defines an embedding from A i to D . 6 / 15

  8. Binary amalgamation can fail. ◮ Let F 1 ∼ = F 2 ∼ = C and F 1 ∩ F 2 = R . Then F 1 and F 2 do not amalgamate as integral domains. ◮ (Kimura, 1957) There are overlapping finite commutative semigroups S 1 , S 2 that do not amalgamate as semigroups. ◮ It follows that there are two finite commutative rings that do not amalgamate as rings. ◮ (Sapir, 1997; Jackson, 2000) There is no algorithm that can decide whether two arbitrary finite semigroups amalgamate. Likewise for finite rings. ◮ There are many papers about binary amalgamation in categories of groups with various extra properties. Some have binary amalgamation. Some don’t. 7 / 15

  9. Integral domains Let F 1 = R [ x ] / ( x 2 + 1) and F 2 = R [ y ] / ( y 2 + 1). Any commutative ring R containing F 1 ∪ F 2 also contains S = R [ x , y ] / ( x 2 + 1 , y 2 + 1). But S has divisors of zero: x ± y are not in the ideal generated by x 2 + 1 , y 2 + 1. But ( x + y )( x − y ) = ( x 2 + 1) − ( y 2 + 1) is in that ideal. 8 / 15

  10. Semigroups Define overlapping commutative semigroups S i = { 0 , a , b , c i } for i = 1 , 2 as follows. · 0 c 2 a b c 1 0 c 2 c 2 b b 0 0 0 0 0 0 0 0 0 a b a b b 0 0 0 a 0 c 1 a a c 1 If some semigroup T contained S 1 ∪ S 2 , then we would reach the contradiction a = c 1 b = c 1 ( ac 2 ) = ( c 1 a ) c 2 = ac 2 = b . 9 / 15

  11. Beyond sunflowers ◮ (H. Neumann, 1948) There are three overlapping groups that do not amalgamate as groups. ◮ (H. Neumann, 1951) Any three overlapping abelian groups amalgamate as abelian groups. ◮ (H. Neumann, 1954) But there are four overlapping abelian groups that do not amalgamate as groups. ◮ Call a ring Boolean if every element x is idempotent: x 2 = x . It is known that sunflowers amalgamate in the category of Boolean rings. But general ternary amalgamation fails: ◮ Let xy = y generate Boolean ring A 1 . ◮ Let yz = z generate Boolean ring A 2 . ◮ Let zx = x generate Boolean ring A 3 . ◮ If A 1 , A 2 , A 3 amalgamated, even as commutative multiplicative semigroups, then x = zx = ( yz ) x = y ( zx ) = yx = xy = y . 10 / 15

  12. Set-theoretic motivations I First, some basics about cardinality: A set S is countable or satisfies | S | ≤ ℵ 0 if there is a strict linear order < S such that every proper initial segment { x | x < S y } is finite. A set satisifes | S | ≤ ℵ n +1 if there is a strict linear order < S such that every proper initial segment I = { x | x < S y } satisfies | I | ≤ ℵ n . 11 / 15

  13. Set-theoretic motivations II A Boolean ring is projective if is a retract of a free Boolean ring. A family F of countable subsets of a set S is a club if every countable T ⊂ S is contained in some E ∈ F and F is closed with respect to unions of countable chains. A subring A of a ring B is relatively complete if, for every principal ideal I of B , the ideal I ∩ A of A is also principal. Theorem (M., 2016) A Boolean ring B satisfying | B | ≤ ℵ d is projective iff there it has the ( d + 1) -ary Freese-Nation property, that is, has a club F of subsets such that the any subring A of B generated by the union of at most d elements of F is relatively complete. 12 / 15

  14. Set-theoretic motivations III A family F of sets is directed if for all X , Y ∈ F there exists Z ∈ F such that X ∪ Y ⊂ Z . Lemma If F is directed, every X ∈ F is countable, and | � F| ≥ ℵ 3 , then there X 1 , X 2 , X 3 ∈ F that are not a sunflower. I wanted to prove that the d -ary and ( d + 1)-ary Freese-Nation properties are inequivalent. I succeeded, but for d ≥ 3, my proof involved cooking up a tricky Boolean algebra of cardinality ℵ d as a union of a directed family of countable Boolean algebras. So, I had to find a safe harbor, avoiding all obstructions to higher-order Boolean amalgamation. . . 13 / 15

  15. Higher-order pushouts Henceforth assume that C , a category consisting of a class of algebraic structures and all homomorphisms between them, has the following closure properties. ◮ If A ∈ C and A ∼ = B , then B ∈ C . ◮ All finite diagrams in C have colimits. Say A 1 , . . . , A n ∈ C are C -overlapping if � i ∈ s A i ∈ C for all s ⊂ { 1 , . . . , n } . Given C -overlapping A 1 , . . . A n ∈ C , define the pushout of A 1 , . . . , A n to be the colimit of the diagram consisting of the morphisms id: � i ∈ t A i → � i ∈ s A i for s ⊂ t . Typically, there is canonical pushout of A 1 , . . . A n . It is the algebra B ∈ C freely generated by the set of elements � i A i and the set of relations � i R i where R i is the set of all relations true of A i . 14 / 15

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