We’re starting a new series on commutative algebra. This has been in the works for way too long, and eventually we just decided to push ahead with it anyway. Most of the articles will be short, and we’ll try to illustrate the concepts with examples and diagrams whenever possible. Hopefully this will turn out well.
Since the notes are rather disorganized, repetitions will undoubtedly occur. For instance, we may introduce concept X in an earlier installation only to reintroduce it later as a special case of a more general machinery. However, we strive at clarity in explaining the concepts. Of course, the reader will be the best judge of how successful we are.
This is a summary of things you should already be familiar with, so we will proceed rather quickly.
We assume you have some basic knowledge about rings and ideals. Here, all rings are assumed to be commutative with 1. For a ring , any ideal gives us a ring quotient . Ideals also allow us to talk about “modulo” relations.
If is an ideal, then for , we write if their difference .
Note that this generalizes the concept of in modular arithmetic. Thanks to the definition of an ideal, we immediately get:
- If and , then
- If , then for all we have .
Proof : exercise.
This illustrates a general maxim here: that ideals generalize the concept of individual elements of a ring. We will see many more examples later. The most startling case of this is the example of a Dedekind domain, where we have unique factorization of ideals into prime ideals, instead of unique factorization of elements into primes.
Also, let us properly define a subring.
A subring of the ring is a subset satisfying:
- is a subgroup of ;
- if , then .
The trivial ring cannot be a subring of any other ring except itself, because of the first condition. However, it can be a ring quotient since for any ring , the ideal gives a trivial ring for the quotient .
Recall that we have the following construction of ideals.
Let be any ring.
- If is a collection of ideals of , their intersection is also an ideal of .
- If are ideals, their product is the set of all finite sums , where and .
Beginning students in algebra are often a bit bewildered by the presence of finite sums in the product of two ideals. But the definition is sensible, for the product satisfies
For this reason, we often write the ideal (whole ring) as . Aesthetically this is pleasing for we get .
Generating an Ideal
The fact that ideals are closed under arbitrary intersections is important, for we can take any subset and generate the smallest ideal containing it.
Consider the collection of all ideals containing . Note that is non-empty since . The intersection of all gives
the ideal generated by set . If , we also denote by .
We will use the computer science notation `:=’ to refer to definition. Thus means the notation on the LHS is defined to be on the RHS.
An ideal generated by one element is easy to describe:
More generally, one sees that is given by the set of all finite sums of the form where and . We can only restrict ourselves to finite sums since we are dealing with algebra here and infinite sums are not well-defined. [One can consider infinite sums in, say, functional analysis, but that is another story for another day.]
Sums of Ideals
Now suppose is a collection of ideals of . If , the ideal generated by S is called the sum of the ideals . It is clear that the sum can also be defined as follows.
The ideal is the collection of all finite sums , where
As expected, multiplication is distributive over addition.
Integral Domains and Fields
Next, we have the following.
An element of ring is a zero-divisor if there exists , such that .
Note that the trivial ring has no zero-divisor by definition. Otherwise for non-trivial ring , the element is a zero-divisor since and . Be forewarned that the trivial ring can be a pitfall for the unwary.
The ring is an integral domain (or just domain) if it is non-trivial, and the only zero-divisor in is .
Note that we have explicitly excluded the case of the trivial ring. Examples of integral domains include
- , where .
Clearly any subring of an integral domain is an integral domain.
An element of a ring is a unit if there exists such that .
Note that the set of units of a ring A forms a group under multiplication. We denote this by and call it the unit group of A. For example, and . Finally, we define:
The ring is said to be a field if it is non-trivial and any non-zero element of is a unit.
For example, are all fields and is not a field. The following is standard, whose proof we skip.
- Any field is an integral domain.
- Any finite integral domain is a field.
A ring A is a field if and only if it has exactly two ideals: 0 and A. [The trivial ring has only one ideal.]
Prime and Maximal Ideals
Given an ideal of ring , one would like to know when is an integral domain or a field.
- is an integral domain if and only if and
- is a field if and only if and, for any ideal containing , we have or .
For the first case, we say that is a prime ideal of A. For the second case, we say that is a maximal ideal of A. By definition, the whole ring is neither prime nor maximal. This should not come as a surprise, since in our definition of prime numbers, we explicitly exclude 1.
We will not prove the above theorem, but the proof of the first result is immediate. For the second result, we use the fact that a field has exactly two ideals, and that the ideals of the ring quotient are in bijection with ideals of containing .
Pictorially, the lattices of ideals correspond as follows (where the arrows represent inclusion):
Furthermore, the prime and maximal ideals coincide, i.e. is a prime (resp. maximal) ideal of the ring if and only if is a prime (resp. maximal) ideal of the ring .
Thus, taking the quotient corresponds to “chopping off” ideals from lower branches in the lattice.
While declaring the operations on modulo did you mean instead of $latex x \equiv y”?
Also at the end of the post : ” For a more concrete example, here is the correspondence between ideals of \mathbb Z/(30) and ideals of \mathbb Z containing (30). ” , but no description is given thereafter.
Thanks fixed it. I also removed the example on since it’s a bit too easy.