Joseph Muscat Universal Algebras. 1 March 2013

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Joseph Muscat 2015 1 Universal Algebras 1 Operations joseph.muscat@um.edu.mt 1 March 2013 A universal algebra is a set X with some operations : X n X and relations 1 X m. For example, there may be specific constants c (n = 0), functions x x (n = 1), and binary operations (x, y) x y (n = 2), etc. An n-ary operation is sometimes written as x y... instead of (x, y,...), and an m-ary relation as x y... instead of (x, y,...). Elements are indistinguishable when (..., x,...) = (..., y,...), (..., x,...) (..., y,...). A subalgebra is a subset closed under all the operations x, y,... Y x y... Y (The relations are obviously inherited.) A, If A i are subalgebras, then i A i is a subalgebra. [A], the subalgebra generated by A, is the smallest subalgebra containing [A] := { Y X : A Y, Y is a subalgebra } Hence A Y [A] Y (for any subalgebra Y ). A B is the largest subalgebra contained in the algebras A and B; [[A B ] is the smallest containing them. The collection of subalgebras form a complete lattice. For any subsets, A [A], [[[A]]] = [A] A B [A] [B ], [A] [B ] = [A B ], [A B ] [A] [B ] The map A [A] is thus a closure map on the lattice of subsets of X, with the closed sets being the subalgebras. 1 Relations are not usually included in the definition of universal algebras.

Joseph Muscat 2015 2 Proof: Let x, y,... [A], then for any sub-algebra Y A, x y Y, so x y [A]. Hence A B [B ] gives [A] [B ]. In particular, [A] [A], so [[A]]] [A] [[[A]]]. A, B A B so [[A] [B ] [A B ]; A, B [A] [B ], so [A B ] [A] [B ]. When the number of operations is finite, the generated subalgebra can be constructed recursively as [[A] = n N B n where B 0 := A, B n + := B n (B n ). Proof: that B n [A] and A = B 0 B n are obvious (by induction); if x,..., y n B n then x B n1,..., y B nk, so r, x,..., y B r and (x,..., y) B r + n B n; so [A] = n B n. Hence if A is countable, so is [A]. The free algebra generated by A is that algebra in which x y... are distinct from each other, for any x, y,... A, and there are no relations. 2 Morphisms The morphisms (also called homomorphisms) between compatible universal algebras (i.e., with the same type of operations and relations) are those functions φ : X Y which preserve all the operations and relations φ(x y...) = φ(x) φ(y)... x y... φ(x) φ(y)... For the special constants, functions, and binary operations, this means φ(c X ) = c Y, φ(x ) = φ(x), φ(x y) = φ(x) φ(y). An algebra with its morphisms forms a category. When φ(x) x, a morphism is an isomorphism when it is bijective (since φ(φ 1 (x) φ 1 (y)...) = x y..., so φ 1 is a morphism). The monomorphisms are the 1-1 morphisms; the epimorphisms are the onto morphisms. Proposition 1

Joseph Muscat 2015 3 For a morphism φ : X Y, If A is a subalgebra of X, then so is φa For any subset S X, φ[s ] = [φs ] If B is a subalgebra of Y, then so is φ 1 B Proof: If φ(x), φ(y),... φa, then φ(x) φ(y) = φ(x y ) φa since A is a subalgebra. Let x, y,... φ 1 B, i.e., φ(x), φ(y),... B, then φ(x y ) = φ(x) φ(y) B, hence x y φ 1 B. Finally, if φa C (a subalgebra of Y ), then A [A] φ 1 C, so [φa] = φ[a]. Thus if morphisms agree on a set S, then they are equal on [S ]. Products: X Y can be given an algebra structure by defining the operations (x 1, y 1 ) (x 2, y 2 )... := (x 1 x 2..., y 1 y 2...) (x 1, y 1 ) (x 2, y 2 )... := (x 1 x 2...) and (y 1 y 2...) More generally, X A is an algebra with (f g...)(a) := f(a) g(a)..., f g... := f(a) g(a)... a A. There is also a coproduct (or free product). An algebra is said to be decomposable when X = Y Z with Y, Z = X. Quotients: An equivalence relation on X which is invariant under the operations and relations is called a congruence (or stable relation), i.e., x 1 x 2, y 1 y 2,... (x 1 y 1...) (x 2 y 2...) and x 1 y 1... x 2 y 2... The operations and relations can then be extended to act on the set X/ of equivalence classes [x] [y]... := [x y...] [x] [y]... := x y... i.e., the mapping π : x [x] is a morphism X X/.

Joseph Muscat 2015 4 For example, indistinguishable elements form a congruent relation, that can be factored away. An algebra X can be analyzed by looking for its congruence relations and then simplifying to get X/ ; this process can be continued until perhaps an algebra is reached that has only trivial congruent relations (x y x = y or x y True), called simple: it has only trivial quotients. Simple algebras are the building blocks of finitary-type algebras. Any morphism φ : X Y which preserves a congruence relation x y φ(x) φ(y), induces a morphism X/ Y, [x] φ(x). The following five Isomorphism theorems apply when the relations satisfy x y... φ(x) φ(y)...: First isomorphism theorem: For any morphism φ : X Y, the relation φ(x) = φ(y) is a congruence (called the kernel of φ), such that the associated quotient space (X/ ker φ) = φx. Proof: That ker φ is a congruence is trivial; so it induces a 1-1 morphism X/ ker φ φx, thus an isomorphism. Third isomorphism theorem: If a congruence 1 is finer than another 2, then 1 induces a congruence ( 2 / 1 ) on X/ 2, and X/ 2 = (X/ 1 )/( 2 / 1 ). Proof: The map X/ 1 X/ 2, [x] [[x]] is a well-defined onto morphism, with kernel ( 2 / 1 ). Second isomorphism theorem: If Y is a subalgebra of X, and is a congruence on X, then is a congruence on Y, and Y/ is isomorphic to the subalgebra Y X/, consisting of all the equivalence classes that contain an element of Y. Proof: The map Y X/, y [y], is a morphism with image Y and kernel. Fourth isomorphism theorem: Given a congruence relation, the subalgebras Y X that satisfy y Y [y] Y, are in correspondence with the subalgebras of X/. Proof: Clearly, Y Z Y Z. Conversely, if Y Z and y Y, then [y] Z, so [y] = [z] for some z Z, thus y z Z. Fifth isomorphism theorem: Given congruences 1, 2 on X 1,X 2, the relation (x 1, x 2 )( 1 2 )(y 1, y 2 ) := (x 1 1 y 1 ) and (x 2 2 y 2 ) on X 1 X 2 is a congruence, and X 1 X 2 X 1 = X 2. 1 2 1 2 Proof: Let φ : X 1 X 2 X1 1 X2 2, (x 1, x 2 ) ([x 1 ], [x 2 ]). This is a morphism with kernel given by ([x 1 ], [x 2 ]) = ([y 1 ], [y 2 ]) x 1 1 y 1 and x 2 2 y 2.

Joseph Muscat 2015 5 There are two senses in which a space X is contained in a larger space Y : 1. Externally, by embedding X in Y, i.e., there is an isomorphism ι : X B Y, denoted X Y because X is, effectively, a subspace of Y ; 2. Internally, by covering X by Y, i.e., there is an onto morphism π : Y X, so X = Y/ ker π; each element of X is refined into several in Y. Then every morphism φ : X Y splits up into three parts with an inner core bijective morphism: X π X/ ker φ im φ ι Y. For example, X Y π X ι X Y. Theorem 2 Every finitely-generated algebra is a quotient of the finitely-generated free algebra. (the same is true if the operations have intrinsic properties, i.e., subalgebras, quotients and products have the same properties) An endomorphism φ : X X induces a sequence of embeddings and partitions: A descending sequence of embedded spaces X im φ im φ 2 n im φ n An ascending sequence of partitions, where x n y φ n (x) = φ n (y), 0 ker φ ker φ 2 n ker φ n 1. φ is said to be of finite descent down to n when im φ n = im φ n+1 (= im φ n+2 = ), i.e., φ is onto im φ n. In this case, every element can be represented, modulo n, by some element in im φ n. Proof: φ n (x) = φ n+1 (y), so x n φ(y) n n φ n (z). 2. φ is of finite ascent up to n when ker φ n = ker φ n+1 (= ker φ n+2 = ), i.e., φ is 1-1 on im φ n. Then each n-equivalence class contains at most one element of im φ n. Proof: φ n (x) n φ n (y) implies x n+n y, so x n y, i.e., φ n (x) = φ n (y). 3. If φ has finite ascent and descent, then the two sequences have the same length. Thus every element can be represented modulo n by a unique element in im φ n, and φ is bijective on im φ n Proof: The ascent cannot be longer than the descent else φ n (x 1 ) = φ n (y 1 ) = φ n+1 (y 2 ) and φ n+1 (x 1 ) = φ n+1 (y 1 ), so x 1 n y 1 but x 1 n+1 y 1, and x 2 n+1 y 2 but x 2 n+2 y 2, etc. The descent cannot be longer than the ascent else if φ m (x) = φ m+1 (y) then x m φ(y), so x n φ(y).

Joseph Muscat 2015 6 3 Compatibility An operation is associative when (x, y,...) exists for every finite number of terms, such that (..., (x, y,...), u,...) = (..., x, y,..., u,...) For example, x y z = (x y) z; so the operation reduces to three basic ones such that 0 := (), (x), (x, y) 0 x = (x), (x y) z = x (y z). Note that (x) = (y) is a stable equivalence relation (with respect to ), with related elements being indistinguishable algebraically; but can assume (x) = x by taking the quotient space and renaming; in this case, 0 x = x. An operation is distributive over another when (..., (x, y,...), z,...) = ( (..., x, z,...), (..., y, z,...),...) For 0-,1-,2-operations, this means (0) 1 2 (0) (0 = 1) (0 = 0 ) (0 x = 0 = x 0) 1 (1 = 1) x = x (x y) = x y = x y 2 (x + y) = x + y x z + y z = (x + y) z z x + z y = z (x + y) Two operations commute when ( (x 1, y 1,...), (x 2, y 2,...),...) = ( (x 1, x 2,...), (y 1, y 2,...)) For 1-,2-operations, this means 0 1 2 0 0 = 1 1 x = x (x + y) = x + y 2 (x 1 + y 1 ) (x 2 + y 2 ) = x 1 x 2 + y 1 y 2 Commuting 2-operations with identities must actually be the same, and must be commutative x y = y x and associative. Proof: The identities are the same: 1 = 11 = (1 + 0)(0 + 1) = 10 + 01 = 0 + 0 = 0. ab = (a + 0)(0 + b) = a0 + 0b = a + b so the operations are the same, and (ab)(cd) = (ac)(bd). Hence ab = (1a)(b1) = (1b)(a1) = ba and a(bc) = (a1)(bc) = (ab)c. In particular if an operation with identity commutes with itself, it must be associative and commutative x y = y x.

Joseph Muscat 2015 7 3.1 Examples 1. The simplest example is a set with a single constant 0, sometimes called the field with one element. 2. A set with a constant and a 1-operation; properties may be 0 = 0, x = x. 3. A set with a constant, a 1-operation, and a 2-operation: e.g. 0 x = x, (x y) = y x.