Abstract Algebra/The hierarchy of rings

Commutative ringsEdit

Definition 11.1:

A ring   with multiplication   is called commutative if and only if   for all  .

Examples 11.2:

  • The whole numbers   are commutative.
  • The matrix ring   of  -by-  real matrices with matrix multiplication and component-wise addition is not commutative for  .

In commutative rings, a left ideal is a right ideal and thus a two-sided ideal, and a right ideal also.

Integral domainsEdit

Definition 11.3:

An integral domain is defined to be a commutative ring (that is, we assume commutativity by definition) such that whenever   ( ), then   or  .

We can characterize integral domains in another way, and this involves the so-called zero-divisors.

Definition 11.4:

Let  

Thus, a ring is an integral domain iff it has no zero divisors.

Unique factorisation domainsEdit

Theorem 11.?:

Suppose that   is a commutative ring

Principal ideal domainsEdit

Due to its importance in algebra, we'll briefly give the definition of noetherian rings, which is a fairly exhaustive class of rings for which many useful properties hold. The theory of noetherian rings is well-studied, powerful and extensive, and we'll only study it in detail in the wikibook on Commutative Algebra. The reason that we give the definition here is that principal ideal domains are noetherian rings, which will imply that they are, in fact, unique factorisation domains.

Definition 11.?:

Let   be a commutative ring.   is called noetherian iff for every sequence of ideals   of   such that

 

there exists an   such that  .

This condition can be interpreted to state that every ascending chain of ideals stabilizes. Noetherian rings are named in honour of Emmy Noether.

Theorem 11.?:

Every PID is Noetherian.

Proof:

We observed earlier that the set of all ideals of a ring is inductive, with an explicit description of. If therefore we are given an ascending chain of ideals

Theorem 11.?:

Every PID is a UFD.

Proof:

Let   be a PID, and let  .

Euclidean domainsEdit

Example 11.? (Gaussian integers):

We have already seen that   is a Euclidean domain. Now consider the ring

 

with addition and multiplication induced by that of  . We'll see in the exercises that this is indeed a commutative ring with identity. Furthermore, on it we define a Euclidean function as thus:

 

This is indeed a Euclidean function, the units of   are   and furthermore we may precisely describe the prime elements of   and set them in relation to the prime elements of  :

  1. If   is a prime in  , then either it is already a prime in  , or there is a prime   in the Gaussian primes such that  .
  2. If   is a Gaussian prime, then set  . Either we have that   is a prime in  , or  , where   is a prime in  .
  3. In 1., if  , the former case happens if and only if   and the latter if and only if  .

Proof:

First, the proof of multiplicativity of   is relegated to the exercises, that is, you'll show in the exercises that

 .

Then we have to prove that division with remainder holds. Let thus   and   be elements of  .

Due to  ,   are units. Any other unit would have to have the form  , where  . Let   be its inverse. Then  , a contradiction.

Finally, let's prove the statements about the relation of the Gaussian primes to the integer primes.

  1. Since   is a Euclidean domain, we have a decomposition of   into prime elements of  , say  , where   is a unit in  . If  , we are done. If  , observe that  , and since   is prime, uniqueness of prime factorisation in the integers implies that at most two of   are not one and those that are are either   or  . If one is  , there is exactly one prime factor of   in  , which is absurd since   is obviously not irreducible. If ,

ExercisesEdit

  1. Prove that the Gaussian integers as defined above do form a commutative ring with identity. Use your knowledge on complex numbers (cf. the corresponding chapter in the wikibook on complex analysis).