# Unique Factorization

Through this article and the next few ones, we will explore unique factorization in rings. The inspiration, of course, comes from ℤ. Here is an application of unique factorization. Warning: not all steps may make sense to the reader at this point of time.

Sample Problem.Find all integer solutions to .

**Solution (Sketch)**

Factor the equation to obtain where . Use the fact that satisfies unique factorization; deduce that and must be coprime (since *x*, *y* are odd). Hence and for some with . This gives . Now we factor in all possible ways and solve a pair of simultaneous equations in *a* and *b*. E.g.

which gives ♦

Before starting, here are some preliminary definitions.

Throughout this article, *A* denotes an integral domain.

Definition.Recall that A is a unit if it is invertible under product.

- Elements are
associated(written as ) if there is a unit such that .- For we say x
dividesy (written as ) if there exists such that .- Let , not a unit. We say it is
reducibleif we can find which are non-unit such that . Otherwise, isirreducible.

All the above conditions can be expressed in terms of ideals.

*x*and*y*are associated if and only if (*x*) = (*y*) as ideals of*A*. Thus, the relation is an equivalence relation.*x*divides*y*if and only if as ideals. Thus and if and only if .*x*is irreducible if and only if when (*x*) = (*y*)(*z*) as a product of ideals, either (*x*) = (*y*) or (*x*) = (*z*). In particular, if*x*is irreducible, so are all its associates.

The descriptions in terms of principal ideals are much cleaner.

## Examples

1. Suppose . The only units are {-1, +1, –*i*, +*i*}. So each non-zero element has 4 associates, e.g.

are associates, are associates.

Note that 1 + 2*i* and 1 – 2*i* are not associates. On the other hand, 1 + *i* and 1 – *i* are associates. (Verify this!)

2. Suppose . The unit group is infinite since it contains for all . Thus every non-zero element has infinitely many associates.

# Factoring Into Irreducibles

Let be non-zero, non-unit. If *x* is reducible, we can factor it as *x* = *yz* where *y*, *z* are non-zero and non-unit. Again, if either of them is reducible, we factor them further. If this process terminates, we obtain *x* as a factor of irreducibles.

Unfortunately, there are cases where it does not.

**Example**

Let . Now can be factorized indefinitely:

Here is a condition which ensures that factorization terminates.

Proposition 1.Partially order the principal ideals of A by reverse inclusion:

.

If this poset is noetherian, then every non-unit can be factored as a product of irreducibles.

**Note**

By proposition 1 here, the above condition is equivalent to either of the following.

- Any non-empty collection of principal ideals of
*A*has a maximal element. - In any chain of principal ideals of
*A*, there exists*n*such that .

One also says *A* satisfies **ascending chain condition** (**a.c.c.**) on the set of principal ideals.

**Proof**

Let *S* be the collection of all non-unit which cannot be factored as a product of irreducibles. If *S* is non-empty, the set of principal ideals (*x*) for has a maximal element (*z*) with respect to ⊆.

Since *z* cannot be factored as a product of irreducibles, it is not irreducible. Write *z* = *xy* where *x*, *y* are non-unit and non-zero. Then and so by maximality of (*z*) we have . Hence *x* and *y* can be expressed as a product of finitely many irreducibles; since *z* = *xy* so does *z*, a contradiction. ♦

But how does one check the above abstract condition for a ring *A*? There are two answers to this.

- Most rings we are interested in actually satisfy a.c.c. on
*all ideals*. Called noetherian rings, we will see later that they are everywhere. - More immediately, we will give a criterion for unique factorizability which also implies a.c.c. on principal ideals.

# UFDs

Finally, here is our main object of interest.

Definition.An integral domain A is called a

unique factorization domain(UFD) if every proper principal ideal can be factored as a product of irreducible principal ideals.uniquely up to permutation.

Here is a non-example. Suppose . In this ring, we have

It is easy to see that no two of these elements are associates. Thus to see that unique factorization fails, it remains to show all of them are irreducible. For that, take the **norm** function where An easy calculation shows:

- ,
- for all
*x*, - .

Now the above elements have norms

If any of them had a proper factor, such a factor would have norm 2 or 3. But it is easy to see no element of norm 2 or 3 exists.

**Exercise A**

Prove that is not a UFD.

# Prime Elements

We end this section by stating a necessary and sufficient condition for unique factorization to hold.

Definition.Let . We say x is

primeif is a prime ideal of A.

Thus *x* is prime if and only if:

- whenever
*yz*is a multiple of*x*, either*y*or*z*is a multiple of*x*.

Lemma 1.A prime element is irreducible.

**Proof**

Suppose *x* = *yz*, where *y*, *z* are non-zero. Then *yz* is a multiple of *x* so either *y* or *z* is a multiple of *x*. Assume the former, so *y* is a multiple of *x*, but since *x* = *yz* we have *x* is a multiple of *y* also. Thus *x* and *y* are associates and *z* is a unit. ♦

Finally, the main result we want is as follows.

Theorem 1.Let A be a domain which satisfies a.c.c. on principal ideals.

Then A is a unique factorization domain if and only if all irreducible elements are prime.

**Proof**

(⇒) Let *A* be a UFD and be irreducible. To show *x* is prime, suppose and ; we need to show . Write for some . Write *y*, *z* and *a* as a product of irreducibles. By unique factorization, since *x* does not occur in the factoring of *y*, it must occur in the factoring of *z*. Thus .

(⇐) Suppose all irreducibles are prime. For any , suppose we have factorizations where are all irreducible, and hence prime. Write this in terms of principal ideals:

Now since , one of the terms, say must be in . Hence is a multiple of ; by irreducibility we have . So and are associates and upon division we have

Repeat until we run out of terms on one side so we get (1). The other side must also be empty so we have established unique factorization. ♦

In the next article, we will find many examples of UFDs based on this criterion.

**Example**

Let us show that is not a UFD by finding an irreducible element which is not prime: 2. By the norm argument above, 2 is irreducible but it is not prime because

is not an integral domain.

**Exercise B**

Find all prime ideals of which contain 2 or 3.

[Hint (highlight to read): prime ideals of *A* containing 2 correspond to prime ideals of *A*/(2).]