Past exam of the model theory and Gödel's incompleteness theorems course of the University of Oxford / 2009 / B / 5 / c

For each fixed finite tuple length $n$, use the finite conjunctionIt asserts that no nontrivial scalar combination vanishes, exactly linear independence. It is an $L$-formula because the field is finite and every scalar multiplication is named. Scalar coefficients are indices of function symbols, not variables ranging over a separate field sort. There is no single finite formula with arbitrarily many free variables; the assertion is a formula for each prescribed finite $n$.

Use the vector-space axioms: addition is an abelian group operation with identity zero; ${\lambda }_1(x)=x$; ${\lambda }_a{\lambda }_b(x)={\lambda }_{ab}(x)$; ${\lambda }_a(x+y)={\lambda }_a(x)+{\lambda }_a(y)$; and ${\lambda }_{a+b}(x)={\lambda }_a(x)+{\lambda }_b(x)$, with ${\lambda }_0(x)=0$. Group inverses may be expressed by ${\lambda }_{-1}$, so no extra inverse symbol is needed.

Add the dimension-unbounded schemeOver a finite field, a finite-dimensional space has finitely many Elements, while every infinite-dimensional space is infinite. Hence this scheme axiomatizes exactly the infinite underlying vector spaces requested. In this finite-field case the vector-space axiom list itself is finite; the infinitude of the additional dimension scheme is substantive.

No. Universal sentences are preserved under substructures. Every infinite $K$-vector space has its zero subspace as an $L$-substructure, since zero is named and every scalar multiplication preserves it. This one-element structure is not infinite and is not a model of $T$. Thus the model class is not closed under substructures and has no universal axiomatization, even relative to the vector-space axioms.

No. If a finite conjunction $\varphi$ axiomatized the infinite spaces, it would follow from the vector-space axioms together with all the $D_n$. Compactness would make it follow from finitely many of those axioms and dimension requirements. Let $N$ be at least every dimension mentioned there. The finite-dimensional space $K^N$ satisfies all vector-space axioms and those finitely many $D_n$, hence would satisfy $\varphi$. But it is a finite set and so cannot be a model of the purported axiomatization. This contradiction proves failure of finite axiomatizability, including finite axiomatizability relative to vector spaces.

For quantifier elimination for infinite underlying vector spaces, we first justify that formulas cannot make finer distinctions than their linear-equation data. In any infinite underlying vector space, eliminate an existential variable from a disjunction of conjunctions of linear equations and inequations. If an equation has a nonzero coefficient of that variable, solve for it and substitute. If none does, the inequations forbid only finitely many values, leaving an available value in the infinite domain; the remaining equations concern only the other variables. Terms are linear combinations, so induction gives quantifier elimination for $T$ and completeness, since closed terms reduce to zero.

For one variable there are two possible complete types, zero and nonzero. Both are isolated, and every definable set is their union. ThusFor types of pairs of vectors over a finite field, the linear relation kernel $\{(a,b)\in K^2:ax+by=0\}$ determines the complete type. If $q=|K|$, the possibilities are one rank-zero pair, $q+1$ rank-one directions, and one rank-two independent type. Concretely these are $(0,0)$; $x\ne 0$ with $y={\lambda }_a(x)$ for each $a\in K$; $x=0,y\ne 0$; and the independent pairs. Their isolating formulas are finite because $K$ is finite. HenceEvery class is the union of a selection of these atoms. Basis extension also shows that pairs with the same relation kernel are conjugate by a linear automorphism within any infinite-dimensional model.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.