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Homelattice of ideals
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lattice of ideals
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:
Proposition 1.
$L(R)$ is a complete lattice.
Proof.
For any collection $S=\{J_{i}\mid i\in I\}$ 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, $ab\in J_{i}$ and so $ab\in\bigwedge S$. 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 welldefined, 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:
Corollary 1.
$L(R)$ is an algebraic lattice.
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 twosided 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 twosided 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.
Mathematics Subject Classification
06B35 no label found14K99 no label found16D25 no label found11N80 no label found13A15 no label found Forums
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