A question on Join homomorphism and Ideals On page 287 of the book Mathematical Methods in Linguistcs, by Barbara Partee, Alice Ter Meulen and Robert E. Wall (Dordrecht, Kluwer Academic Press, 1993), I find the following theorem, which they consider to establish, in their own words "a connection between join homomorphisms and ideals" (Ideals are referred here to lattice ideals; I am not sure whether they are equivalent to ring ideals. Could anyone clarify me that?):
THEOREM 11.4 $I$ is a proper ideal of the lattice $L$ if and only if there is a join homomorphism $G$ from $L$ onto the two element chain $C = \{0,1\}$ such that $I = G^{-1} (0)$, i.e. $I$ consists of those $x$ for which $G(x) = 0$
I am particulary interested in knowing:
1) Whether the connection stated in the theorem relates to some specific object of ring theory or to other results
2) How right and left ideals can be implemented within that theorem?
Thanks in advance (I hope the editing of the theorem comes up correctly)
 A: The ideal in a lattice meant in the theorem is definitely the order theoretic notion of an ideal.  It has nothing to do with ring theory apart from some similarity in the definition.  Regarding (2): Left and right ideals are notions from (non-commutative) ring theory, so that they don't apply here.
In plain English the theorem tells you roughly the following.  Say you have some subset $I$ of a lattice $L$ (a partially ordered set with unique meets and joins) and you want to know if it forms an ideal in $L$, that is, it is closed under going down and forming joins.  The Theorem says that this happens exactly if you can find a lattice map (homomorphism) from $L$ to the two element lattice $C$ which respects joins and whose preimage of the lower element $0\in C$ is the set $I$.  The theorem is not hard to prove.  It is basically unwraping definitions.
A: For question 1, this link seems to provide an answer (that's rather extensive for planetmath). 
With regards to the second question, I'm not sure what you're aiming at. If you consider a lattice algebraically, with binary operations $\wedge$ and $\vee$, your structure will be commutative. What would it, for instance, mean if $a\wedge b\neq b\wedge a$? I therefore wouldn't expect any difference between the notions of ideal, left ideal, right ideal and two-sided ideal.
A: I think the connection should be interpreted in the following way. We can give the following equivalent definition of an ideal $I$ in a lattice $L$.

Proposition. A subset $I$ of a lattice $L$ is an ideal if and only if the following two criteria are met:
  
  
*
  
*For $a \in L$ and $b\in I$ one has $a \wedge b \in I$.
  
*For $a, b \in I$ one has $a \vee b \in I$.
  
  
  Proof. On the one hand, let $I \subseteq L$ be an ideal. For $a \in L$ and $b \in I$ we have $a \wedge b \leq b$, hence $a \wedge b \in I$. Furthermore, $I$ is closed under joins.
Conversely, suppose that $I \subseteq L$ satisfies the above criteria. Let $a \in L$ and $b \in I$ be given with $a \leq b$. Note that we have $a \wedge b = a$, as well as $a \wedge b \in I$. Thus, we have $a \in I$. Furthermore, $I$ is closed under joins. This proves that $I$ is an ideal.

Now this looks a lot like an ideal in a ring. An ideal in a ring has even more structure, as it also contains the additive identity and additive inverses. However, a lattice doesn't generally have any sort of identity (let alone inverses), so this is the closest we can get.
Additionally, just like in the case of rings, the ideals are precisely those sets that occur as kernels of homomorphisms, in a sense that I shall make more precise now. In order to talk about kernels, we need a zero object. We say that a lattice $M$ is bounded below if it has a smallest element $0$, that is, if $0 \in M$ satisfies $0 \leq m$ for all $m \in M$. For arbitrary $m \in M$ we have $m \wedge 0 = 0$ and $m \vee 0 = m$.

Proposition. Let $f : L \to M$ be a join homomorphism between lattices $L$ and $M$. Then $L$ is order-preserving.
Proof. Let $a,b \in L$ be given with $a \leq b$. Then we have $a \vee b = b$, hence
  $$ f(b) = f(a \vee b) = f(a) \vee f(b). $$
  Therefore we have $f(a) \leq f(b)$.

Now we prove the connection between ideals and kernels.

Proposition. Let $f : L \to M$ be a join homomorphism between lattices $L$ and $M$, where $M$ is bounded below. If $0 \in M$ is in the image of $f$, then the set $I := f^{-1}(0) \subseteq L$ is an ideal in $L$.
Proof. Since $0$ is in the image of $f$, the set $I$ is non-empty.
  
  
*
  
*Suppose first that $a \in L$ and $b \in I$ are given with $a \leq b$. Since $f$ is order-preserving, we have $f(a) \leq f(b) = 0$. On the other hand, we also have $0 \leq f(a)$, because $0$ is the smallest element. We conclude that $f(a) = 0$ must hold as well, so we have $a \in I$.
  
*Secondly, suppose that $a,b\in I$ are given. Then we have
  $$ f(a \vee b) = f(a) \vee f(b) = 0 \vee 0 = 0. $$
  Therefore we also have $a \vee b \in I$.

The theorem from your question shows that every ideal can be realised as the kernel of some homomorphism. The same holds for ideals in rings: a subset $I \subseteq R$ is an ideal if and only if it is the kernel of a ring homomorphism $f : R \to S$. (The same holds for, e.g., subspaces of a vector space and normal subgroups of a group.)
