can we "represent" this finite topology as submodules of $R^X$? Specific Question Let $X:= \{a,b,c,d\}$ equipped with the topology with base $c,d,abd,abc$ – where out of convenience we identify $x$ with $\{x\}$ and leave out union signs. Is there a ring $R$ containing a field $k$ (preferably $\mathbb R$ or $\mathbb C$, because I think I have a better mental picture of these than for other fields) and an $R$-module structure on $k^{abcd}$ such that the $k^S$, $S\subseteq X$, which are $R$-submodules are precisely the ones where $S$ is closed?
Context I was trying to revisit my knowledge of finite topologies and non-$T_1$ spaces, and found $X$ to be a nice example of a topology which is $T_0$ but for which the specialization preorder is not a lattice.  I then remembered that in representation theory of groups, one often can take a finite structure, like a finite group $G$, andd turn it into an interesting $RG$-Module (via the permutation action). One could do something similar with finite topological spaces: attach a copy of $k$ to each element, and you've got a $k$ vector space. To capture other structure, it is natural to ask whether we can identify the closure operator induced by the topology with something related to this vector space. Of course, acting with e.g. a linear operator $T$ yields a closure operator in terms of invariant subspaces: indeed, these are the submodules of $k^X$ when viewed as a $k[T]$-module in this way. For instance: if our topology was generated by $a,ab,abc$, a corresponding operator would map $a\mapsto b \mapsto c \mapsto d\mapsto 0$, since the invariant closures corresponding to $k^{a,b,c,d}$ are$k^S$ with $S=abcd,bcd,cd,d$, respectively – and these are precisely the closed sets.
My ”gut feeling“ says that we could try to realize the specialization preorder as some sort of monoid action of $M$ and construct a Ring $M[k]$ out of this – whatever that would be – to just copy the action on $X$ to $k^X$. Unfortunately, my knowledge of „actions“ stops when our arrows are not invertible anymore (i.e., we have a preorder instead of (the category-theoretic isomorphism quotient of) a groupoid).
 A: More generally, let $C$ be any category.  Then you can define a $k$-algebra $k[C]$ which has as a basis the morphisms of $C$, with multiplication given by composition in $C$ or $0$ when the composition is not defined (because the domain and codomain do not match up).  Given a functor $F:C\to \mathtt{Vect}_k$, we then get an action of $k[C]$ on the direct sum $V$ of the values of $F$ on all the objects of $C$, with morphisms of $C$ acting via $F$ on the summand of $V$ corresponding to their domain and acting by $0$ otherwise.
In particular, suppose $F$ is the constant functor with value $k$, so $V=k^{\oplus O}$ where $O$ is the set of objects of $C$.  Then I claim the $k[C]$-submodules of $V$ are exactly the subsets of the form $k^{\oplus S}$ where $S\subseteq O$ has the property that if $s\in S$ and there is a morphism $s\to t$, then $t\in S$.  It is clear that these subsets are submodules.  Conversely, suppose $M$ is a submodule of $V$.  Note that the identity map on an object $o\in O$ acts on $V$ as the projection onto the summand corresponding to $o$.  It follows that if $m\in M$, so are each of its projections, and thus $M=k^{\oplus S}$ where $S$ is the set of $s$ such that $M$ contains an element whose $s$-coordinate is nonzero.  But now any morphism $s\to t$ maps the summand corresponding to $s$ to the summand corresponding to $t$ isomorphically, so if $s\in S$ then $t$ must be in $S$ as well.
In particular, if $C$ is a preorder, this gives a module structure on $k^{\oplus C}$ whose submodules are exactly sets of the form $k^{\oplus S}$ where $S$ is an upper set of $C$.  Taking the specialization preorder of an arbitrary topological space (or its opposite, depending on your conventions), this gives a module structure where the submodules correspond to the specialization-closed sets.  For a finite space, specialization-closed sets are the same as closed sets.
