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A first-order model class is finitely axiomatizable in a fixed language if a finite set of sentences, equivalently one finite conjunction, has exactly that model class. Relative finite axiomatizability permits a fixed background theory and asks for only finitely many additional axioms. The distinction matters in an infinite scalar language, where even vector-space structure itself needs infinitely many named-symbol constraints.

No finite theory in the full language containing one function symbol for each scalar of an infinite field can axiomatize exactly its nonzero vector spaces. It mentions only finitely many symbols; changing an unused scalar function to violate preservation of zero leaves those finite sentences unchanged. Relative to the vector-space axiom schema, nonzeroness is instead the single axiom $\exists x(x\ne 0)$. Absolute finite axiomatizability and finite additional axiomatizability must be distinguished.

A finite axiomatization of a class of structures is a finite set of first-order sentences whose first-order models are exactly that class. The existence of such a set is finite axiomatizability.

The class of torsion-free abelian groups has an infinite first-order axiomatization but no finite axiomatization. Add $\forall x(nx=0\to x=0)$ for every positive integer $n$ to the abelian group axioms. If an equivalent finite theory existed, compactness would make it follow from finitely many of these conditions. A cyclic group of prime order larger than the finitely many integers mentioned satisfies those conditions but has torsion, a contradiction.

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.

The cardinality of a vector space is central: here “infinite” must continue to refer to the underlying set. Over a countably infinite field, every nonzero finite-dimensional space is already infinite. Thus the model class now consists of all nonzero $K$-vector spaces, not only infinite-dimensional ones.

Since independence of a pair is not definable over an infinite field, for (i), independence of a single vector is still $x\ne 0$, but independence of tuples of length at least two is not definable by one formula in this one-sorted language. Quantifier elimination follows from the linear-equation argument in the preceding part for all these infinite domains. Any candidate two-variable formula therefore involves only finitely many linear equations. Take an independent pair, and compare it with $(x,{\lambda }_{a}x)$, where $x\ne 0$ and $a\in K$ avoids the finitely many ratios making one of those nontrivial equations vanish. Both pairs satisfy the same listed atomic formulas, contradicting a proposed definition of independence. Longer tuple independence would define pair independence by existentially completing a pair inside an infinite-dimensional model, giving the same obstruction.

For (ii), the vector-space axiom schema together with $\exists x(x\ne 0)$ axiomatizes the infinite underlying spaces. For (iii), there is again no universal axiomatization, because the zero subspace is a substructure.

For finite axiomatizability with infinitely many scalar symbols, in (iv) there is no finite axiomatization in the full language: any finite axiom set mentions only finitely many scalar symbols. In a nonzero vector space alter an unused nonzero ${\lambda }_a$ to send zero to a nonzero vector, retaining all mentioned interpretations. The finite sentences keep their truth values, but a vector-space axiom is now false. Relative to the already imposed vector-space axioms, the extra condition is finitely axiomatized by the single nonzero-vector sentence. These absolute and relative answers must be distinguished.

For (v), the one-variable algebra still has two atoms and four Elements. In two variables the isolated atoms are $(0,0)$ and the countably many rank-one directions, represented by $x\ne 0,y={\lambda }_{a}x$ and by $x=0,y\ne 0$. There is also a nonisolated complete type of an independent pair. A formula using finitely many linear equations either picks finitely many isolated atoms, or contains that independent type and all but finitely many isolated atoms. Therefore $F_2/E_2(T)$ is the finite-cofinite Boolean algebra on a countably infinite set of atoms, with the independent type the extra nonisolated point in its Stone space. In particular this algebra is countably infinite, rather than the full power set of its atoms.

For (vi), dimension can be $1,2,3,\dots$ or ${\aleph }_0$ while the domain is countable, producing pairwise nonisomorphic countable models. Thus $T$ is not countably categorical. For an uncountable $\kappa$, a vector space over this countable field has cardinality $\kappa$ exactly when its dimension is $\kappa$, since in general a nonzero space has size $\max (|K|,\dim V)$ when that maximum is infinite. Basis bijections then give categoricity in every uncountable cardinal. Quantifier elimination also shows that these different positive dimensions are elementarily equivalent; independence of an arbitrarily long tuple is not being silently treated as definable over this infinite field.