Abelian groups ℵ0-categorical over a subgroup

AbstractAn abelian group A is said to be ℵ0-categorical over its subgroup B when there is a unique countable model of the theory of A with distinguished subgroup B for any possible choice of countable dinstinguish subgroup. We give necessary and sufficient conditions for an abelian group to be ℵ0-categorical over one of its subgroups. Furthermore we give an axiomatization of such theories in terms of some first-order invariants and show that these invariants can have any value as long as they satisfy some minor conditions. With these results we obtain a new proof of Hodges' decomposition theorem (Corollary 2.6). Finally, in the case of torsion-free abelian groups we conclude that A is ℵ0-categorical over its subgroup BIT iff B=mA for some integer m

Equivariant Zariski Structures

A new class of noncommutative $k$-algebras (for $k$ an algebraically closed field) is defined and shown to contain some important examples of quantum groups. To each such algebra, a first order theory is assigned describing models of a suitable corresponding geometric space. Model-theoretic results for these geometric structures are established (uncountable categoricity, quantifier elimination to the level of existential formulas) and that an appropriate dimension theory exists, making them Zariski structures

Covers of Multiplicative Groups of Algebraically Closed Fields of Arbitrary Characteristic

We show that algebraic analogues of universal group covers, surjective group homomorphisms from a $\mathbb{Q}$-vector space to $F^{\times}$ with "standard kernel", are determined up to isomorphism of the algebraic structure by the characteristic and transcendence degree of $F$ and, in positive characteristic, the restriction of the cover to finite fields. This extends the main result of "Covers of the Multiplicative Group of an Algebraically Closed Field of Characteristic Zero" (B. Zilber, JLMS 2007), and our proof fills a hole in the proof given there.Comment: Version accepted by the Bull. London Math. So