Then there exist two binary operations + and · such that: (1) (R

HUA-CHIEH LI

In this note, our ring is always a commutative ring. In other words, suppose that R is a ring. Then there exist two binary operations + and · such that:

(1) (R, +) is an abelian group;

(2) (a· b) · c = a · (b · c) for all a, b, c ∈ R;

(3) a· (b + c) = a · b + a · c for all a, b, c ∈ R;

(4) a· b = b · a for all a, b ∈ R.

Moreover, we say R is an integral domain if R satisfies the following extra conditions:

• there exists 1 ∈ R such that 1 · a = a · 1 = a for all a ∈ R;

• if a = 0 and b = 0 in R, then a · b = 0.

1. Euclidean Domain

LetN be the set of nonnegative integers and R a ring. We say that R is a Euclidean Ring if there is a function φ : R\ {0} → N such that: if a, b ∈ R and b = 0, then there exist q, r ∈ R such that a = bq + r with either r = 0 or φ(r) < φ(b).

A Euclidean ring which is an integral domain is called a Euclidean domain.

Example 1.1. The Ring Z of integers with φ(n) = |n| is a Euclidean domain.

Proof. For x∈ Q, denote [x] the greatest integer less than or equal to x. Given a, b ∈ Z, we claim that there exist q, r ∈ Z such that a = bq + r with r = 0 or |r| < |b|.

We first consider the case that b > 0. Let q = [a/b] and r = a− b[a/b]. Then a = bq + r.

It remains to show that 0≤ r < b. We have that a

b − 1 <
a b

 ≤ a b . Multiplying all terms of this inequality by −b, we obtain

b− a > −b
a b

 ≥ −a

and hence

0≤ a − b
a b



< b , which is precisely 0 ≤ r < b as desired.

For the case b < 0, use the similar argument above for a and −b. We find that there exist q and r ∈ Z such that a = (−b)q + r with r = 0 or r < |b| = −b; so −q and r have the

desired properties.

Example 1.2. If F is a field, then the ring of polynomials in one variable F [x] is a Euclidean domain with φ(f ) = deg(f ).

1

Proof. Given f , g ∈ F [x] with g = 0, if deg(f) < deg(g), then let q = 0 and r = f. If deg(f )≥ deg(g), then we proceed by induction on deg(f).

If deg(f ) = 0, then deg(g) = 0. Thus f and g are in F . Let q = f · g⁻¹ and r = 0. We have f = gq + r with r = 0 as desired.

Assume now that the property for Euclidean domain is true for polynomials of degree less than n = deg(f ). Suppose

f =

n i=0

a_ixⁱ, g =

m i=0

b_ixⁱ, with a_n = 0, b_m = 0.

Let f₁ = f − (anb⁻¹_m x^m−n)g. It is clear that deg(f₁) ≤ n − 1. By the induction hypothesis there are polynomials q₁ and r₁ such that f₁ = gq₁ + r₁ with r₁ = 0 or deg(r₁) < deg(g).

Therefore, let q = a_nb⁻¹_m x^n−m+ q
and r = r₁. Then

f = f₁+ (a_nb⁻¹_m x^m−n)g = g(q₁+ a_nb⁻¹_m x^m−n) + r₁ = gq + r

with r = 0 or deg(r) < deg(g) as desired.

Recall that the set of complex numbers C consists of elements of the form x + yi, with x, y ∈ R where i satisfies i² = −1. For α = x + yi ∈ C, we define the norm of α by N (α) = x²+ y². Given α = x + yi and β = u + vi, we have that αβ = (xu− yv) + (xv + yu)i and

N (αβ) = (xu− yv)²+ (xv + yu)² = (x²+ y²)(u²+ v²) = N (α)N (β).

Example 1.3. Let Z[i] = {a + bi | a, b ∈ Z} be a subset of complex numbers. Z[i] is an integral domain called the domain of Gaussian integers. Moreover,Z[i] is a Euclidean domain with φ(a + bi) = N (a + bi) = a²+ b².

Proof. Z[i] is clearly closed under addition and substraction. Moreover, if a+bi, c+di ∈ Z[i], then

(a + bi)(c + di) = (ac− bd) + (ad + bc)i ∈ Z[i].

ThusZ[i] is closed under multiplication and is a ring. Since Z[i] is contained in the complex numbers it is an integral domain.

It is clear that the norm defines a map from Z[i] to N. Let α = a + bi, β = c + di ∈ Z[i]

and suppose that β = 0. Consider α

β = a + bi

c + di = ac + bd

c²+ d² + bc− ad

c²+ d² i = s + ti.

Choose integers m, n ∈ Z such that |s − m| ≤ 1/2 and |t − n| ≤ 1/2. Set δ = m + ni and γ = α− βδ. Then δ, γ ∈ Z[i] and either γ = 0 or

φ(γ) = φ(β(α

β − δ)) = φ(β)φ(α

β − δ)) = φ(β)((s − m)²+ (t− n)²)≤ 1

2φ(β) < φ(β).

Exercise 1. Let ω = (−1 +√

−3)/2 and Z[ω] = {a + bω | a, b ∈ Z}. Show that Z[ω] is a Euclidean domain.

Example 1.4. Let θ = (1 +√

−19)/2 and Z[θ] = {a + bθ | a, b ∈ Z}. Z[θ] is an integral domain but is not a Euclidean domain.

Proof. Z[θ] is clearly closed under addition and substraction. Moreover, θ² = θ− 5. Hence, if a + bθ, c + dθ ∈ Z[θ], then

(a + bθ)(c + dθ) = ac + (ad + bc)θ + bdθ² = (ac− 5bd) + (ad + bc + bd)θ ∈ Z[θ].

ThusZ[θ] is closed under multiplication and is a ring. Since Z[θ] is contained in the complex numbers it is an integral domain.

Suppose that Z[θ] is a Euclidean domain with φ : Z[θ] \ {0} → N satisfies the Euclidean domain property. Let α∈ Z[θ] be an element such that

φ(α) = min{φ(λ) | λ = 0, 1, −1, λ ∈ Z[θ]}.

By the Euclidean domain property, there exist δ, γ ∈ Z[θ] such that 2 = αδ + γ with γ = 0 or φ(γ) < φ(α). However, by the definition of α, this implies that γ = 0, 1 or −1. In other words, αδ = 1, 2 or 3.

Recall that if β = a + bθ∈ Z[θ], then N(β) = a²+ ab + 5b² ∈ N. Moreover, suppose β = 0, 1 or −1. If a = 0 then N(β) = 5b² ≥ 5 and if b = 0 then N(β) = a² ≥ 4. If ab > 0, then

N (β) = a²+ ab + 5b² = (a− b)²+ 4b²+ 3ab≥ 4b²+ 3ab≥ 7 and if ab < 0, then

N (β) = a²+ ab + 5b² = (a + b)²+ 4b²− ab ≥ 4b²− ab ≥ 5.

In conclusion, if β ∈ Z[θ] \ {0, 1, −1} then N(β) ∈ N and N(β) ≥ 4.

Since N (αδ) = 1, 4 or 9, and N (αδ) = N (α)N (δ), we have that N (α)| 1, N(α)| 4 or N (α)| 9. The discussion above shows that N(α) = 1, 2, 3. Hence we have that N(α) = 4 or N (α) = 9.

The Euclidean domain property shows that there exist δ
and γ
∈ Z[θ] such that θ = αδ
+ γ
with either γ
= 0 or φ(γ
) < φ(α). Again, the definition of α implies that αδ
= θ, θ−1 or θ+1. Taking norms, we have N(α)|N(θ), N(α)|N(θ−1) or N(α)|N(θ+1). However, N (θ) = 5, N (θ− 1) = 5 and N(θ + 1) = 7. Neither one of them can be divided by 4 or 9.

We get a contradiction. Hence Z[θ] is not a Euclidean domain.
Definition 1.5. A nonzero element a of a ring R is said to divide an element b ∈ R (notation:

a| b) if there exists x ∈ R such that b = ax. Elements a, b of R are said to be associates (notation: a≈ b) if a | b and b | a.

Let S be a nonempty subset of R. An element d ∈ R is a greatest common divisor of S provided:

(1) d| a for all a ∈ S;

(2) if c| a for all a ∈ S, then c | d.

In general, greatest common divisors do not always exist. For example, in the ring 2Z of even integers, 2 has no divisor at all, whence 2, 4 has no greatest common divisor. Even when a greatest common divisor exists, it need not be unique. However, any two greatest common divisors of S are clearly associates by property (2). Furthermore any associate of a greatest common divisor of S is easily seen to be a greatest common divisor of S.

In the following we provide some basic properties of greatest common divisor.

Lemma 1.6. Let R be a ring and a, b, c ∈ R. Suppose that d is a greatest common divisor of a, b.

(1) Suppose that c = aq + b for some q∈ R. Then d is a greatest common divisor of a, c.

(2) Suppose that d
is a greatest common divisor of d, c. Then d
is a greatest common divisor of a, b, c.

Proof. (proof of (1)) We first show that d divides a and c. We know d divides a by definition.

Since d| a and d | b, we have a = dx and b = dy for some x, y ∈ R. Hence c = dxq + dy = d(xq + y). This shows that d| c.

Suppose e ∈ R such that e | a and e | c. Then there exist u, v ∈ R such that a = eu and c = ev. Hence b = c− aq = e(v − uq). This shows that e | b. Since e divides a and b, by the

definition of greatest common divisors, we have e| d.

Exercise 2. Prove (2) of Lemma 1.6.

Example 1.7 (The Euclidean Algorithm). Let a, b ∈ Z. By Example 1.1, there exist q1, r₁ ∈ Z such that

a = bq₁+ r₁, 0≤ r1 <|b| . If r₁ > 0, there exist q₂, r₂ ∈ Z such that

b = r₁q₂+ r₂, 0≤ r₂ < r₁. If r₂ > 0, there exist q₃, r₃ ∈ Z such that

r₁ = r₂q₃+ r₃, 0≤ r3 < r₂.

Continue this process. Then r_n = 0 for some n∈ N. If n > 1 then r_n−1 is a greatest common divisor of a, b. If n = 1, then b is a greatest common divisor of a, b.

Proof. Note that r₁ > r₂ > . . . . If r_n = 0 for all n ∈ N, then r₁, r₂, r₃, . . . is an infinite, strictly decreasing sequence of positive integers, which is impossible. So r_n = 0 for some n.

If r₁ = 0, then a = bq₁. So b| a and of course b | b. If c divides both a and b, then of course c| b. Hence b is a greatest common divisor of a, b.

Now suppose r_n = 0 for n > 1. Then r_n−2 = r_n−1q_n (we set r₀ = b). By the argument above, we have that r_n−1 is a greatest common divisor of r_n−2, r_n−1. However, r_n−3 = r_n−2q_n−1+ r_n−1 (we set r₋₁ = a). By Lemma 1.6 (1), we have r_n−1 is a greatest common divisor of r_n−2, r_n−3. Continue this argument inductively. We have that r_n−1 is a greatest

common divisor of a, b.

Exercise 3. Suppose R is a Euclidean domain and a₁, . . . , a_n∈ R. Show that there exists a greatest common divisor of a₁, . . . , a_n.

2. Principle Ideal Domain

Given a ring R, a subring I of R is an ideal provided rx ∈ I for r ∈ R, x ∈ I. A principal ideal ring is a ring in which every ideal is principle. In other words, for every ideal I of R, there exists x∈ I such that if λ ∈ I, λ = rx for some r ∈ R. A principle ideal ring which is an integral domain is called a principle ideal domain

Example 2.1. Z is a principle ideal domain.

Proof. Given a nonzero ideal I of Z. Consider n ∈ Z such that

|n| = min {|x| : x ∈ I \ {0}} .

Given a∈ I, by Example 1.1, there exist h, r ∈ Z such that a = nh + r with either r = 0 or

|r| < |n|. Since r = a − nh ∈ I, by the definition of n, we conclude that r = 0 and hence

a = nh. In other words, I = (n).

Using similar argument we can show the following:

Theorem 2.2. Every Euclidean ring is a principle ideal ring.

Exercise 4. Prove Theorem 2.2.

From Theorem 2.2, the polynomial ring F [x] in Example 1.2 and the Gaussian integers Z[i] in Example 1.3 are principle ideal domains.

In general, to prove a ring is a principle ideal ring is not easy. We can imitate the proof of Theorem 2.2 to show certain rings are principle ideal rings.

Theorem 2.3. Let R be a ring. Suppose that there is a function φ : R \ {0} → N such that given α, β ∈ R, β = 0, if β does not divide α then there exist γ, δ ∈ R such that αγ − βδ = 0 and

φ(αγ− βδ) < φ(β).

Then R is a principle ideal ring.

Proof. Let I be a nonzero ideal of R. Let β ∈ I be an element with the property that φ(β) = min{φ(x) : x ∈ I \ {0}} .

We claim that I = (β). Given α∈ I, suppose that β does not divide α. By the hypothesis, there exist δ, γ ∈ R such that αγ − βδ = 0 and φ(αγ − βδ) < φ(β). Since αγ − βδ ∈ I and αγ − βδ = 0, this contradicts the assumption of β. Therefore β divides every element of

I.

Example 2.4. Let θ = (1 +√

−19)/2 and Z[θ] = {a + bθ | a, b ∈ Z}. Z[θ] is a principle ideal domain.

Proof. Let φ(α) = N (α) for all α∈ Z[θ] \ {0} . We will show that Z[θ] satisfies the condition in Theorem 2.3.

Given α, β ∈ Z[θ] with β = 0, if β does not divide α then a case by case consideration will lead to elements γ, δ ∈ Z[θ] such that

0 < N

α βγ − δ



< 1, whence αγ− βδ = 0 and N(αγ − βδ) < N(β).

Write

α

β = s + tθ, with s, t∈ Q.

(1) t ∈ Z: In this case, s ∈ Z. Let n ∈ Z such that |s − n| ≤ 1/2 and take γ = 1, δ = n + tθ. Now,

0 < N

α βγ− δ



= N (s− n) ≤ 1 4 < 1.

(2) s∈ Z:

(a) 5t ∈ Z: Let m ∈ Z such that |t − m| ≤ 1/2. In fact, because 5t ∈ Z, we have

|t − m| ≤ 2/5. Take γ = 1 and δ = s + mθ. Now 0 < N

α βγ− δ



= N ((t− m)θ) ≤ 4

25 × 5 < 1.

(b) 5t∈ Z: Consider

(s + tθ)(1− θ) = s − sθ + tθ − tθ² = s− sθ + tθ − tθ + 5t = s + 5t − sθ.

Let n∈ Z such that |s + 5t − n| ≤ 1/2 and take γ = 1 − θ, δ = n − sθ. Now, 0 < N

α βγ− δ



= N (s + 5t− n) ≤ 1 4 < 1.

(3) s, t∈ Z:

(a) 2s, 2t∈ Z: Consider

(s + tθ)θ = sθ + tθ− 5t = −5t + (s + t)θ.

Since s + t∈ Z, letting n ∈ Z such that |−5t − n| ≤ 1/2, we can take γ = θ and δ = n + (s + t)θ. Now,

0 < N

α βγ− δ



= N (−5t − n) ≤ 1 4 < 1.

(b) 2s ∈ Z and 2t ∈ Z: Let n ∈ Z such that |2s − n| ≤ 1/2. Take γ = 2 and δ = n + 2tθ. Now,

0 < N

α βγ− δ



= N (2s− n) ≤ 1 4 < 1.

(c) 2s∈ Z and 2t ∈ Z: When 10t ∈ Z, let m ∈ Z such that |2t − m| ≤ 1/2. In fact, because 5× 2t ∈ Z, we have |2t − m| ≤ 2/5. Take γ = 2 and δ = 2s + mθ. Now

0 < N

α βγ− δ



= N ((2t− m)θ) ≤ 4

25× 5 < 1.

If 10t∈ Z, then consider

(s + tθ)(2− 2θ) = 2s − 2sθ + 2tθ − 2tθ² = 2s + 10t− 2sθ.

Let n∈ Z such that |2s + 10t − n| ≤ 1/2 (note that 10t ∈ Z) and take γ = 2−2θ, δ = n− 2sθ. Now,

0 < N

α βγ− δ



= N (2s + 10t− n) ≤ 1 4 < 1.

(d) 2s∈ Z and 2t ∈ Z: Let m ∈ Z such that |t − m| ≤ 1/2. If |t − m| ≤ 1/3, letting n∈ Z such that |s − n| ≤ 1/2, then we can take γ = 1 and δ = n + mθ. Now, 0 < N

α βγ− δ



= N ((s− n) + (t − m)θ) ≤ 1 4 +1

6 +1

9 × 5 = 35 36 < 1.

If 1/3 < |t − m| < 1/2, then 2/3 < |2t − 2m| < 1. Let m
∈ Z such that

|2t − m
| ≤ 1/2. Then we have |2t − m
| < 1/3. Let n
∈ Z such that |2s − n
| ≤ 1/2. Take γ = 2 and δ = n
+ m
θ. Now,

0 < N

α βγ− δ



= N ((2s− n
) + (2t− m
)θ) < 1 4 +1

6 +1

9 × 5 = 35 36 < 1.

Remark 2.5. The converse of Theorem 2.2 is false sinceZ[θ] is a principle ideal domain that is not a Euclidean domain (Example 1.4).

Example 2.6. Let Z[x] be the ring of polynomials over Z. Then Z[x] is an integral domain but is not a principle ideal domain.

Proof. Considering the leading coefficients of f (x) and g(x), we can easily conclude that if f (x)= 0 and g(x) = 0 in Z[x], then f(x)g(x) = 0.

To show thatZ[x] is not a principle ideal domain, we consider the ideal I generated by 2 and x (i.e. I = (2, x)). We first claim that I = Z[x]. Otherwise there exist u(x), v(x) ∈ Z[x]

such that 1 = 2u(x) + xv(x). Substitute x = 0 into the identity. We have that 1 = 2u(0) which is absurd because u(0)∈ Z.

Now, suppose that there exists f (x) ∈ Z[x] such that (f(x)) = I. In other words, there exist g(x) ∈ Z[x] and h(x) ∈ Z[x] such that 2 = g(x)f(x) and x = h(x)f(x). From 2 = g(x)f (x), we conclude that f (x)∈ Z. Because I = Z[x], f(x) can not be a unit, whence f (x) = ±2. On the other hand, by x = h(x)f(x), we have h(x) = ax + b for some a, b ∈ Z.

Since ±2a = 1 for all a ∈ Z, we get a contradiction.

Exercise 5. Suppose that R is an integral domain. Suppose further that there exists a ∈ R such that a = 0 and a is not a unit in R. Prove that R[x] the polynomial ring over R is an integral domain but is not a Euclidean domain.

Finally we provide some basic properties of principle ideal rings.

Proposition 2.7. Every principle ideal ring is a ring with identity.

Proof. Since R itself is an ideal of R, R = (a) for some a ∈ R. Consequently, a ∈ R, so a = ea = ae for some e ∈ R. If b ∈ R, then b = xa for some x ∈ R. Therefore, be = (xa)e = x(ae) = xa = b, whence e is the identity of R.
Exercise 6. Prove that every Euclidean ring is a ring with identity without using the fact that every Euclidean ring is a principle ideal ring.

Proposition 2.8. If R is a principle ideal ring, given a₁, . . . , a_n ∈ R, then a greatest common divisor of {a1, . . . , a_n} exists.

Proof. Consider I = (a₁, . . . , a_n) the ideal generated by a₁, . . . , a_n. Since R is a principle ideal ring, there exists d ∈ R such that I = (d). We claim that d is a greatest common divisor of {a₁, . . . , a_n}.

First, since a_i ∈ I = (d), there exist ri ∈ R such that ai = r_id for i = 1, . . . , n. Hence d| ai

for i = 1, . . . , n.

Second, since (a₁, . . . , a_n) = (d), there exist λ_i ∈ R such that d =_n

i=1λ_ia_i. Suppose that c| ai for i = 1, . . . , n. There exist γ_i ∈ R such that ai = γ_ic for i = 1, . . . , n. This implies that d =_n

i=1(λ_iγ_i)c, whence c| d.

Recall that a ring is Noetherian if it satisfies the ascending chain condition on ideals. It can be proved that R is Noetherian if and only if every ideal of R is finitely generated. We do not need this fact here. However, we can show that a principle ideal ring is Noetherian.

Lemma 2.9. If R is a principle ideal ring and

I₁ ⊆ I₂ ⊆ · · · ⊆ I_n ⊆ · · ·

is a chain of ideals in R, then for some n∈ N, I_j = I_n for all j ≥ n.

Proof. Let I =∪i∈NI_i. We claim that I is an ideal of R. If b, c∈ I, then we have b ∈ Ii and c∈ Ij for some i, j ∈ N. Without loss of generality, we can assume that i ≥ j. Consequently I_j ⊆ Ii, and hence b, c ∈ Ii. Therefore, b− c ∈ Ii ⊆ I. Similarly, if r ∈ R and b ∈ I, then b ∈ Ii for some i ∈ N, whence rb ∈ Ii ⊆ I. Therefore, I is an ideal of R. By hypothesis I is principle, say I = (a). Since a ∈ I, we have a ∈ I_n for some n ∈ N. Hence (a) ⊆ I_n. Therefore, for every j ≥ n,

(a)⊆ I_n ⊆ I_j ⊆ (a),

whence I_j = I_n.

Exercise 7. Suppose that R is a principle ideal ring. Let a₁, . . . , a_n, . . . be (infinitely many) elements in R. Prove that there exists a greatest common divisor of {a₁, . . . , a_n, . . .} .

3. Unique Factorization Domain

3.1. General Properties. The Fundamental Theorem of Arithmetic says that any positive integer n > 1 can be written uniquely in the form n = p^t₁¹· · · p^t_r^r, where p₁ < · · · < p_r are primes and t_i > 0 for all i. In this section we study those integral domains in which an analogue of the fundamental theorem of arithmetic holds.

In Z, a prime number p has the following properties:

(1) If p = ab then a or b is a unit.

(2) If p| ab then p | a or p | b.

For arbitrary ring, these are two different properties.

Definition 3.1. Let R be a ring with identity. An element π ∈ R is irreducible provided that π is not a unit and if π = ab for some a, b ∈ R then a or b is a unit.

An element p ∈ R is prime provided that p is not a unit and if p | ab then p | a or p | b.

Example 3.2. In the ring Z/6Z = {¯0, ¯1, ¯2, ¯3, ¯4, ¯5}, ¯2 is prime but it is not irreducible.

Proof. ¯2 does not divide ¯1· ¯1 = ¯5 · ¯5 = ¯1, ¯1 · ¯3 = ¯3 · ¯3 = ¯3 · ¯5 = ¯3, and ¯1 · ¯5 = ¯5. Hence ¯2 is prime. On the other hand, ¯2 is not irreducible because ¯2 = ¯2· ¯4 and neither ¯2 nor ¯4 are

units in Z/6Z.

Example 3.3. In the ring Z[√

10] = 

a + b√

10 : a, b∈ Z

, 2 is irreducible but it is not prime.

Proof. Recall that the map N : Z[√

10] → Z given by N (a + b√

10) = a² − 10b² has the properties that N (αβ) = N (α)N (β) for all α, β ∈ Z[√

10] andN (α) = ±1 if and only if α is a unit.

Suppose that there exist α and β inZ[√

10] which are not units such that 2 = αβ. Then we have 4 =N (2) = N (α)N (β). Since α = a+b√

10 is not a unit, we haveN (α) = a²−10b² =

±2. This shows that a² ≡ ±2 (mod 5). However, neither 2 nor −2 is a quadratic residue modulo 5. We get a contradiction. Hence 2 is irreducible.

On the other hand, since 2·3 = 6 = (4+√

10)(4−√

10), we have that 2| (4+√

10)(4−√ 10).

Suppose that 2| (4 +√

10) or 2| (4 −√

10). By taking N , we have that 4 | 6 in Z, which is absurd. Hence 2 is not prime in Z[√

10].

From examples above, we know that in general prime elements and irreducible elements are distinct. However in some cases, they are related.

Lemma 3.4. Let R be an integral domain. Then every prime element of R is irreducible.

Proof. Suppose that p is prime. If p = ab, then either p| a or p | b; say p | a. Thus there exist x ∈ R such that a = px. Therefore, p = ab = pxb, and hence p(1 − xb) = 0. Since R is an integral domain, this implies that 1 = xb. Therefore, b is a unit. Hence p is irreducible.
We include an important property for irreducible elements of an integral domain which is familiar for the integer ring Z.

Lemma 3.5. Let R be an integral domain. The only divisors of an irreducible element of R are its associates and the units of R.

Proof. If π is irreducible and d| π, then because π = dx for some x ∈ R, this implies that either d or x is a unit. The second case implies that d and π are associates.

Exercise 8. Let R be an integral domain. Suppose that a, b ∈ R are associates.

(1) Prove that there exists an unit u∈ R such that a = ub.

(2) Prove that a is irreducible if and only if b is irreducible.

(3) Prove that a is prime if and only if b is prime.

Definition 3.6. An integral domain R is a unique factorization domain provided that:

(1) Every nonzero element a ∈ R which is not a unit can be written as a = α1· · · αn

with α_i irreducible.

(2) If a = α₁· · · αn = β₁· · · βm with α_i, β_j irreducible, then n = m and for some permutation σ of {1, 2, . . . , n}, αi and β_σ(i) are associates for every i.

Remark 3.7. From the definition, every irreducible element in a unique factorization domain is necessary prime. Consequently, prime elements and irreducible elements coincide in a unique factorization domain by Lemma 3.4.

Example 3.8. The polynomial ring F [x] over a field F is a unique factorization domain.

Proof. Because every nonzero constant is a unit, we show first that every nonconstant poly- nomial can be written as a product of finitely many irreducible polynomial. It is to see that polynomials of degree 1 are irreducible. assume that we have proved the result for all polynomials of degree less than n and that deg(f ) = n. If f is irreducible, we are done.

Otherwise f = gh where 1 ≤ deg(g), deg(h) < n. By the induction assumption both g and h can be written as products of finitely many irreducible polynomials. Thus so is f .

Next, we show that every irreducible polynomial is prime. Suppose that π is an irreducible polynomial and π|fg. Consider the ideal (f, π). Since F [x] is a principle ideal domain (c.f.

Theorem 2.2), we have (f, π) = (d) for some d ∈ F [x]. π ∈ (d) implies that d | π, and hence by Lemma 3.5, (f, π) = (1) or (π). If (f, π) = (π), then π| f. If (f, π) = 1, then there exist l, h∈ F [x] such that lπ + hf = 1. Thus lπg + hfg = g. Since π divides the left-hand side of this equation, π must divide g.

Finally if f = π₁· · · πn= p₁· · · pm with π_i, p_j irreducible, then since π₁ is prime, π₁ divides some p_j; say p₁. On the other hand, since p₁ is irreducible and π₁is not a unit, by Lemma 3.5 π₁ and p₁ are associates; say uπ₁ = p₁ for some unit u of R. Hence π₂· · · πn = (up₂)· · · pm. By Exercise 8, up₂ is also irreducible, the proof of uniqueness is now completed by a routine

inductive argument.

Exercise 9. Let R be an integral domain.

(1) Prove that p is a prime element in R if and only if (p) is a prime ideal of R.

(2) Suppose that R is a principle ideal domain. Prove that π is irreducible in R if and only if (π) is a maximal ideal of R.

(3) Suppose that R is a principle ideal domain. Prove that an element in R is prime if and only if it is irreducible.

(4) Show that Z[√

10] is not a principle ideal domain.

In general, to show a ring is a unique factorization domain we only have to show the following:

(1) using the irreducibility to show that in the specific ring every nonzero element which is not a unit can be written as a product of finitely many irreducible elements;

(2) show that in the specific ring every irreducible element is prime. Then the proof of uniqueness can be completed by a routine inductive argument as in the proof of Example 3.8.

Theorem 3.9. Every principle ideal domain is a unique factorization domain.

Proof. Suppose that R is a principle ideal domain. We claim first that if a ∈ R, a = 0 and a is not a unit, then a can be written as a product of finitely many irreducible elements.

If a can not be written as a product of finitely many irreducible elements, then a is not irreducible and hence a = a₁b₁ for some a₁, b₁ ∈ R which are not units. By assumption, one of the a₁ or b₁ can not be written as a product of finitely many irreducible elements; say a₁. Then a₁ = a₂b₂ for some a₂, b₂ ∈ R which are not units and a₂ can not be written as a product of finitely many irreducible elements. Continuing in this way, we construct infinitely many a_i with a_i = a_i+1b_i+1 where all the a_i and b_i ∈ R are not units. Since a = a₁b₁ and b₁ is not a unit, we have that (a)
(a₁). Similarly, we have (a_i)
(a_i+1). In other words we have a nonstop ascending chain of ideals

(a)
(a1)
· · ·
(ai)
· · · , contradicting Lemma 2.9.

For the uniqueness, exercise 9 says that every irreducible element of R is prime. This completes the proof.

Exercise 10. Suppose that R is a unique factorization domain. Let S be a set of primes in R such that every prime in R is associate to a prime in S and no two primes in S are associate.

(1) If a∈ R, a = 0, show that we can uniquely write a = u

p∈S

p^vp(a),

where u is a unit and v_p(a) are nonnegative integers which are positive only for finitely many p∈ S.

(2) Prove that v_p(ab) = v_p(a) + v_p(b) for all p∈ S and a, b ∈ R.

(3) Given a₁, . . . , a_n ∈ R, prove that there exists a greatest common divisor of a₁, . . . , a_n. By Theorem 3.9, we know that Z[i] and Z[^1+√₂⁻¹⁹] are unique factorization domains. The converse of Theorem 3.9 is not always true. For example, we know thatZ[x] is not a principle ideal domain (c.f. Example 2.6), but we will show later that Z[x] is a unique factorization domain.

3.2. Factorization in Polynomial Rings. In the rest of this section, we devote entirely to show that if R is a unique factorization domain, then R[x], the polynomial ring over R is also a unique factorization domain.

Let F be the quotient field of R. In other words, every element of F can be written as a/b for some a, b ∈ R with b = 0. Our strategy is using the fact that F [x] is a unique factorization domain to show that R[x] is a unique factorization domain.

Let f = _n

i=0a_ixⁱ be a nonzero polynomial in R[x]. Since R is a unique factorization domain, by Exercise 10 (3), a greatest common divisor of the coefficients a₀, a₁, . . . , a_n exists.

We call it a content of f and denotes it by C(f ). Strictly speaking, C(f ) is ambiguous since greatest common divisors are not unique. But any two contents of are necessarily associates.

We shall write b≈ c whenever b and c are associates in R. If f ∈ R[x] and C(f) is a unit in R, then f is said to be primitive.

Lemma 3.10. Let R be a unique factorization domain. a ∈ R and f, g ∈ R[x].

(1) C(af )≈ aC(f). In particular, f = C(f)f1 with f₁ primitive in R[x].

(2) (Gauss) C(f g) ≈ C(f)C(g). In particular, the product of primitive polynomials in R[x] is also primitive.

Proof. (1) Suppose that f =_n

i=0a_ixⁱ and d = C(f ) which is a greatest common divisor of a₀, a₁, . . . , a_n. Then af =_n

i=0aa_ixⁱand ad is a greatest common divisor of aa₀, aa₁, . . . , aa_n. On the other hand, let b_i = a_i/d∈ R. The greatest common divisor of b0, b₁, . . . , b_nis a unit.

Hence f = d_n

i=0b_ixⁱ = C(f )f₁ with f₁ =_n

i=0b_ixⁱ primitive.

(2) f = C(f )f₁ and g = C(g)g₁ with f₁, g₁ primitive, by (1). Consequently C(f g) ≈ C(f )C(g)C(f₁g₁). Hence it suffices to prove that if f and g are primitive then f g is primitive (i.e. C(f g) is a unit). If f = _n

i=0a_ixⁱ and g = _m

j=0b_jx^j, then f g = _n+m

k=0 c_kx^k with c_k=

i+j=ka_ib_j. If C(f g) is not a unit, then since R is a unique factorization domain, there exists a prime element p ∈ R such that p | C(fg). That is, p | ck for all k. Since C(f ) is a unit, p  C(f). Hence there is an integer s such that p | ai for i < s and p  as. Similarly there is an integer t such that p| bj for j < t and p bt. Consider

c_s+t = a₀b_s+t+ a₁b_s+t−1+· · · + as−1b_t+1+ a_sb_t+ a_s+1b_t−1+· · · + as+tb₀.

p divides every term on the right-hand side of the equation except the term a_sb_t. Hence p cs+t. This is a contradiction. Therefore f g is primitive.
Now for study the irreducible elements in R[x], we first notice that if α∈ R is irreducible in R, then α is also irreducible in R[x]. Indeed, if α = f₁f₂ for f₁, f₂ ∈ R[x], then comparing the degrees of both side we have f₁, f₂ ∈ R. Since α is irreducible in R, either f₁ or f₂ is a unit in R and hence a unit in R[x].

Next, we compare elements in R[x] and elements in F [x]. Suppose f =_n

i=0a_ixⁱ ∈ F [x].

We can write a_i = α_iβ_i⁻¹ for some α_i, β_i ∈ R and βi = 0. Let β = _n

i=0β_i. We have βa_i = α_iγ_i for some γ_i ∈ R and hence βf = _n

i=0α_iγ_ixⁱ ∈ R[x]. In other word, every f ∈ F [x] can always be written as f = ab⁻¹f₁ with a, b∈ R, b = 0 and f1 primitive in R[x].

Lemma 3.11. Let f be a primitive polynomial in R[x] and g ∈ R[x]. Then f divides g in R[x] if and only if f divides g in F [x].

Proof. If f| g in R[x], then g = fh for some h ∈ R[x] ⊆ F [x]. Hence f | g in F [x].

On the other hand, if f| g in F [x], then g = fh for some h ∈ F [x]. Because h = ab⁻¹h₁ with a, b ∈ R, b = 0 and h₁ primitive in R[x], we have that bg = af h₁. Taking contents on both side, by Lemma 3.10 we have

bC(g)≈ C(bg) ≈ C(afh₁)≈ aC(f)C(h₁)≈ a,

because C(f ) and C(h₁) are units in R. Hence ab⁻¹ ∈ R. In other words, h = ab⁻¹h₁ ∈ R[x]

and hence f| g in R[x].

Lemma 3.12. Let f be a primitive polynomial in R[x]. Then f is irreducible in R[x] if and only if f is irreducible in F [x].

Proof. Suppose f is irreducible in F [x] and f = gh with g, h∈ R[x]. Then one of g and h is a unit in F [x]; say g and hence g is a constant. Thus C(f )≈ gC(h). Since C(f) is a unit in R, g must be a unit in R and hence in R[x]. Therefore, f is irreducible in R[x].

Conversely, if f is irreducible in R[x] and f = gh with g, h ∈ F [x]. We can write g = ab⁻¹g₁ with a, b ∈ R, b = 0 and g1 primitive in R[x] and h = cd⁻¹h₁ with c, d ∈ R, d = 0 and h1

primitive in R[x]. Consequently, bdf = acg₁h₁. Since f and g₁h₁ are primitive, bd≈ bdC(f) ≈ C(bdf) ≈ C(acg₁h₁)≈ acC(g₁h₁)≈ ac.

Thus bd and ac are associates and this implies that acb⁻¹d⁻¹ = α ∈ R is a unit. Hence f = αg₁h₁ in R[x]. By hypothesis, one of g₁, h₁ is a unit in R[x]; say g₁. Hence g₁ is a constant and so is g = ab⁻¹g₁. This implies that f is irreducible in F [x].
Exercise 11. Let f be a primitive polynomial in R[x]. Prove that f is prime in R[x] if and only if f is prime in F [x].

Theorem 3.13. If R is a unique factorization domain, then the polynomial ring R[x] is also a unique factorization domain.

Proof. Given f ∈ R[x], we can write f as f = C(f)f₁ with f₁ primitive in R[x]. Since C(f ) ∈ R and R is a unique factorization domain, if C(f) is not a unit, we can write C(f) as a product of finitely many irreducible elements in R. Theses elements are also irreducible in R[x]. Hence it is sufficient to show that every primitive polynomial of positive degree in R[x] can be written as a product of finitely many irreducible elements in R[x]. Suppose f is a primitive polynomial in R[x]. Since F [x] is a unique factorization domain (c.f. Example 3.8) which contains R[x], f = p₁· · · pn with each p_i irreducible in F [x]. Writing p_i = a_ib⁻¹_i q_i with a_i, b_i ∈ R, bi = 0 and qi primitive in R[x]. Clearly each q_i is irreducible in F [x] and hence is irreducible in R[x] by Lemma 3.12. Let a = a₁· · · an and b = b₁· · · bn. Then bf = aq₁· · · qn. Because C(f ) and C(q₁· · · qn) are units in R, it follows that a and b are associates in R.

Thus a = bu with u a unit in R. Therefore f = uq₁· · · qn with uq₁ and q₂, . . . , q_n irreducible in R[x].

To show the uniqueness, as in the proof of Theorem 3.9, we only have to show that every irreducible polynomial in R[x] is prime. Suppose f is irreducible in R[x]. If f ∈ R, then by R is a unique factorization domain, f is prime in R. If f| gh for some g, h ∈ R[x], then lf = gh for some l∈ R[x]. By Lemma 3.10, we have

f C(l)≈ C(lf) ≈ C(gh) ≈ C(g)C(h).

This implies that f| C(g)C(h) in R and hence f | C(g) or f | C(h). Therefore, f | g or f | h in R[x]. Therefore, f is prime in R[x]. Now suppose that f is a polynomial of positive degree in R[x]. f is irreducible in R[x] implies that f is a primitive polynomial in R[x]. Lemma 3.12 says that f is irreducible in F [x] and hence f is prime in F [x] because F [x] is a unique factorization domain. By Exercise 11, f is prime in R[x].
Corollary 3.14. If R is a unique factorization domain, then the polynomial ring over R in n indeterminates, R[x₁, . . . , x_n] is also a unique factorization domain.

Proof. By Theorem 3.13, R[x₁] is a unique factorization domain. Since R[x₁, . . . , x_n] = R[x₁, . . . , x_n−1][x_n], the proof is now completed by a routine inductive argument.

Department of Mathematics, National Taiwan Normal University, Taipei, Taiwan, R.O.C.

E-mail address: [email protected]