I second William's comment that the concept of "base" means different things in different areas of math. So let me focus on the concept of "base" thought as "minimal set of elements that generate a structure".
There is a nice and simple categorical approach to this concept. Categorically, to give an algebraic structure, is to give a monad on a category $T \colon \mathcal{C} \rightarrow \mathcal{C}$. An algebra for a monad is an object in the category $\mathit{Alg}(T)$ of the Eilenberg-Moore resolution for the monad. This resolution comes equipped with a pair of adjoint functors: the forgetful functor $U \colon \mathit{Alg}(T) \rightarrow \mathcal{C}$ and its left adjoint "the free functor" $F \colon \mathcal{C} \rightarrow \mathit{Alg}(T)$ such that $T = U \circ F$.
Now, the crucial thing is that we may compose these functors in the other direction obtaining a comonad $D = F \circ U$ on the category $\mathit{Alg}(T)$. As it is the case with every comonad, $D$ has its own coresolution in the category of the Eilenberg-Moore coalgebras $\mathit{coAlg}(D)$. Therefore, we may consider coalgebras over algebras. It turns out that in many cases such induced coalgebras over algebras correpond to the usual notion of base. For example: coalgebras over vector spaces decompose vectors into coefficients according to a base; coalgebras over algebras for the power-set monad give decomposition of elements of an atomic (complete) lattice into their descending atoms.
It may help to work out a simple example. Let us consider $\mathcal{C} = \mathbf{Set}$ together with a finite sequences monad:
$$T(X) = X^*$$
The Eilenberg-Moore resolution for $T$ gives the category of monoids $\mathbf{Mon}$. The forgetful functor $U \colon \mathbf{Mon} \rightarrow \mathbf{Set}$ assigns to a monoid its carrier:
$$U(\langle M, \bullet, \iota\rangle) = M$$
and the free functor $F \colon \mathbf{Set} \rightarrow \mathbf{Mon}$ gives a monoid action via concatenation:
$$F(X) = \langle X^*, \circ, [\;]\rangle$$
The induced comonad $D = F \circ U$ associates with every monoid the free monoid build upon the same carrier:
$$D(\langle M, \bullet, \iota\rangle) = \langle M^*, \circ, [\;]\rangle$$
There is the counit of the comonad $\epsilon \colon \langle M^*, \circ, [\;]\rangle \rightarrow \langle M, \bullet, \iota\rangle$ defined by folding with monoid's multiplication:
$$\epsilon([a_1, a_2, \dotsc, a_n]) = a_1 \bullet a_2 \bullet \cdots \bullet a_n$$
and the comultiplication of the comonad $\delta \colon \langle M^*, \circ, e\rangle \rightarrow \langle M^{**}, \circ, e\rangle$ which decomposes words on singletons:
$$\delta([a_1, a_2, \dotsc, a_n]) = [[a_1], [a_2], \dotsc, [a_n]]$$
By definition, a coalgebra over a monoid $\langle M, \bullet, \iota\rangle$ is a monoid homomorphism $h \colon \langle M, \bullet, \iota\rangle \rightarrow \langle M^*, \circ, [\;]\rangle$ such that:
- $\epsilon \circ h = \mathit{id}$, that is: $h(r)_1 \bullet h(r)_2 \bullet \cdots \bullet h(r)_n = r$, where $h(r)_k$ is the $k$-th element of sequence $h(r)$
- $D(h) \circ h = \delta \circ h$, that is: $h(h(r)_k) = [h(r)_k]$
These conditions say that coalgebras over monoids are tantamount to finite decompositions of elements on indecomposable elements.
