A ring with multiplication is called commutative if and only if for all .
- 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.
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.
Thus, a ring is an integral domain iff it has no zero divisors.
Unique factorisation domains[edit | edit source]
Suppose that is a commutative ring
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.
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.
Every PID is Noetherian.
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
Every PID is a UFD.
Let be a PID, and let .
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 :
- If is a prime in , then either it is already a prime in , or there is a prime in the Gaussian primes such that .
- If is a Gaussian prime, then set . Either we have that is a prime in , or , where is a prime in .
- In 1., if , the former case happens if and only if and the latter if and only if .
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.
- 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 ,
- 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).