# lattice of ideals

###### 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, $a-b\in J_{i}$ 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:

###### 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.

 Title lattice of ideals Canonical name LatticeOfIdeals Date of creation 2013-03-22 16:59:40 Last modified on 2013-03-22 16:59:40 Owner CWoo (3771) Last modified by CWoo (3771) Numerical id 13 Author CWoo (3771) Entry type Definition Classification msc 06B35 Classification msc 14K99 Classification msc 16D25 Classification msc 11N80 Classification msc 13A15 Related topic SumOfIdeals Related topic LatticeIdeal Related topic IdealCompletionOfAPoset