We also have the complete book here. Clasen and M. Kearnes and E.

## Simple Word Problems in Universal Algebras

Kiss, The Shape of Congruence Lattices , pdf. Willard, An overview of modern universal algebra , pp. Andretta, K. Kearnes and D. Zambella, Lecture Notes in Logic, vol. Press, Freese and O. Articles and Notes J. II Math.

## Characterization of protomodular varieties of universal algebras

Birkhoff, On the structure of abstract algebras , Proc. Cmabridge Phil.

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Freese, Computing congruences efficiently , Algebra Universalis, 59 , Freese, Notes on the Birkhoff Construction , pdf. Freese and M. Valeriote, On the complexity of some Maltsev conditions , Internation J. Algebra and Computation, to appear. The categories y,p and N y l R y are isomorphic.

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The structures A and M A, z have the same polynomial functions and the same congruences. The variety of all quasigroups is a modular variety, in fact permutable. Following Jezek and Kepka [13], we call this subvariety the variety of cia-quasi- groups commutative, idempotent, abelian quasigroups—any quasigroup satisfying a is abelian. It is a straightforward verification that k is a congruence on Q X Q and that the diagonal is a congruence class. To define module addition and subtraction, choose any z g Q to be the zero element.

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This has been known since as a special case of "Toyoda's Theorem" [18]. Similarly, the product in the ring corresponds to the product of rational numbers in F use induction on the length of a term. Thus we see that the ring R y — Z[i]. We would now like to apply Theorem 3. The key is to show that the information contained in the module N y is not needed. This will follow from Theorem 3.

Let y be an abelian variety of finite type. Then yis deductive if and only if y has the subalgebra condition and R y is deductive. Let A be any non tri vial member of y and let k be the congruence on A X A that contains the diagonal as a class. Thus by Theorem 3. Let SI be a subquasivariety of y.

## Maths - Universal Algebra - Martin Baker

It is a somewhat tedious, but not difficult, exercise to show that SM is closed under the formation of submodules, products and ultraproducts use the corresponding property of J together with 3. Furthermore, S is closed under homomorphic images if and only if SlMis so closed. D Returning to cia-quasigroups, let us apply Theorem 3. The ring is not Artinian, but as remarked earlier, Example 2.

On the other hand, Z[j] is a PID and, by 2. Translating back to quasigroups, every proper subvariety of cia-quasi- groups is a deductive variety. We now apply Theorems 2. As an example: for any natural number n, the variety of abelian groups of exponent n is directly representable. We require the following properties of directly representable varieties from McKenzie [14, Theorems 5. Then yis permutable hence modular and every subdirectly irreducible algebra in yis finite and either simple or abelian.

Thus we have the following characterization. Let y be a directly representable variety of finite type, yis deductive if and only if: 1 y is equationally complete, or 2 y has the subalgebra condition and yah is deductive. Let y be deductive.

Then by 3. Thus y has no nontrivial abelian algebras, so y must be semisimple. By Corollary 3. On the other hand, if y has the subalgebra condition, then alternative 2 obviously holds. For the converse, let y be equationally complete. Let M be a finite, nontrivial algebra of y of smallest cardinality. M is simple, has no proper, nontrivial subalgebras and generates y, which is permutable.

Therefore by [16, Theorems 1. Since M can be embedded into every direct power of M, and since V is locally finite, M is primitive. Therefore by Theorem 3. Now suppose y has the subalgebra condition and yah is deductive. By Theorem 3. There are two cases. Case 1. B is abelian. Then 6 d [1,1] in Con A. By assumption, yah is deductive, so B g IS C by 3. Case 2. B is nonabelian. Then B is simple, so 0 is a coatom of Con A. D Example. Thus y is not deductive. Since the order of Q is prime, Q is simple.

A is isomorphic to the group Z2, so A is simple and abelian. By [3, Theorem 1. Q is nonabelian since it is simple, but has a subalgebra and y has the subalgebra condition since Q and A have idempotents. The subvariety yah is generated by A, so the associated ring is Z2.