Let $R$ be a ring. Consider the set $L(R)$ of all left ideals of $R$ . Order this set by inclusion, and we have a partially ordered set. In fact, we have the following:

Proof. For any collection $S=\lbrace J_i\mid i\in I\rbrace$ of (left) ideals of $R$ ($I$ is an index set), define $$\bigwedge S:=\bigcap S\qquad\mbox{and}\qquad\bigvee S=\sum_i J_i,$$ the sum of ideals $J_i$ . We assert that $\bigwedge S$ is the greatest lower bound of the $J_i$ , and $\bigvee S$ the least upper bound of the $J_i$ , and we show these facts separately

• First, $\bigwedge S$ is a left ideal of $R$ : if $a,b\in \bigwedge S$ , then $a,b\in J_i$ for all $i\in I$ . Consequently, $a-b\in J_i$ and so $a-b\in \bigwedge S$ . Furthermore, if $r\in R$ , then $ra\in J_i$ for any $i\in I$ , so $ra\in \bigwedge S$ also. Hence $\bigwedge S$ is a left ideal. By construction, $\bigwedge S$ is clearly contained in all of $J_i$ , and is clearly the largest such ideal.
• For the second part, we want to show that $\bigvee S$ actually exists for arbitrary $S$ . We know the existence of $\bigvee S$ if $S$ is finite. Suppose now $S$ is infinite. Define $J$ to be the set of finite sums of elements of $\bigcup_i J_i$ . If $a,b\in J$ , then $a+b$ , being a finite sum itself, clearly belongs to $J$ . Also, $-a\in J$ as well, since the additive inverse of each of the additive components of $a$ is an element of $\bigcup_i J_i$ . Now, if $r\in R$ , then $ra\in J$ too, since multiplying each additive component of $a$ by $r$ (on the left) lands back in $\bigcup_i J_i$ . So $J$ is a left ideal. It is evident that $J_i\subseteq J$ . Also, if $M$ is a left ideal containing each $J_i$ , then any finite sum of elements of $J_i$ must also be in $M$ , hence $J\subseteq M$ . This implies that $J$ is the smallest ideal containing each of the $J_i$ . Therefore $S$ exists and is equal to $J$ .

In summary, both $\bigvee S$ and $\bigwedge S$ are well-defined, and exist for finite $S$ , so $L(R)$ is a lattice. Additionally, both operations work for arbitrary $S$ , so $L(R)$ is complete.

From the above proof, we see that the sum $S$ of ideals $J_i$ can be equivalently interpreted as

• the ``ideal'' of finite sums of the elements of $J_i$ , or
• the ``ideal'' generated by (elements of) $J_i$ , or
• the join of ideals $J_i$ .

A special sublattice of $L(R)$ is the lattice of finitely generated ideals of $R$ . It is not hard to see that this sublattice comprises precisely the compact elements in $L(R)$ .

Looking more closely at the above proof, we also have the following:

Proof. As we have already shown, $L(R)$ is a complete lattice. If $J$ is any (left) ideal of $R$ , by the previous remark, each $J$ is the sum (or join) of ideals generated by individual elements of $J$ . Since these ideals are principal ideals (generated by a single element), they are compact, and therefore $L(R)$ is algebraic.

Remarks.

• One can easily reconstruct all of the above, if $L(R)$ is the set of right ideals, or even two-sided ideals of $R$ . We may distinguish the three notions: $l.L(R),r.L(R),$ and $L(R)$ as the lattices of left, right, and two-sided ideals of $R$ .
• When $R$ is commutative, $l.L(R)=r.L(R)=L(R)$ . Furthermore, it can also be shown that $L(R)$ has the additional structure of a quantale.
• There is also a related result on lattice theory: the set $\operatorname{Id}(L)$ of lattice ideals in a upper semilattice $L$ with bottom $0$ forms a complete lattice. For a proof of this, see this entry.
• However, the more general case is not true: the set of order ideals in a poset is a dcpo.