## Generated Ideals and the Lattice of Ideals (Pt. II)

**Point of Post: **This post is a continuation of this one.

But, when is unital we have a different description of . Namely, we know for a unital ring that is the set of all finite sums of the form with , , and . But, since each is an ideal it’s clear that left and right multiplying by -elements doesn’t affect anything and so for unital rings is just the set of all sums of the form such that . In other words, is the set of all finite sums where the summands lie in . This motivates the following definition: let be a ring (not necessarily unital) and be a collection of subrings we define then the *sum of *to be the set

in other words the sum of a collection of subrings is the set of all finite sums where the summands belong to the union of the subrings. What we’ve seen above is that if is unital ring and is a collection of ideals of then is equal to . One would hope that this holds true for not only two-sided ideals but left and right as well (this should be clear since the same argument applies) but also for non-unital rings. Indeed, this is true. To prove this we start with the following theorem:

**Theorem: ***Let be a ring (not necessarily unital) and be a collection of left,right, or two-sided ldeals of . Then, is a left, right, or two-sided ideal of respectively.*

**Proof: **For notational convenience let denoted . Let then be arbitrary. Then, and with for some . We see then that where if and for , and since clearly each belongs to some we may conclude that . Thus, is a subgroup of . We now suppose that each is an ideal since the other two cases are the same. To prove that is an ideal we let be arbitrary and note that

and since each is a two-sided ideal we see that each and belongs to whichever did. Thus, and are (finite) sums of elements from and thus in . Since and were arbitrary the conclusion follows.

With this in mind we prove the following:

**Theorem: ***Let be a ring and be a collection of left, right, or two-sided ideals then is equal to (where, while the result remains the same, the meaning of supremum depends on whether the ideals are left, right, or two-sided in the sense that we are dealing with the lattices , , and respectively–we won’t make explicit note of this in the proof since all the manipulations apply equally well to all three, but keep this in mind).*

**Proof: **Evidently since is a (left, right or two-sided) ideal containing is must contain all finite sums of elements from and thus it must contain . Conversely, we have just proven that is a (left, right, or two sided) ideal which clearly contains and thus by definition it must contain . The conclusion follows.

**Modularity and the Product of Ideals**

You’ll now recall though that I claimed that is a modular lattice. A *modular lattice *is a lattice that has the property that if . For us this shall mean that if with then . Indeed:

**Theorem: ***Let be a ring and . Then,*

*if then equality holds.*

**Proof: **Let then where and . We note firstly that since that . But, since and we also see that and thus .

Suppose now that , we aim to show that . We know from the previous paragraph that

Conversely, if we then know that and where and . But, since as well by assumption we know that and so , thus .

Of course, a similar result holds true for elements of and . Thus, collecting all the facts from this post we get:

**Theorem: ***Let be a ring, then forms a complete modular lattice.*

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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