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4.1 Unique Factorization Domains (UFDs)

Throughout this section R will denote an integral domain (i.e. a commutative ring with identity containing no zero-divisors). Recall that a unit of R is an element that has an inverse with respect to multiplication. If a is any element of R and u is a unit, we can write

a = u(u ^-1a):

This is not considered to be a proper factorization of a. For example we do not consider 5 = 1(5) or 5 = (-1)(-5) to be proper factorizations of 5 in Z. We do not consider to be a proper factorization of x² + 2 in Q[x].

Definition 4.1.1 An element a in an integral domain R is called irreducible if it is not zero or a unit, and if whenever a is written as the product of two elements of R, one of these is a unit.

An element p of an integral domain R is cal led prime if p is not zero or a unit, and whenever p divides ab for elements a; b of R, either p divides a or p divides b.

Note

1. Elements r and s are called associates of each other if s = ur for a unit u of R. So a ∈ R irreducible if it can only be factorized as the product of a unit and one of its own associates.

2. If R is an integral domain, every prime element of R is irreducible. To see this let p ∈ R be prime and suppose that p = rs is a factorisation of p in R. Then since p divides rs, either p divides r or p|s. There is no loss of generality in assuming p divides r. Then r = pa for some element a of R, and p = rs so p = pas. Then p pas = 0 so p(1 - as) = 0 in R. Thus as = 1 since R is an integral domain and p ≠ 0. Then s is a unit and p = rs is not a proper factorisation of p. Hence p is irreducible in R.

It is not true that every irreducible element of an integral domain must be prime, as we will shortly see.

Examples:

1. In Z the units are 1 and 1 and each non-zero non-unit element has two associates, namely itself and its negative. SO 5 and -5 are associates, 6 and -6 are associates, and so on. The irreducible elements of Z are p and -p, for p prime.

2. In Q[x], the units are the non-zero constant polynomials. The associates of a non-zero non-constant polynomial f (x) are the polynomials of the form af (x) where a ∈ Q^x .
So x² + 2 is associate to  .

3. In Z the irreducible elements are the integers p and -p where p is a prime numbers. The prime elements of Z are exactly the irreducible elements - the prime numbers and their negatives.

Definition 4.1.2 An integral domain R is a unique factorization domain if the following conditions hold for each element a of R that is neither zero nor a unit.

1. a can be written as the product of a nite number of irreducible elements of R.

2. This can be done in an essential ly unique way. If a = p₁ p₂ ... p_r and a = q₁ q₂ ... q_s are two expressions for a as a product of irreducible elements, then s = r and q₁ ... q_s can be reordered so that for each i, q_i is an associate of p_i.

Example 4.1.3 Z is a UFD.

(This is the Fundamental Theorem of Arithmetic). 

Example 4.1.4 Let   denote the set of complex numbers of the form  where a and b are integers and  denotes the complex number . We will show that  is not a UFD (it is easily shown to be a ring under the usual addition and multiplication of complex numbers).

Claim:  is not a UFD.

The proof if this claim will involve a number of steps.

1. We define a function  denotes the complex conjugate of a. thus

So φ is multiplicative.

2. Suppose α is a unit of  and let β be its inverse.   are positive integers this means  So φ(α) = 1 whenever α is a unit.

On the other hand  for integers a and b which means b = 0 and a = ±1. So the only units of Z[] are 1 and -1.

3. 3. Suppose φ(α) = 9 for some a ∈ Z[], If a is not irreducible in Z[] then it factorizes as α₁α₂ where a\ and a2 are non-units. Then we must have

Now this would means 3 = c² + 5d² for integers c and d which is impossible. So if φ(α) = 9 then α is irreducible in Z[].

4. Now 9 = 3 x 3 and 9 = (2 + )(2 -) in Z[]. The elements (3, 2 + ) and 2 - ) are all irreducible in Z[] by item 3. above. Furthermore 3 is not an associate of either 2 +  or 2 - as the only units in Z[] are 1 and -1. We conclude that the factorizations of 9 above are genuinely different, and Z[] is not a UFD.

Note that 3 is an example of an element of Z[] that is irreducible but not prime.

Remark: The ring Z[i] = {a + bi : a; b ∈ Z} is a UFD.

Theorem 4 .1.5 Let F be a eld. Then the polynomial ring F [x] is a UFD.

Proof: We need to show that every non-zero non-unit in F[x] can be written as a product of irreducible polynomials in a manner that is unique up to order and associates.
So let f (x) be a polynomial of degree n > 1 in F [x]. If f (x) is irreducible there is nothing to do. If not then  f (x) = g(x)h(x) where g(x)  and h(x) both have degree less than n. If
g(x) or h(x) is reducible further factorization is possible; the process ends after at most n steps with an expression for f (x) as a product of irreducibles.

To see the uniqueness, suppose that

f (x) = p₁ (x)p₂ (x) .... p_r (x) and

f (x) = q₁ (x)q₂ (x) : : : q_s (x)

are two such expressions, with s > r. Then q₁ (x)q₂ (x)... q_s (x) belongs to the ideal (p₁ (x)) of F[x]. Since this ideal is prime (as p₁(x) is irreducible) this means that either q₁(x) ∈ (p₁(x)) or q₂(x)... q_s(x) ∈ (p₁(x)). Repeating this step leads to the conclusion that at least one of the q_i (x) belongs to (p₁(x)). After reordering the q_i(x) if necessary we have q₁(x) ∈ (p₁(x)). Since q₁ (x) is irreducible this means q₁(x) = u₁p₁(x) for some unit u₁. Then

Pl₁(x)p₂(x) ...p_r(x) = u₁p₁(x)q₂(x) ...q_s(x).

Since F [x] is an integral domain we can cancel p₁(x) from both sides to obtain
P₂(x) .. .P₁(x) = u₁q₂(x) . .. q_s(x).

After repeating this step a further r - 1 times we have
1 = u₁u₂ . . . u_r q_{r + 1}(x) . .. q_s(x),
where u₁,... ,u_r are units in F[x] (i.e. non-zero elements of F). This means s = r, since the polynomial on the right in the above expression must have degree zero. We conclude that qi (x),..., q_s (x) are associates (in some order) of p₁ (x),..., pr (x). This completes the proof.