The document Unique Factorization Domain Mathematics Notes | EduRev is a part of the Mathematics Course Algebra for IIT JAM Mathematics.

All you need of Mathematics at this link: Mathematics

**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 ^{-1}a):

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 x^{2} + 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 âˆˆ 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 âˆˆ 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 â‰ 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 âˆˆ Q^{x} .

So x^{2} + 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 = p_{1} p_{2} ... p_{r} and a = q_{1} q_{2} ... q_{s} are two expressions for a as a product of irreducible elements, then s = r and q_{1} ... 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 Ï† is multiplicative.

2. Suppose Î± is a unit of and let Î² be its inverse. are positive integers this means So Ï†(Î±) = 1 whenever Î± is a unit.

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

3. 3. Suppose Ï†(Î±) = 9 for some a âˆˆ Z[], If a is not irreducible in Z[] then it factorizes as Î±_{1}Î±_{2} where a\ and a2 are non-units. Then we must have

Now this would means 3 = c^{2} + 5d^{2} for integers c and d which is impossible. So if Ï†(Î±) = 9 then Î± 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 âˆˆ 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) = p_{1} (x)p_{2} (x) .... p_{r} (x) and

f (x) = q_{1} (x)q_{2} (x) : : : q_{s} (x)

are two such expressions, with s __>__ r. Then q_{1} (x)q_{2} (x)... q_{s} (x) belongs to the ideal (p_{1} (x)) of F[x]. Since this ideal is prime (as p_{1}(x) is irreducible) this means that either q_{1}(x) âˆˆ (p_{1}(x)) or q_{2}(x)... q_{s}(x) âˆˆ (p_{1}(x)). Repeating this step leads to the conclusion that at least one of the q_{i} (x) belongs to (p_{1}(x)). After reordering the q_{i}(x) if necessary we have q_{1}(x) âˆˆ (p_{1}(x)). Since q_{1} (x) is irreducible this means q_{1}(x) = u_{1}p_{1}(x) for some unit u_{1}. Then

Pl_{1}(x)p_{2}(x) ...p_{r}(x) = u_{1}p_{1}(x)q_{2}(x) ...q_{s}(x).

Since F [x] is an integral domain we can cancel p_{1}(x) from both sides to obtain

P_{2}(x) .. .P_{1}(x) = u_{1}q_{2}(x) . .. q_{s}(x).

After repeating this step a further r - 1 times we have

1 = u_{1}u_{2} . . . u_{r} q_{r + 1}(x) . .. q_{s}(x),

where u_{1},... ,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 p_{1} (x),..., pr (x). This completes the proof.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

58 docs

### Principal Ideal Domain

- Doc | 2 pages
### Euclidean Domain

- Doc | 1 pages

- Prime and Maximal Ideals
- Doc | 7 pages
- Rings, Ideals, Quotient Rings - Ring Theory
- Doc | 4 pages