= Solution
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 <first-order model>[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 $x\ne0$, 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 $x\ne0$ 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 <first-order model>[model], giving the same obstruction.
For (ii), the vector-space <axiom> schema together with $\exists x(x\ne0)$ axiomatizes the infinite underlying spaces. For (iii), there is again no universal axiomatization, because the zero subspace is a <substructure of a first-order structure>[substructure].
For <finite axiomatizability with infinitely many scalar symbols>, in (iv) there is no <finite axiomatizability>[finite axiomatization] in the full language: any finite <axiom> <set> mentions only finitely many <scalar multiplication>[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. \b[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 <atom of a Boolean algebra>[atoms] and four <Elements>. In two variables the isolated <atom of a Boolean algebra>[atoms] are $(0,0)$ and the countably many rank-one directions, represented by $x\ne0,y=\lambda_a x$ and by $x=0,y\ne0$. There is also a nonisolated <complete type> of an independent pair. A formula using finitely many <linear equations> either picks finitely many isolated <atom of a Boolean algebra>[atoms], or contains that independent type and all but finitely many isolated <atom of a Boolean algebra>[atoms]. Therefore \b[$F_2/E_2(T)$ is the <finite-cofinite Boolean algebra>[finite-cofinite Boolean algebra] on a countably <infinite set> of <atom of a Boolean algebra>[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 <atom of a Boolean algebra>[atoms].
For (vi), <dimension of a vector space>[dimension] can be $1,2,3,\ldots$ or $\aleph_0$ while the domain is <countable>, producing pairwise nonisomorphic <countable> <first-order model>[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 of a vector space>[dimension] is $\kappa$, since in general a nonzero space has size $\max(|K|,\dim V)$ when that maximum is infinite. <vector space basis>[Basis] <bijections> then give \b[categoricity in every uncountable cardinal]. <Quantifier elimination> also shows that these different positive <dimension of a vector space>[dimensions] are <elementarily equivalent>; independence of an arbitrarily long tuple is not being silently treated as definable over this infinite <field>.
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