AATA Factorization in Integral Domains

Section 18.2 Factorization in Integral Domains

The building blocks of the integers are the prime numbers. If

F

is a field, then irreducible polynomials in

F [x]

play a role that is very similar to that of the prime numbers in the ring of integers. Given an arbitrary integral domain, we are led to the following series of definitions.

🔗

Let

R

be a commutative ring with identity, and let

a

and

b

be elements in

R .

We say that

a

divides

b,

and write

a ∣ b,

if there exists an element

c \in R

such that

b = a c .

A unit in

R

is an element that has a multiplicative inverse. Two elements

a

and

b

R

are said to be associates if there exists a unit

u

R

such that

a = u b .

🔗

Let

D

be an integral domain. A nonzero element

p \in D

that is not a unit is said to be irreducible provided that whenever

p = a b,

either

a

b

is a unit. Furthermore,

p

is prime if whenever

p ∣ a b

either

p ∣ a

p ∣ b .

🔗

Example 18.8.

🔗

It is important to notice that prime and irreducible elements do not always coincide. Let

R

be the subring (with identity) of

Q [x, y]

generated by

x^{2},

y^{2},

and

x y .

Each of these elements is irreducible in

R;

however,

x y

is not prime, since

x y

divides

x^{2} y^{2}

but does not divide either

x^{2}

y^{2} .

🔗

The Fundamental Theorem of Arithmetic states that every positive integer

n > 1

can be factored into a product of prime numbers

p_{1} \dots p_{k},

where the

p_{i}

’s are not necessarily distinct. We also know that such factorizations are unique up to the order of the

p_{i}

’s. We can easily extend this result to the integers. The question arises of whether or not such factorizations are possible in other rings. Generalizing this definition, we say an integral domain

D

is a unique factorization domain, or UFD, if

D

satisfies the following criteria.

Let $a \in D$ such that $a \neq 0$ and $a$ is not a unit. Then $a$ can be written as the product of irreducible elements in $D .$
Let $a = p_{1} \dots p_{r} = q_{1} \dots q_{s},$ where the $p_{i}$ ’s and the $q_{i}$ ’s are irreducible. Then $r = s$ and there is a $π \in S_{r}$ such that $p_{i}$ and $q_{π (j)}$ are associates for $j = 1, \dots, r .$

🔗

Example 18.9.

🔗

The integers are a unique factorization domain by the Fundamental Theorem of Arithmetic.

🔗

Example 18.10.

🔗

Not every integral domain is a unique factorization domain. The subring

Z [\sqrt{3} i] = {a + b \sqrt{3} i}

of the complex numbers is an integral domain (Exercise 16.7.12, Chapter 16). Let

z = a + b \sqrt{3} i

and define

ν : Z [\sqrt{3} i] \to N \cup {0}

ν (z) = | z |^{2} = a^{2} + 3 b^{2} .

It is clear that

ν (z) \geq 0

with equality when

z = 0 .

Also, from our knowledge of complex numbers we know that

ν (z w) = ν (z) ν (w) .

It is easy to show that if

ν (z) = 1,

then

z

is a unit, and that the only units of

Z [\sqrt{3} i]

are

1

and

- 1 .

🔗

We claim that

4

has two distinct factorizations into irreducible elements:

4 = 2 \cdot 2 = (1 - \sqrt{3} i) (1 + \sqrt{3} i) .

🔗

We must show that each of these factors is an irreducible element in

Z [\sqrt{3} i] .

2

is not irreducible, then

2 = z w

for elements

z, w

Z [\sqrt{3} i]

where

ν (z) = ν (w) = 2 .

However, there does not exist an element in

z

Z [\sqrt{3} i]

such that

ν (z) = 2

because the equation

a^{2} + 3 b^{2} = 2

has no integer solutions. Therefore,

2

must be irreducible. A similar argument shows that both

1 - \sqrt{3} i

and

1 + \sqrt{3} i

are irreducible. Since

2

is not a unit multiple of either

1 - \sqrt{3} i

1 + \sqrt{3} i,

4

has at least two distinct factorizations into irreducible elements.

🔗

Subsection Principal Ideal Domains

🔗

Let

R

be a commutative ring with identity. Recall that a principal ideal generated by

a \in R

is an ideal of the form

⟨ a ⟩ = {r a : r \in R} .

An integral domain in which every ideal is principal is called a principal ideal domain, or PID.

🔗

Lemma 18.11.

🔗

Let

D

be an integral domain and let

a, b \in D .

Then

$a ∣ b$ if and only if $⟨ b ⟩ \subset ⟨ a ⟩ .$
$a$ and $b$ are associates if and only if $⟨ b ⟩ = ⟨ a ⟩ .$
$a$ is a unit in $D$ if and only if $⟨ a ⟩ = D .$

Proof.

(1) Suppose that

a ∣ b .

Then

b = a x

for some

x \in D .

Hence, for every

r

D,

b r = (a x) r = a (x r)

and

⟨ b ⟩ \subset ⟨ a ⟩ .

Conversely, suppose that

⟨ b ⟩ \subset ⟨ a ⟩ .

Then

b \in ⟨ a ⟩ .

Consequently,

b = a x

for some

x \in D .

Thus,

a ∣ b .

(2) Since

a

and

b

are associates, there exists a unit

u

such that

a = u b .

Therefore,

b ∣ a

and

⟨ a ⟩ \subset ⟨ b ⟩ .

Similarly,

⟨ b ⟩ \subset ⟨ a ⟩ .

It follows that

⟨ a ⟩ = ⟨ b ⟩ .

Conversely, suppose that

⟨ a ⟩ = ⟨ b ⟩ .

By part (1),

a ∣ b

and

b ∣ a .

Then

a = b x

and

b = a y

for some

x, y \in D .

Therefore,

a = b x = a y x .

Since

D

is an integral domain,

x y = 1;

that is,

x

and

y

are units and

a

and

b

are associates.

(3) An element

a \in D

is a unit if and only if

a

is an associate of

1 .

However,

a

is an associate of

1

if and only if

⟨ a ⟩ = ⟨ 1 ⟩ = D .

🔗

Theorem 18.12.

🔗

Let

D

be a PID and

⟨ p ⟩

be a nonzero ideal in

D .

Then

⟨ p ⟩

is a maximal ideal if and only if

p

is irreducible.

Proof.

Suppose that

⟨ p ⟩

is a maximal ideal. If some element

a

D

divides

p,

then

⟨ p ⟩ \subset ⟨ a ⟩ .

Since

⟨ p ⟩

is maximal, either

D = ⟨ a ⟩

⟨ p ⟩ = ⟨ a ⟩ .

Consequently, either

a

and

p

are associates or

a

is a unit. Therefore,

p

is irreducible.

Conversely, let

p

be irreducible. If

⟨ a ⟩

is an ideal in

D

such that

⟨ p ⟩ \subset ⟨ a ⟩ \subset D,

then

a ∣ p .

Since

p

is irreducible, either

a

must be a unit or

a

and

p

are associates. Therefore, either

D = ⟨ a ⟩

⟨ p ⟩ = ⟨ a ⟩ .

Thus,

⟨ p ⟩

is a maximal ideal.

🔗

Corollary 18.13.

🔗

Let

D

be a PID. If

p

is irreducible, then

p

is prime.

Proof.

Let

p

be irreducible and suppose that

p ∣ a b .

Then

⟨ a b ⟩ \subset ⟨ p ⟩ .

By Corollary 16.40, since

⟨ p ⟩

is a maximal ideal,

⟨ p ⟩

must also be a prime ideal. Thus, either

a \in ⟨ p ⟩

b \in ⟨ p ⟩ .

Hence, either

p ∣ a

p ∣ b .

🔗

Lemma 18.14.

🔗

Let

D

be a PID. Let

I_{1}, I_{2}, \dots

be a set of ideals such that

I_{1} \subset I_{2} \subset \dots .

Then there exists an integer

N

such that

I_{n} = I_{N}

for all

n \geq N .

Proof.

We claim that

I = ⋃_{i = 1}^{\infty} I_{i}

is an ideal of

D .

Certainly

I

is not empty, since

I_{1} \subset I

and

0 \in I .

a, b \in I,

then

a \in I_{i}

and

b \in I_{j}

for some

i

and

j

N .

Without loss of generality we can assume that

i \leq j .

Hence,

a

and

b

are both in

I_{j}

and so

a - b

is also in

I_{j} .

Now let

r \in D

and

a \in I .

Again, we note that

a \in I_{i}

for some positive integer

i .

Since

I_{i}

is an ideal,

r a \in I_{i}

and hence must be in

I .

Therefore, we have shown that

I

is an ideal in

D .

Since

D

is a principal ideal domain, there exists an element

\overset{―}{a} \in D

that generates

I .

Since

\overset{―}{a}

is in

I_{N}

for some

N \in N,

we know that

I_{N} = I = ⟨ \overset{―}{a} ⟩ .

Consequently,

I_{n} = I_{N}

for

n \geq N .

🔗

Any commutative ring satisfying the condition in Lemma 18.14 is said to satisfy the ascending chain condition, or ACC. Such rings are called Noetherian rings, after Emmy Noether.

🔗

Theorem 18.15.

🔗

Every PID is a UFD.

Proof.

Existence of a factorization.

Let

D

be a PID and

a

be a nonzero element in

D

that is not a unit. If

a

is irreducible, then we are done. If not, then there exists a factorization

a = a_{1} b_{1},

where neither

a_{1}

nor

b_{1}

is a unit. Hence,

⟨ a ⟩ \subset ⟨ a_{1} ⟩ .

By Lemma 18.11, we know that

⟨ a ⟩ \neq ⟨ a_{1} ⟩;

otherwise,

a

and

a_{1}

would be associates and

b_{1}

would be a unit, which would contradict our assumption. Now suppose that

a_{1} = a_{2} b_{2},

where neither

a_{2}

nor

b_{2}

is a unit. By the same argument as before,

⟨ a_{1} ⟩ \subset ⟨ a_{2} ⟩ .

We can continue with this construction to obtain an ascending chain of ideals

⟨ a ⟩ \subset ⟨ a_{1} ⟩ \subset ⟨ a_{2} ⟩ \subset \dots .

By Lemma 18.14, there exists a positive integer

N

such that

⟨ a_{n} ⟩ = ⟨ a_{N} ⟩

for all

n \geq N .

Consequently,

a_{N}

must be irreducible. We have now shown that

a

is the product of two elements, one of which must be irreducible.

Now suppose that

a = c_{1} p_{1},

where

p_{1}

is irreducible. If

c_{1}

is not a unit, we can repeat the preceding argument to conclude that

⟨ a ⟩ \subset ⟨ c_{1} ⟩ .

Either

c_{1}

is irreducible or

c_{1} = c_{2} p_{2},

where

p_{2}

is irreducible and

c_{2}

is not a unit. Continuing in this manner, we obtain another chain of ideals

⟨ a ⟩ \subset ⟨ c_{1} ⟩ \subset ⟨ c_{2} ⟩ \subset \dots .

This chain must satisfy the ascending chain condition; therefore,

a = p_{1} p_{2} \dots p_{r}

for irreducible elements

p_{1}, \dots, p_{r} .

Uniqueness of the factorization.

To show uniqueness, let

a = p_{1} p_{2} \dots p_{r} = q_{1} q_{2} \dots q_{s},

where each

p_{i}

and each

q_{i}

is irreducible. Without loss of generality, we can assume that

r < s .

Since

p_{1}

divides

q_{1} q_{2} \dots q_{s},

by Corollary 18.13 it must divide some

q_{i} .

By rearranging the

q_{i}

’s, we can assume that

p_{1} ∣ q_{1};

hence,

q_{1} = u_{1} p_{1}

for some unit

u_{1}

D .

Therefore,

a = p_{1} p_{2} \dots p_{r} = u_{1} p_{1} q_{2} \dots q_{s}

p_{2} \dots p_{r} = u_{1} q_{2} \dots q_{s} .

Continuing in this manner, we can arrange the

q_{i}

’s such that

p_{2} = q_{2}, p_{3} = q_{3}, \dots, p_{r} = q_{r},

to obtain

u_{1} u_{2} \dots u_{r} q_{r + 1} \dots q_{s} = 1 .

In this case

q_{r + 1} \dots q_{s}

is a unit, which contradicts the fact that

q_{r + 1}, \dots, q_{s}

are irreducibles. Therefore,

r = s

and the factorization of

a

is unique.

🔗

Corollary 18.16.

🔗

Let

F

be a field. Then

F [x]

is a UFD.

🔗

Example 18.17.

🔗

Every PID is a UFD, but it is not the case that every UFD is a PID. In Corollary 18.31, we will prove that

Z [x]

is a UFD. However,

Z [x]

is not a PID. Let

I = {5 f (x) + x g (x) : f (x), g (x) \in Z [x]} .

We can easily show that

I

is an ideal of

Z [x] .

Suppose that

I = ⟨ p (x) ⟩ .

Since

5 \in I,

5 = f (x) p (x) .

In this case

p (x) = p

must be a constant. Since

x \in I,

x = p g (x);

consequently,

p = \pm 1 .

However, it follows from this fact that

⟨ p (x) ⟩ = Z [x] .

But this would mean that

3

is in

I .

Therefore, we can write

3 = 5 f (x) + x g (x)

for some

f (x)

and

g (x)

Z [x] .

Examining the constant term of this polynomial, we see that

3 = 5 f (x),

which is impossible.

🔗

Subsection Euclidean Domains

🔗

We have repeatedly used the division algorithm when proving results about either

Z

F [x],

where

F

is a field. We should now ask when a division algorithm is available for an integral domain.

🔗

Let

D

be an integral domain such that there is a function

ν : D ∖ {0} \to N_{0} = N \cup {0}

satisfying the following conditions.

If $a$ and $b$ are nonzero elements in $D,$ then $ν (a) \leq ν (a b) .$
Let $a, b \in D$ and suppose that $b \neq 0 .$ Then there exist elements $q, r \in D$ such that $a = b q + r$ and either $r = 0$ or $ν (r) < ν (b) .$

🔗

Then

D

is called a Euclidean domain and

ν

is called a Euclidean valuation.

🔗

Example 18.18.

🔗

Absolute value on

Z

is a Euclidean valuation.

🔗

Example 18.19.

🔗

Let

F

be a field. Then the degree of a polynomial in

F [x]

is a Euclidean valuation.

🔗

Example 18.20.

🔗

Recall that the Gaussian integers in Example 16.12 of Chapter 16 are defined by

Z [i] = {a + b i : a, b \in Z} .

🔗

We usually measure the size of a complex number

a + b i

by its absolute value,.

| a + b i | = \sqrt{a^{2} + b^{2}};

however,

\sqrt{a^{2} + b^{2}}

may not be an integer. For our valuation we will let

ν (a + b i) = a^{2} + b^{2}

to ensure that we have an integer.

🔗

We claim that

ν (a + b i) = a^{2} + b^{2}

is a Euclidean valuation on

Z [i] .

Let

z, w \in Z [i] .

Then

ν (z w) = | z w |^{2} = | z |^{2} | w |^{2} = ν (z) ν (w) .

Since

ν (z) \geq 1

for every nonzero

z \in Z [i],

ν (z) \leq ν (z) ν (w) .

🔗

Next, we must show that for any

z = a + b i

and

w = c + d i

Z [i]

with

w \neq 0,

there exist elements

q

and

r

Z [i]

such that

z = q w + r

with either

r = 0

ν (r) < ν (w) .

We can view

z

and

w

as elements in

Q (i) = {p + q i : p, q \in Q},

the field of fractions of

Z [i] .

Observe that

\begin{aligned} z w^{- 1} & = (a + b i) \frac{c - d i}{c^{2} + d^{2}} \\ = \frac{a c + b d}{c^{2} + d^{2}} + \frac{b c - a d}{c^{2} + d^{2}} i \\ = (m_{1} + \frac{n_{1}}{c^{2} + d^{2}}) + (m_{2} + \frac{n_{2}}{c^{2} + d^{2}}) i \\ = (m_{1} + m_{2} i) + (\frac{n_{1}}{c^{2} + d^{2}} + \frac{n_{2}}{c^{2} + d^{2}} i) \\ = (m_{1} + m_{2} i) + (s + t i) \end{aligned}

🔗

Q (i) .

In the last steps we are writing the real and imaginary parts as an integer plus a proper fraction. That is, we take the closest integer

m_{i}

such that the fractional part satisfies

| n_{i} / (a^{2} + b^{2}) | \leq 1 / 2 .

For example, we write

\begin{aligned} \frac{9}{8} & = 1 + \frac{1}{8} \\ \frac{15}{8} & = 2 - \frac{1}{8} . \end{aligned}

🔗

Thus,

s

and

t

are the “fractional parts” of

z w^{- 1} = (m_{1} + m_{2} i) + (s + t i) .

We also know that

s^{2} + t^{2} \leq 1 / 4 + 1 / 4 = 1 / 2 .

Multiplying by

w,

we have

z = z w^{- 1} w = w (m_{1} + m_{2} i) + w (s + t i) = q w + r,

🔗

where

q = m_{1} + m_{2} i

and

r = w (s + t i) .

Since

z

and

q w

are in

Z [i],

r

must be in

Z [i] .

Finally, we need to show that either

r = 0

ν (r) < ν (w) .

However,

ν (r) = ν (w) ν (s + t i) \leq \frac{1}{2} ν (w) < ν (w) .

🔗

Theorem 18.21.

🔗

Every Euclidean domain is a principal ideal domain.

Proof.

Let

D

be a Euclidean domain and let

ν

be a Euclidean valuation on

D .

Suppose

I

is a nontrivial ideal in

D

and choose a nonzero element

b \in I

such that

ν (b)

is minimal for all

a \in I .

Since

D

is a Euclidean domain, there exist elements

q

and

r

D

such that

a = b q + r

and either

r = 0

ν (r) < ν (b) .

But

r = a - b q

is in

I

since

I

is an ideal; therefore,

r = 0

by the minimality of

b .

It follows that

a = b q

and

I = ⟨ b ⟩ .

🔗

Corollary 18.22.

🔗

Every Euclidean domain is a unique factorization domain.

🔗

Subsection Factorization in $D [x]$

🔗

One of the most important polynomial rings is

Z [x] .

One of the first questions that come to mind about

Z [x]

is whether or not it is a UFD. We will prove a more general statement here. Our first task is to obtain a more general version of Gauss’s Lemma (Theorem 17.14).

🔗

Let

D

be a unique factorization domain and suppose that

p (x) = a_{n} x^{n} + \dots + a_{1} x + a_{0}

🔗

D [x] .

Then the content of

p (x)

is the greatest common divisor of

a_{0}, \dots, a_{n} .

We say that

p (x)

is primitive if

gcd (a_{0}, \dots, a_{n}) = 1 .

🔗

Example 18.23.

🔗

Z [x]

the polynomial

p (x) = 5 x^{4} - 3 x^{3} + x - 4

is a primitive polynomial since the greatest common divisor of the coefficients is

1;

however, the polynomial

q (x) = 4 x^{2} - 6 x + 8

is not primitive since the content of

q (x)

2 .

🔗

Theorem 18.24. Gauss’s Lemma.

🔗

Let

D

be a UFD and let

f (x)

and

g (x)

be primitive polynomials in

D [x] .

Then

f (x) g (x)

is primitive.

Proof.

Let

f (x) = \sum_{i = 0}^{m} a_{i} x^{i}

and

g (x) = \sum_{i = 0}^{n} b_{i} x^{i} .

Suppose that

p

is a prime dividing the coefficients of

f (x) g (x) .

Let

r

be the smallest integer such that

p ∤ a_{r}

and

s

be the smallest integer such that

p ∤ b_{s} .

The coefficient of

x^{r + s}

f (x) g (x)

c_{r + s} = a_{0} b_{r + s} + a_{1} b_{r + s - 1} + \dots + a_{r + s - 1} b_{1} + a_{r + s} b_{0} .

Since

p

divides

a_{0}, \dots, a_{r - 1}

and

b_{0}, \dots, b_{s - 1},

p

divides every term of

c_{r + s}

except for the term

a_{r} b_{s} .

However, since

p ∣ c_{r + s},

either

p

divides

a_{r}

p

divides

b_{s} .

But this is impossible.

🔗

Lemma 18.25.

🔗

Let

D

be a UFD, and let

p (x)

and

q (x)

be in

D [x] .

Then the content of

p (x) q (x)

is equal to the product of the contents of

p (x)

and

q (x) .

Proof.

Let

p (x) = c p_{1} (x)

and

q (x) = d q_{1} (x),

where

c

and

d

are the contents of

p (x)

and

q (x),

respectively. Then

p_{1} (x)

and

q_{1} (x)

are primitive. We can now write

p (x) q (x) = c d p_{1} (x) q_{1} (x) .

Since

p_{1} (x) q_{1} (x)

is primitive, the content of

p (x) q (x)

must be

c d .

🔗

Lemma 18.26.

🔗

Let

D

be a UFD and

F

its field of fractions. Suppose that

p (x) \in D [x]

and

p (x) = f (x) g (x),

where

f (x)

and

g (x)

are in

F [x] .

Then

p (x) = f_{1} (x) g_{1} (x),

where

f_{1} (x)

and

g_{1} (x)

are in

D [x] .

Furthermore,

\deg f (x) = \deg f_{1} (x)

and

\deg g (x) = \deg g_{1} (x) .

Proof.

Let

a

and

b

be nonzero elements of

D

such that

a f (x), b g (x)

are in

D [x] .

We can find

a_{1}, b_{1} \in D

such that

a f (x) = a_{1} f_{1} (x)

and

b g (x) = b_{1} g_{1} (x),

where

f_{1} (x)

and

g_{1} (x)

are primitive polynomials in

D [x] .

Therefore,

a b p (x) = (a_{1} f_{1} (x)) (b_{1} g_{1} (x)) .

Since

f_{1} (x)

and

g_{1} (x)

are primitive polynomials, it must be the case that

a b ∣ a_{1} b_{1}

by Gauss’s Lemma. Thus there exists a

c \in D

such that

p (x) = c f_{1} (x) g_{1} (x) .

Clearly,

\deg f (x) = \deg f_{1} (x)

and

\deg g (x) = \deg g_{1} (x) .

🔗

The following corollaries are direct consequences of Lemma 18.26.

🔗

Corollary 18.27.

🔗

Let

D

be a UFD and

F

its field of fractions. A primitive polynomial

p (x)

D [x]

is irreducible in

F [x]

if and only if it is irreducible in

D [x] .

🔗

Corollary 18.28.

🔗

Let

D

be a UFD and

F

its field of fractions. If

p (x)

is a monic polynomial in

D [x]

with

p (x) = f (x) g (x)

F [x],

then

p (x) = f_{1} (x) g_{1} (x),

where

f_{1} (x)

and

g_{1} (x)

are in

D [x] .

Furthermore,

\deg f (x) = \deg f_{1} (x)

and

\deg g (x) = \deg g_{1} (x) .

🔗

Theorem 18.29.

🔗

D

is a UFD, then

D [x]

is a UFD.

Proof.

Let

p (x)

be a nonzero polynomial in

D [x] .

p (x)

is a constant polynomial, then it must have a unique factorization since

D

is a UFD. Now suppose that

p (x)

is a polynomial of positive degree in

D [x] .

Let

F

be the field of fractions of

D,

and let

p (x) = f_{1} (x) f_{2} (x) \dots f_{n} (x)

by a factorization of

p (x),

where each

f_{i} (x)

is irreducible. Choose

a_{i} \in D

such that

a_{i} f_{i} (x)

is in

D [x] .

There exist

b_{1}, \dots, b_{n} \in D

such that

a_{i} f_{i} (x) = b_{i} g_{i} (x),

where

g_{i} (x)

is a primitive polynomial in

D [x] .

By Corollary 18.27, each

g_{i} (x)

is irreducible in

D [x] .

Consequently, we can write

a_{1} \dots a_{n} p (x) = b_{1} \dots b_{n} g_{1} (x) \dots g_{n} (x) .

Let

b = b_{1} \dots b_{n} .

Since

g_{1} (x) \dots g_{n} (x)

is primitive,

a_{1} \dots a_{n}

divides

b .

Therefore,

p (x) = a g_{1} (x) \dots g_{n} (x),

where

a \in D .

Since

D

is a UFD, we can factor

a

u c_{1} \dots c_{k},

where

u

is a unit and each of the

c_{i}

’s is irreducible in

D .

We will now show the uniqueness of this factorization. Let

p (x) = a_{1} \dots a_{m} f_{1} (x) \dots f_{n} (x) = b_{1} \dots b_{r} g_{1} (x) \dots g_{s} (x)

be two factorizations of

p (x),

where all of the factors are irreducible in

D [x] .

By Corollary 18.27, each of the

f_{i}

’s and

g_{i}

’s is irreducible in

F [x] .

The

a_{i}

’s and the

b_{i}

’s are units in

F .

Since

F [x]

is a PID, it is a UFD; therefore,

n = s .

Now rearrange the

g_{i} (x)

’s so that

f_{i} (x)

and

g_{i} (x)

are associates for

i = 1, \dots, n .

Then there exist

c_{1}, \dots, c_{n}

and

d_{1}, \dots, d_{n}

D

such that

(c_{i} / d_{i}) f_{i} (x) = g_{i} (x)

c_{i} f_{i} (x) = d_{i} g_{i} (x) .

The polynomials

f_{i} (x)

and

g_{i} (x)

are primitive; hence,

c_{i}

and

d_{i}

are associates in

D .

Thus,

a_{1} \dots a_{m} = u b_{1} \dots b_{r}

D,

where

u

is a unit in

D .

Since

D

is a unique factorization domain,

m = s .

Finally, we can reorder the

b_{i}

’s so that

a_{i}

and

b_{i}

are associates for each

i .

This completes the uniqueness part of the proof.

🔗

The theorem that we have just proven has several obvious but important corollaries.

🔗

Corollary 18.30.

🔗

Let

F

be a field. Then

F [x]

is a UFD.

🔗

Corollary 18.31.

🔗

The ring of polynomials over the integers,

Z [x],

is a UFD.

🔗

Corollary 18.32.

🔗

Let

D

be a UFD. Then

D [x_{1}, \dots, x_{n}]

is a UFD.

🔗

Remark 18.33.

🔗

It is important to notice that every Euclidean domain is a PID and every PID is a UFD. However, the converse of each of these statements fails. There are principal ideal domains that are not Euclidean domains, and there are unique factorization domains that are not principal ideal domains (

Z [x]

🔗

Subsection Historical Note

🔗

Karl Friedrich Gauss, born in Brunswick, Germany on April 30, 1777, is considered to be one of the greatest mathematicians who ever lived. Gauss was truly a child prodigy. At the age of three he was able to detect errors in the books of his father’s business. Gauss entered college at the age of 15. Before the age of 20, Gauss was able to construct a regular

17

-sided polygon with a ruler and compass. This was the first new construction of a regular

n

-sided polygon since the time of the ancient Greeks. Gauss succeeded in showing that if

N = 2^{2^{n}} + 1

was prime, then it was possible to construct a regular

N

-sided polygon.

🔗

Gauss obtained his Ph.D. in 1799 under the direction of Pfaff at the University of Helmstedt. In his dissertation he gave the first complete proof of the Fundamental Theorem of Algebra, which states that every polynomial with real coefficients can be factored into linear factors over the complex numbers. The acceptance of complex numbers was brought about by Gauss, who was the first person to use the notation of

i

for

\sqrt{- 1} .

🔗

Gauss then turned his attention toward number theory; in 1801, he published his famous book on number theory, Disquisitiones Arithmeticae. Throughout his life Gauss was intrigued with this branch of mathematics. He once wrote, “Mathematics is the queen of the sciences, and the theory of numbers is the queen of mathematics.”

🔗

In 1807, Gauss was appointed director of the Observatory at the University of Göttingen, a position he held until his death. This position required him to study applications of mathematics to the sciences. He succeeded in making contributions to fields such as astronomy, mechanics, optics, geodesy, and magnetism. Along with Wilhelm Weber, he coinvented the first practical electric telegraph some years before a better version was invented by Samuel F. B. Morse.

🔗

Gauss was clearly the most prominent mathematician in the world in the early nineteenth century. His status naturally made his discoveries subject to intense scrutiny. Gauss’s cold and distant personality many times led him to ignore the work of his contemporaries, making him many enemies. He did not enjoy teaching very much, and young mathematicians who sought him out for encouragement were often rebuffed. Nevertheless, he had many outstanding students, including Eisenstein, Riemann, Kummer, Dirichlet, and Dedekind. Gauss also offered a great deal of encouragement to Sophie Germain (1776–1831), who overcame the many obstacles facing women in her day to become a very prominent mathematician. Gauss died at the age of 78 in Göttingen on February 23, 1855.

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Section 18.2 Factorization in Integral Domains

Example 18.8.

Example 18.9.

Example 18.10.

Subsection Principal Ideal Domains

Lemma 18.11.

Proof.

Theorem 18.12.

Proof.

Corollary 18.13.

Proof.

Lemma 18.14.

Proof.

Theorem 18.15.

Proof.

Existence of a factorization.

Uniqueness of the factorization.

Corollary 18.16.

Example 18.17.

Subsection Euclidean Domains

Example 18.18.

Example 18.19.

Example 18.20.

Theorem 18.21.

Proof.

Corollary 18.22.

Subsection Factorization in D[x]

Example 18.23.

Theorem 18.24. Gauss’s Lemma.

Proof.

Lemma 18.25.

Proof.

Lemma 18.26.

Proof.

Corollary 18.27.

Corollary 18.28.

Theorem 18.29.

Proof.

Corollary 18.30.

Corollary 18.31.

Corollary 18.32.

Remark 18.33.

Subsection Historical Note

Subsection Factorization in $D [x]$