We know that the collection of all ideals of a given ring form a rig. In fact, they also form a lattice. We put the partial order of inclusion on ideals, so 
is below 
if 
.
To show that this poset is a lattice we have to show that pairwise greatest lower bounds and least upper bounds exist. Lower bounds are easy: the intersection 
of two ideals is again an ideal. By definition, any ideal contained in both and is contained in .
Upper bounds are a little trickier, since we can’t just take the union of two ideals. That would work for subsets of a given set, but in general the union of two ideals isn’t an ideal. Instead, we take their sum. Clearly 
and 
. Also, if 
is another ideal containing both and , then contains all linear combinations of elements of and . But 
is the set of all such linear combinations. Thus 
, and is the least upper bound of and .
This lattice is related to the divisibility preorder. Given a commutative unital ring 
and two elements 
, recall that 
if there is an 
so that 
. Then every multiple 
of 
is also a multiple 
of 
. Thus we see that the principal ideal 
is contained in the principal ideal 
. On the other hand, if 
we can see that for some , so . In particular, two elements are associated if and only if they generate the same principal ideal.
Notice that this correspondence reverses the direction of the order. If is below in the divisibility ordering, then is above in the ideal ordering. Thus the “greatest common divisor” of two ideals is actually now the least ideal containing both of them. The language of ideals, however, is far more general than that of divisibility. We now need to recast most of what we know about divisibility from our experience with natural numbers into these more general ring-theoretic terms.
May 15, 2007
Posted by John Armstrong |
Ring theory |
8 Comments
