Let R be a

ring. An

element b of R has an

**infinite divisor chain** iff there is a

infinite sequence (a

_{i}) in R with:

- a
_{1} divides b
- a
_{i+1} divides a_{i} for i > 0, i an integer
- Any a
_{i} is only associated to a finite number of elements a_{j}

Where is this definition is important ?

Let's look at an unique factorization domain.

In an UFD, every irreducible element is a prime.

Proof: Let be p irreducible and p divide x y, which means b p = x y. We must show that p divides x or y. Let x = x_{1}...x_{n},
y = y_{1}...y_{m} factorizations of x and y. Then x_{1}...x_{n}y_{1}...y_{m} is a factorization of b p. Because p is irreducible and any factorization is unique, p must be associated to a x_{i} or y_{j}. Therefore p divides x or y.

Now arises the question "If R is not an UFD, are the irreducible elements which are not primes ?" or "Are UFD and prime follows from irreducible equivalent ?" Note that "irreducible -> prime" makes factorizations unique.

The answer is no, because there is the possibility that you have reducible elements without an factorization into irreducible elements. Sounds weird, but is true: an element might have an infinite divisor chain.

Let's give an example:

Let R be the set of holomorphic functions on **C**. (here: the function must be holomorphic at every point of **C**).

R is a ring be the standard addition and multiplication of functions. R is commutative and an integral domain. It can be shown, that any irreducible element is prime.

By the Weierstrass approximation theorem there are holomorphic functions which are infinite products of functions of the form (z-n)g(z), (g(z) is never 0 on **C**). These products can be divided with these (z-n) (throwing (z-n)g(z) out of the product gives again a holomorphic function). This gives obviously an infinite divisor chain. An infinite divisor chain obviously contradicts the properties of an UFD.

I had to look this example up and there seems to be no easier one.