I want to build some structure and I think a semilattice is the best way to do so, but don't know for sure.
I have some (non-empty) set $X$ with some distinguished elements $a,b,c,d$ and $e$. I know that the free join-semilattice $F(X)$ exists and is by definition the collection of finite subsets of $X$ ordered by inclusion.
Now I want to keep the lattice but imposing the relations
- $a$ is the least element, that is, for any other $x\in F(X)$ we have $a\le x$ or $a\vee x=x$
- $e$ is the greatest element, that is, for any other $x\in F(X)$ we have $x\le e$ or $x\vee e=e$
- We also have $b\vee c\vee d=e$
Is there something like a quotient by the normal (?) sublattice generated by the relations like in the free group construction to make a group presentation? Can I say something like "Let $L$ be the smallest join-semilattice such that [conditions]"?? I hope the intention is clear.
Also I'm wondering if the free (abelian) monoid could do the job better, because I don't know what to do about the empty set being a part of the semilattice (and a least element). Can I not use the element $a$ and still force a greates element?
Thanks!
A semilattice is a commutative and idempotent monoid, so you can indeed work with monoids, or with commutatice monoids. In this setting, the least element is the identity of the monoid. You can indeed define presentations, but it is not defined by a sublattice, but by a congruence and you don't get the smallest but the largest semilattice satsfying the given relations.
In your case, the lattice $L$ would be the quotient of the free monoid $\{b,c,d,e\}^*$ by the following relations: $xy=yx$ and $x^2 = x$ for all $x, y \in \{b,c,d,e\}^*$, $eb=ec=ed=e$ and $bcd=e$.
Of course $L$ is also the quotient of the free abelian monoid generated by $\{b,c,d,e\}$ by the following relations (in additive notation): $x+x =x$ for all $x$, $e + b = e + c = e + d = e$ and $b + c + d = e$.
Finally, setting $X = \{a, b, c, d, e\}$, $L$ is the quotient of the free $\vee$-semilattice $F(X)$ by the relations $a \vee b = b$, $a \vee c = c$, $a \vee d = d$, $a \vee e =e$, $e \vee a = e \vee b = e \vee c =e \vee d= e$ and $b \vee c \vee d = e$.
It would actually be even simpler to start with the free bounded semilattice on $X$, in which the top and bottom elements are already given.