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Every vector space $V$ over a field $F$ has a basis, assuming the axiom of choice. More generally, any linearly independent subset $A\subseteq V$ extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

Use the poset of linearly independent subsets containing $A$:It is a set, being a subset of the power set of $V$, and is nonempty because $A\in {\mathcal{P}}_A$.

A short proof outline supplies both essential steps:

Taking $A=\varnothing$ proves existence, including $V=\{0\}$, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail: $\{v\}$ and $\{2v\}$ for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.

A vector space with scalar field $\mathbb{Q}$ is equivalently a torsion-free divisible Abelian group. An embedding of such groups automatically respects rational scalar multiplication because solutions to $nb=ma$ are unique.

A ring is von Neumann regular if every element $a$ has an element $b$ with $a=aba$. In a commutative ring, this is $a=a^{2}b$. An arbitrary direct product of rings whose factors are fields has this property by taking inverses in each nonzero coordinate. Every localization at a prime ideal of a commutative von Neumann regular ring is a field.

The ring with a single element, which is both its additive and multiplicative identity. If a unital ring has $0=1$, then every element $r$ equals $r1=r0=0$, so it is this ring. Under the usual definition it is excluded from integral domains and fields. The quotient ring $\mathbb{Z}/(1)$ is an example.