category-theory
A subobject of $X$ in category theory is defined as a certain equivalence class of monomorphisms to $X$.
What I don’t understand about this definition, is that this doesn’t actually define an object in the category (as far as I know). It seems at first sight to me to be a type error? We are looking for an object, not an equivalence class of arrows? How does this define an object?
The question is ancient, but IMHO the most convincing answer is missing:
The definition of "subobject in a category $\mathcal{C}$" is chosen in such a way that it generalize the notion of $k$-vector subspaces (when $\mathcal{C} = \mathsf{Vect}_k$), the notion of subsets (when $\mathcal{C} = \mathsf{Set}$), the notion of subrings (when $\mathcal{C} = \mathsf{Ring}$), and lots of other classical "sub-something" notions that appear throughout mathematics (probably not all of them, though). If you were to define "subobjects of $A \in \mathcal{C}$" to mean "monomorphisms with codomain $A$", then the notion of a subobject would not generalize all of these classical notions, because you get too many different subobjects that correspond to the same "sub-something". For instance, in $\mathsf{Set}$, the monomorphisms $\emptyset \to \left\{1\right\}$, $\left\{1\right\} \to \left\{1\right\}$ and $\left\{2\right\} \to \left\{1\right\}$ would be three different subobjects of the object $\left\{1\right\}$, but there are only two subsets of the set $\left\{1\right\}$. So this would be a bad definition.
However, if you define subobjects of $A \in \mathcal{C}$ to be isomorphism classes of monomorphisms with codomain $A$ (where "isomorphism" is to be correctly interpreted: an isomorphism between two monomorphisms $\alpha : S \to A$ and $\alpha^{\prime} : S^{\prime} \to A$ means a morphism $s : S \to S^{\prime}$ satisfying $\alpha = \alpha^{\prime} \circ s$), then, in all of the examples listed above, the subobjects of $A$ are in a canonical bijection with the "sub-somethings" (i.e., the $k$-vector subspaces, or the subsets, or the subrings). For instance, in $\mathsf{Set}$, the subobjects of a set $A$ are the isomorphism classes of monomorphisms with codomain $f$. The isomorphism class of such a monomorphism $f : S \to A$ can be identified with the subset $f\left(S\right)$ of $A$. Thus, the subobjects of $A$ are in bijection with the subsets of $A$ here. The same construction works for rings and for $k$-vector spaces.
The good definition of a subobject also has the advantage (compared with the bad definition) that the subobjects of a given object $A \in \mathcal{C}$ often form a set (as opposed to just a class). I don't personally find this vital; it is not usually true in constructive mathematics anyway, and I don't believe that a definition is necessarily bad just because it sometimes returns proper classes.
WARNING: A few details below only work for fields, or for right division rings, but no substantial changes are necessary, so I'm probably going to leave fixing this to the reader.
The idea is roughly this. Work in some category $\mathcal C$, to which we'll add properties as we need them. One way to express the notion of division ring was given in the comments by Derek: a division ring is a ring $\mathcal R$ (that is, a ring object: an object $\mathcal R$ equipped with maps $\times,+:\mathcal R\times \mathcal R\to \mathcal R,0,1:*\to \mathcal R$ satisfying the usual axioms) in which the union of the subobjects $0$ (a map out of a terminal object is always a monomorphism) and $\mathcal R^\times$ is all of $\mathcal R$. So, we have to have unions of subobjects available in $\mathcal C$.
We also have to be able to define $\mathcal R^\times$, which is supposed to satisfy some sentence like $\mathcal R^\times=\{x\in \mathcal R:\exists x'\in \mathcal R:xx'=1\}$. This isn't too hard. We have a subobject $H\subset \mathcal R\times\mathcal R$ which should be defined as $\{(x,x')\in \mathcal R\times \mathcal R:xx'=1\}$. To define this categorically, we simply take the pullback of the multiplication map $\times:\mathcal R\times \mathcal R\to \mathcal R$ along the unit map $1:*\to \mathcal R$. Having defined $H$, we can construct $\mathcal R^\times$ as the image of the first projection map $H\to \mathcal R$. So, in $\mathcal C$ we need every morphism to have an image which is a subobject.
However, this definition is a bit strong. For instance, if your category is the topos of sheaves on an interesting topological space $X$, and $\mathcal R$ is the sheaf of continuous functions valued in some topological ring $R$, then $\mathcal R^\times(U)$ will be the elements of $\mathcal R(U)$ which are everywhere nonzero in $U$. So it'll be almost impossible to make $\mathcal R$ a division ring in the above sense: you'd be asking that every section of the sheaf be either everywhere or nowhere zero. This is the reason for the other notions of field in the nLab article linked by Derek.
A particularly nice alternative is to say that the complement of $\mathcal R^\times$ is precisely $0$. This sounds like the same thing, of course, but logic internal to a category is rarely classical, so it's actually not. More formally, we define the complement $\left(\mathcal R^\times\right)'$ to be the largest subobject of $\mathcal R$ whose intersection with $\mathcal R^\times$ is empty (initial.) To define the complement directly in this way, $\mathcal C$ needs to admit arbitrary unions of subobjects; it's possible to get away with less than this by characterizing the complement differently, but in so we need $\mathcal C$ to be a "Heyting" category. In the example of sheaves on $X$ valued in a topological ring $R$, this is the subsheaf of functions $f:U\to R$ such that $f^{-1}((R^\times)')$ is dense: $f$ is almost everywhere valued in nonunits. It's much easier to make $\mathcal R$ a division ring under this definition! We merely need to know that a continuous $R$-valued function which is a non-unit on a dense subset is actually everywhere zero. For this it suffices that $R$ itself be a division ring in which $0$ is a closed subset, as is usually the case.
It may be worth observing that in our example, the union of $\mathcal R^\times$ with its complement is not all of $\mathcal R$: it doesn't include those $R$-valued functions whose zeroset is nonempty but nowhere dense. This non-Boolean aspect of the logic is, again, characteristic of most categories. Unfortunately, it means that even though there are many "division rings" in many categories, not all theorems of elementary linear algebra hold. For instance, we cannot prove that every finitely generated $\mathcal R$-module admits a basis, which would imply in particular that every finite-dimensional vector bundle (viewed as a $C^0(\mathbb{R})$-module) was a direct sum of line bundles.
Best Answer
The object is the domain of the monomorphism; this is why IMO the correct definition should just be "a monomorphism," period. Taking equivalence classes doesn't buy you much, really.
If you haven't, it's worth working through special cases of this definition. In $\text{Set}$ you get subsets, in $\text{Grp}$ you get subgroups, and so on and so forth.