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Home , Domain, Euclidean domain, Integral domain, Principal ideal, Principal ideal domain, Principal ideal ring

Note. From now on all rings are commutative rings with identity 1 6= 0 unless stated otherwise.

27 Principal and Euclidean rings

27.1 Definition. If R is a and S is a subset of R then denote

hSi = the smallest ideal of R that contains S

We say that hSi is the ideal of R generated by the set S.

27.2 Note. We have

hSi = {b1a1 + . . . bkak | ai ∈ I, bi ∈ R, k ≥ 0}

27.3 Definition. An ideal I C R is finitely generated if I = ha1, . . . , ani for some a1, . . . , an ∈ R.

An ideal I C R is a if I = hai for some a ∈ R.

27.4 Definition. A ring R is a principal ideal (PID) if it is an domain (25.5) such that every ideal of R is a principal ideal.

27.5 Proposition. The ring of Z is a PID.

Proof. Let I C Z. If I = {0} then I = h0i, so I is a principal ideal. If I 6= {0} then let a be the smallest such that a > 0 and a ∈ I. We will show that I = hai.

110 Since a ∈ I we have hai ⊆ I. Conversely, if b ∈ I then we have

b = qa + r for some q, r ∈ Z, 0 ≤ r ≤ a − 1. This gives r = b − qa, so r ∈ I. Since a is the smallest positive element of I, this implies that r = 0. Therefore b = qa, and so b ∈ hai.

27.6 Proposition. If F is a field then F is a PID.

Proof. If I C F and 0 6= a ∈ I then for every b ∈ F we have b = (ba−1)a ∈ I

And so I = F. As a consequence the only ideals of F are {0} = h0i and F = h1i.

27.7 Proposition. If F is a field then the ring of F[x] is a PID.

Proof. Let I ∈ R. If I = {0} then I = h0i. Otherwise let 0 6= p(x) be a such that p(x) ∈ I and deg p(x) ≤ deg q(x) for all q(x) ∈ I − {0}. Check that I = hp(x)i.

27.8 Note. Z[x] is not a PID. E.g. the ideal h2, xi is not a principal ideal of Z[x] (check!).

27.9 Definition. A is an R equipped with a function N : R − {0} −→ N = {0, 1,... } such that

111 1) N(ab) ≥ N(a) for all a, b ∈ R − {0} 2) for any a, b ∈ R, a 6= 0 there exist q, r ∈ R such that

b = qa + r

and either r = 0 or N(r) < N(a). The function N is called the norm function on R.

27.10 Examples.

1) Z is a Euclidean domain with the norm function given by the .

2) Any field F is a Euclidean domain with N(a) = 0 for all a ∈ F − {0}.

3) If F is a field then F[x] is a Euclidean domain with N(p(x)) = deg p(x) for p(x) ∈ F[x], p(x) 6= 0.

Note: Z[x] is not a Euclidean domain with N(p(x)) = deg p(x). E. g. there are no q(x), r(x) ∈ Z[x] such that either r(x) = 0 or deg r(x) < 1 and that x = 2q(x) + r(x)

4) The ring of Gaussian integers is the Z[i] ⊆ C given by

Z[i] := {a + bi ∈ C | a, b ∈ Z}

Z[i] is a Euclidean domain with N(a + bi) = a2 + b2 = (a + bi)(a + bi)

(exercise).

5) Define √ √ Z[ −5] := {a + b 5i | a, b ∈ Z} √ Exercise: Z[ −5] is not a Euclidean domain.

112 27.11 Theorem. If R is a Euclidean domain then R is a PID.

Proof. Let ICR, I 6= {0}. Choose a ∈ I such that a 6= 0 and that N(a) ≤ N(b) for all b ∈ I − {0}. Check that hai = I.

27.12 Note.√ It is not true that every PID is a Euclidean domain. Take e.g. 1 19 α = 2 + 2 i and let Z[α] = {a + bα | a, b ∈ Z} Then Z[α] is a PID, but it is not a Euclidean domain. See J. C. Wilson, A that is not a euclidean ring, Magazine, 46 (1) (1973), 34-38. (Note: this link requires a JSTOR access)

113 28 Prime ideals and maximal ideals

28.1 Definition. Let R be a ring.

1) An ideal I C R is a if I 6= R and for any a, b ∈ R we have ab ∈ I iff either a ∈ I or b ∈ I

2) An ideal I C R is a if I 6= R and for any J C R such that I ⊆ J ⊆ R we have either J = I or J = R.

28.2 Examples.

1) The zero ideal {0} ∈ R is a prime ideal iff R is an integral domain, and it is a maximal ideal iff R is a field.

2) Recall that if I C Z then I = nZ for some n ≥ 0. Check:

(nZ is a prime ideal) iff (nZ is a maximal ideal) iff (n is a )

3) hxi C Z[x] is prime ideal (check!) but it is not a maximal ideal:

hxi ⊆ h2, xi C Z[x]

28.3 Proposition. Let I C R 1) The ideal I is a prime ideal iff R/I is an integral domain.

2) The ideal I is a maximal ideal iff R/I is a field.

28.4 Corollary. Every maximal ideal is a prime ideal.

114 Proof. If I C R is a maximal ideal then R/I is a field. In particular R/I is an integral domain, and so I is a prime ideal.

Proof of Proposition 28.3.

1) Assume that I C R is a prime ideal. For a + I, b + I ∈ R/I we have (a + I)(b + I) = ab + I

Thus if (a + I)(b + I) = 0 + I then ab + I = 0 + I i.e. ab ∈ I. Since I is a prime ideal we get that either a ∈ I (and so a + I = 0 + I) or b ∈ I (and so b + I = 0 + I). Therefore R/I is an integral domain.

The other implication follows from a similar argument.

2) Assume that I C R is a maximal ideal. Let a + I ∈ R/I, a + I 6= 0 + I. We want to show that there exists b + I ∈ R/I such that

(a + I)(b + I) = 1 + I

Take the ideal J = hai + I. We have

I ⊆ J ⊆ R and I 6= J (since a 6∈ I). Since I is a maximal ideal we must have J = R. In particular 1 ∈ J, so 1 = ab + c for some b ∈ R, c ∈ I. This gives

1 + I = (ab + c) + I = ab + I = (a + I)(b + I)

Conversely, assume that R/I is a field, and let J C R be an ideal such that I ⊆ J ⊆ R

115 We will show that either J = I or J = R.

Take the canonical epimorphism π : R → R/I. Since π(J) is an ideal of R/I (check!) and R/I is a field we have either π(J) = {0 + I} or π(J) = R/I. Also, since I ⊆ J, we have π−1(π(J)) = J. If follows that either J = π−1({0 + I}) = I or J = π−1(R/I) = R

28.5 Examples. 1) Since an ideal nZ of Z is prime (and maximal) iff n is a prime number therefore Z/nZ is an integral domain (and in fact a field) iff n is a prime number.

2) Take x2 + 1 ∈ R[x]. We have an epimorphism of rings f : R[x] −→ C f(p(x)) = p(i) Check: Ker(f) = hx2 + 1i. By the First Theorem (26.13) 2 ∼ 2 we get R[x]/hx + 1i = C. Since C is a field this shows that hx + 1i is a maximal ideal of R[x].

28.6 Note. For I,J C R define

IJ := {a1b1 + ... + akbk | ai ∈ I, bi ∈ J, k ≥ 0} Check: IJ is an ideal of R.

28.7 Proposition. Let I C R, I 6= R. The ideal I is a prime ideal iff for any ideals J1,J2 such that J1J2 ⊆ I we have either J1 ⊆ I or J2 ⊆ I.

Proof. Exercise.

116 29 Zorn’s Lemma and maximal ideals

Goal:

29.1 Theorem. If R is a ring, I C R, and I 6= R then there exists a maximal ideal J C R such that I ⊆ J.

29.2 Definition. A (or poset) is a set S equipped with a binary relation ≤ satisfying 1) x ≤ x for all x ∈ S (reflexivity) 2) if x ≤ y and y ≤ z then x ≤ z (transitivity) 3) if x ≤ y and y ≤ x then y = x (antisymmetry).

29.3 Example. If A is a set and S is the set of all subsets of A then S is a poset with ordering given by the inclusion of subsets.

29.4 Definition. A linearly ordered set is a poset (S, ≤) such that for any x, y ∈ S we have either x ≤ y or y ≤ x.

29.5 Definition. If (S, ≤) is a poset then an element x ∈ S is a maximal element of S if we have x ≤ y only for y = x.

29.6 Example. Let S be the set of all proper subsets of a set A ordered with respect to the inclusion. For every a ∈ A the set A − {a} is a maximal element of S.

117 29.7 Note. If (S, ≤) is a poset and T ⊆ S then T is also a poset with ordering inherited from S.

29.8 Definition. Let (S, ≤) is a poset and let T ⊆ S. An upper bound of T is an element x ∈ S such that y ≤ x for all y ∈ T .

29.9 Definition. If (S, ≤) is a poset. A chain in S is a subset T ⊆ S such that T is linearly ordered.

29.10 Zorn’s Lemma. If (S, ≤) is a non-empty poset such that every chain in S has an upper bound in S then S contains a maximal element.

Proof of Theorem 29.1. Let I 6= R be an ideal of R, and let S be the set of all ideals J C R such that I ⊆ J and J 6= R. Notice that S 6= ∅ since I ∈ S. The set S is a poset ordered with respect to inclusion of ideals. We will show that every chain in S has an upper bound in S. Let

T = {Ji}i∈A S be a chain in S. Check: JT := i∈A Ji is an ideal of R. Moreover, I ⊆ JT . Finally, we have JT 6= R. Indeed, otherwise 1 ∈ JT , and so 1 ∈ Ji for some i ∈ A. This would give Ji = R, which contradicts our assumptions.

It follows that JT ∈ S. Since Ji ⊆ JT for all i ∈ A, we obtain that JT is an upper bound of the chain T .

By Zorn’s Lemma (29.10) there is a maximal element J ∈ S. This means, in particular, that J is an ideal of R such I ⊆ J. Moreover, let K C R be any ideal such that K 6= R and J ⊆ K. We have I ⊆ J ⊆ K which means that K ∈ S. Maximality of J is S implies then that J = K. This shows that J is a maximal ideal of R.

118 29.11 Corollary. Every ring contains a maximal ideal.

Proof. If R is a ring then by Theorem 29.1 there exists a maximal ideal I C R such that I contains the zero ideal {0} C R.

119 30 Unique domains

Motivation:

30.1 Fundamental Theorem of Arithmetic. If n ∈ Z, n > 1 then n = p1p2 · ... · pk where p1, . . . , pk are primes. Moreover, this decomposition is unique up to re- ordering of factors.

Goal. Extend this to other rings.

30.2 Definition. Let R be an integral domain. An element a ∈ R is irreducible if a 6= 0, a is not a and if a = bc for some b, c ∈ R then either b or c is a unit.

30.3 Examples.

1) n ∈ Z is irreducible iff n = ±p where p is a prime number.

2) A field has no irreducible elements.

3) Take p(x) ∈ R[x], p(x) = x2 + 1. Then p(x) is irreducible in R[x].

4) Take p(x) ∈ C[x], p(x) = x2 + 1. Then p(x) is not irreducible in C[x]: p(x) = (x − i)(x + i)

30.4 Note. If a ∈ R is an and u ∈ R is a unit then ua is irreducible.

120 30.5 Definition. Let R be an integral domain. Elements a, b ∈ R are associates if a = ub for some unit u ∈ R. We write: a ∼ b.

30.6 Examples.

1) If m, n ∈ Z then m ∼ n iff m = ±n.

2) Check: units in R[x] are non-zero polynomials of degree 0. It follows that if p(x), q(x) ∈ R[x] then f(x) ∼ g(x) iff f(x) = ag(x) for some a ∈ R − {0}.

30.7 Definition. A unique factorization domain (UFD) is an integral domain R that satisfies the following conditions:

1) if a ∈ R is a non-zero, non-unit element then

a = b1 · ... · bk

for some irreducible elements b1, . . . , bk ∈ R

2) if b1, . . . , bk, c1, . . . , cl are irreducible elements such that

b1 · ... · bk = c1 · ... · cl

then k = l and for some permutation σ : {1, . . . , k} → {1, . . . , k} we have b1 ∼ cσ(1), ... , bk ∼ cσ(k).

30.8 Examples.

1) Z is a UFD by the Fundamental Theorem of Arithmetic (30.1).

2) If F is a field then F is a UFD since all non-zero elements of F are units.

121 √ Recall. Z[ −5] is the subring of C given by √ √ Z[ −5] = {a + b 5i | a, b ∈ Z}

√ 30.9 Proposition. Z[ −5] is not a UFD.

√ √ Proof. For a + b 5i ∈ Z[ −5] define √ √ √ 2 2 N(a + b 5i) = (a + b 5i)(a + b 5i) = a + 5b ∈ N Notice that 1) N(α) = 1 iff α = ±1 2) N(α) = 0 iff α = 0 √ 3) N(α) 6= 3 for all α ∈ Z[ −5] 4) N(αβ) = N(α)N(β)

√ Observation 1. The only units in Z[ −5] are 1 and −1. As a consequence for any α, β we have α ∼ β iff α = ±β.

√ Indeed, if α ∈ Z[ −5] is a unit then N(α)N(α−1) = N(αα−1) = N(1) = 1

Therefore N(α) = 1, and so α = ±1.

√ Observation 2. If α ∈ Z[ −5] is an element such that N(α) = 9 then α is irreducible.

Indeed, if α = ββ0 then

N(β)N(β0) = N(α) = 9

122 Therefore N(β) must be either 1 (and so β is a unit), 3 (impossible), or 9 (and then N(β0) = 1, i.e. β0 is a unit) .

√ Take 9 ∈ Z[ −5]. We have √ √ 3 · 3 = 9 = (2 + 5i)(2 − 5i) √ √ √ By Observation 2 the elements 3, 2 + 5i , 2 − 5i are irreducible in Z[ −5]. On the other hand by Observation 1 we obtain √ √ 3 6∼ (2 + 5i), 3 6∼ (2 − 5i) √ As a consequence 9 does not have a unique factorization in Z[ −5]

123