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G and T ∈ S. Prove that this is a group action. (d) Prove p \|\|OT \| for some T ∈ S. (e) Let {T1,..., Tu} be an orbit such that p \|u and H = {g ∈ G : gT1 = T1}. Prove that H is a subgroup of G and show that \|G\| = u\|H\|. (f) Show that pk divides \|H\| and pk ≤ \|H\|. (g) Show that \|H\| = \|OT \| ≤ pk; conclude that therefore pk = \|H\|. 26. Let G be a group. Prove that G = aba−1b−1 : a, b ∈ G is a normal subgroup of G and G/G is abelian. Find an example to show that {aba−1b−1 : a, b ∈ G} is not necessarily a group. A Project Order Number Order Number Order Number Order Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15??????????? 5?? 1 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 14 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 1 51 1? 1 14 1? 2 14 1? 1 4 * 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 2 1 52? 5??? 15 2? 2? 1 13 Table 15.1. Numbers of distinct groups G, \|G\| ≤ 60 242 CHAPTER 15 THE SYLOW THEOREMS The main objective of ﬁnite group theory is to classify all possible ﬁnite groups up to isomorphism. This problem is very diﬃcult even if we try to classify the groups of order less than or equal to 60. However, we can break the problem down into several intermediate problems. 1. Find all simple groups G ( \|G\| ≤ 60). Do not use the Odd Order Theorem unless you are prepared to prove it. 2. Find the number of distinct groups G, where the order of G is n for n = 1,..., 60. 3. Find the actual groups (up to isomorphism) for each n. This is a challenging project that requires a working knowledge of the group theory you have learned up to
this point. Even if you do not complete it, it will teach you a great deal about ﬁnite groups. You can use Table 15.2 as a guide. References and Suggested Readings [1] Edwards, H. “A Short History of the Fields Medal,” Mathematical Intelligencer 1 (1978), 127–29. [2] Feit, W. and Thompson, J. G. “Solvability of Groups of Odd Order,” Paciﬁc Journal of Mathematics 13 (1963), 775–1029. [3] Gallian, J. A. “The Search for Finite Simple Groups,” Mathematics Magazine 49(1976), 163–79. [4] Gorenstein, D. “Classifying the Finite Simple Groups,” Bulletin of the Amer- ican Mathematical Society 14 (1986), 1–98. [5] Gorenstein, D. Finite Groups. AMS Chelsea Publishing, Providence RI, 1968. [6] Gorenstein, D., Lyons, R., and Solomon, R. The Classiﬁcation of Finite Simple Groups. American Mathematical Society, Providence RI, 1994. Sage Sage will compute a single Sylow p-subgroup for each prime divisor p of the order of the group. Then, with conjugacy, all of the Sylow p-subgroups can be enumerated. It is also possible to compute the normalizer of a subgroup. 16 Rings Up to this point we have studied sets with a single binary operation satisfying certain axioms, but often we are more interested in working with sets that have two binary operations. For example, one of the most natural algebraic structures to study is the integers with the operations of addition and multiplication. These operations are related to one another by the distributive property. If we consider a set with two such related binary operations satisfying certain axioms, we have an algebraic structure called a ring. In a ring we add and multiply such elements as real numbers, complex numbers, matrices, and functions. 16.1 Rings A nonempty set R is a ring if it has two closed binary operations, addition and multiplication, satisfying the following conditions. 1. a + b = b + a for a, b ∈ R. 2. (a + b) + c = a + (b + c) for a, b, c �
� R. 3. There is an element 0 in R such that a + 0 = a for all a ∈ R. 4. For every element a ∈ R, there exists an element −a in R such that a + (−a) = 0. 5. (ab)c = a(bc) for a, b, c ∈ R. 6. For a, b, c ∈ R, a(b + c) = ab + ac (a + b)c = ac + bc. 243 244 CHAPTER 16 RINGS This last condition, the distributive axiom, relates the binary operations of addition and multiplication. Notice that the ﬁrst four axioms simply require that a ring be an abelian group under addition, so we could also have deﬁned a ring to be an abelian group (R, +) together with a second binary operation satisfying the ﬁfth and sixth conditions given above. If there is an element 1 ∈ R such that 1 = 0 and 1a = a1 = a for each element a ∈ R, we say that R is a ring with unity or identity. A ring R for which ab = ba for all a, b in R is called a commutative ring. A commutative ring R with identity is called an integral domain if, for every a, b ∈ R such that ab = 0, either a = 0 or b = 0. A division ring is a ring R, with an identity, in which every nonzero element in R is a unit; that is, for each a ∈ R with a = 0, there exists a unique element a−1 such that a−1a = aa−1 = 1. A commutative division ring is called a ﬁeld. The relationship among rings, integral domains, division rings, and ﬁelds is shown in Figure 16.1. Rings Commutative Rings Rings with Identity Integral Domains Division Rings Fields Figure 16.1. Types of rings Example 1. As we have mentioned previously, the integers form a ring. In fact, Z is an integral domain. Certainly if ab = 0 for two integers a and b, either a = 0 or b = 0. However, Z is not a ﬁeld. There is no integer that is the multiplicative inverse of 2, since 1/2 is not an integer. The only
integers with multiplicative inverses are 1 and −1. Example 2. Under the ordinary operations of addition and multiplication, all of the familiar number systems are rings: the rationals, Q; the real numbers, R; and the complex numbers, C. Each of these rings is a ﬁeld. 16.1 RINGS 245 Example 3. We can deﬁne the product of two elements a and b in Zn by ab (mod n). For instance, in Z12, 5 · 7 ≡ 11 (mod 12). This product makes the abelian group Zn into a ring. Certainly Zn is a commutative ring; however, it may fail to be an integral domain. If we consider 3 · 4 ≡ 0 (mod 12) in Z12, it is easy to see that a product of two nonzero elements in the ring can be equal to zero. A nonzero element a in a ring R is called a zero divisor if there is a nonzero element b in R such that ab = 0. In the previous example, 3 and 4 are zero divisors in Z12. Example 4. In calculus the continuous real-valued functions on an interval [a, b] form a commutative ring. We add or multiply two functions by adding or multiplying the values of the functions. If f (x) = x2 and g(x) = cos x, then (f + g)(x) = f (x) + g(x) = x2 + cos x and (f g)(x) = f (x)g(x) = x2 cos x. Example 5. The 2 × 2 matrices with entries in R form a ring under the usual operations of matrix addition and multiplication. This ring is noncommutative, since it is usually the case that AB = BA. Also, notice that we can have AB = 0 when neither A nor B is zero. Example 6. For an example of a noncommutative division ring, let 1 = 1 0 0 1 0 1 −i, where i2 = −1. These elements satisfy the following relations: i2 = j2 = k2 = −1 ij = k jk = i ki = j ji = −k kj = −i ik = −j. Let H consist of elements of the form a + bi + cj + dk, where a, b, c, d are real
numbers. Equivalently, H can be considered to be the set of all 2 × 2 matrices of the form α β −β α, 246 CHAPTER 16 RINGS where α = a + di and β = b + ci are complex numbers. We can deﬁne addition and multiplication on H either by the usual matrix operations or in terms of the generators 1, i, j, and k: (a1 + b1i + c1j + d1k) + (a2 + b2i + c2j + d2k) = (a1 + a2) + (b1 + b2)i + (c1 + c2)j + (d1 + d2)k and where (a1 + b1i + c1j + d1k)(a2 + b2i + c2j + d2k) = α + βi + γj + δk, α = a1a2 − b1b2 − c1c2 − d1d2 β = a1b2 + a1b1 + c1d2 − d1c2 γ = a1c2 − b1d2 + c1a2 − d1b2 δ = a1d2 + b1c2 − c1b2 − d1a2. Though multiplication looks complicated, it is actually a straightforward computation if we remember that we just add and multiply elements in H like polynomials and keep in mind the relationships between the generators i, j, and k. The ring H is called the ring of quaternions. To show that the quaternions are a division ring, we must be able to ﬁnd an inverse for each nonzero element. Notice that (a + bi + cj + dk)(a − bi − cj − dk) = a2 + b2 + c2 + d2. This element can be zero only if a, b, c, and d are all zero. So if a+bi+cj+dk = 0, (a + bi + cj + dk) a − bi − cj − dk a2 + b2 + c2 + d2 = 1. Proposition 16.1 Let R be a ring with a, b ∈ R. Then 1. a0 = 0a = 0; 2. a(−b) =
(−a)b = −ab; 3. (−a)(−b) = ab. 16.1 RINGS 247 Proof. To prove (1), observe that a0 = a(0 + 0) = a0 + a0; hence, a0 = 0. Similarly, 0a = 0. For (2), we have ab + a(−b) = a(b − b) = a0 = 0; consequently, −ab = a(−b). Similarly, −ab = (−a)b. Part (3) follows directly from (2) since (−a)(−b) = −(a(−b)) = −(−ab) = ab. Just as we have subgroups of groups, we have an analogous class of substructures for rings. A subring S of a ring R is a subset S of R such that S is also a ring under the inherited operations from R. Example 7. The ring nZ is a subring of Z. Notice that even though the original ring may have an identity, we do not require that its subring have an identity. We have the following chain of subrings: Z ⊂ Q ⊂ R ⊂ C. The following proposition gives us some easy criteria for determining whether or not a subset of a ring is indeed a subring. (We will leave the proof of this proposition as an exercise.) Proposition 16.2 Let R be a ring and S a subset of R. Then S is a subring of R if and only if the following conditions are satisﬁed. 1. S = ∅. 2. rs ∈ S for all r, s ∈ S. 3. r − s ∈ S for all r, s ∈ S. Example 8. Let R = M2(R) be the ring of 2 × 2 matrices with entries in R. If T is the set of upper triangular matrices in R; i.e., T = a b 0 c : a, b, c ∈ R, then T is a subring of R. If A = a b 0 c and B = a 0 b c are in T, then clearly A − B is also in T. Also, aa ab + bc AB = 0 cc is in T. 248 CHAPTER 16 RINGS 16.2 Integral Domains and Fields Let us brieﬂy recall some deﬁn
itions. If R is a ring and r is a nonzero element in R, then r is said to be a zero divisor if there is some nonzero element s ∈ R such that rs = 0. A commutative ring with identity is said to be an integral domain if it has no zero divisors. If an element a in a ring R with identity has a multiplicative inverse, we say that a is a unit. If every nonzero element in a ring R is a unit, then R is called a division ring. A commutative division ring is called a ﬁeld. Example 9. If i2 = −1, then the set Z[i] = {m + ni : m, n ∈ Z} forms a ring known as the Gaussian integers. It is easily seen that the Gaussian integers are a subring of the complex numbers since they are closed under addition and multiplication. Let α = a + bi be a unit in Z[i]. Then α = a − bi is also a unit since if αβ = 1, then αβ = 1. If β = c + di, then 1 = αβαβ = (a2 + b2)(c2 + d2). Therefore, a2 + b2 must either be 1 or −1; or, equivalently, a + bi = ±1 or a + bi = ±i. Therefore, units of this ring are ±1 and ±i; hence, the Gaussian integers are not a ﬁeld. We will leave it as an exercise to prove that the Gaussian integers are an integral domain. Example 10. The set of matrices, with entries in Z2 forms a ﬁeld. Example 11. The set Q( of an element a + b 2 in Q( √ 2 ) = {a + b √ 2 ) is √ √ 2 : a, b ∈ Q} is a ﬁeld. The inverse a a2 − 2b2 + −b a2 − 2b2 √ 2. We have the following alternative characterization of integral domains. Proposition 16.3 (Cancellation Law) Let D be a commutative ring with identity. Then D is an integral domain if and only if for all nonzero elements a ∈ D with ab = ac, we have b = c. 16.2 INTEGRAL DOMAINS AND FIELDS
249 Proof. Let D be an integral domain. Then D has no zero divisors. Let ab = ac with a = 0. Then a(b − c) = 0. Hence, b − c = 0 and b = c. Conversely, let us suppose that cancellation is possible in D. That is, suppose that ab = ac implies b = c. Let ab = 0. If a = 0, then ab = a0 or b = 0. Therefore, a cannot be a zero divisor. The following surprising theorem is due to Wedderburn. Theorem 16.4 Every ﬁnite integral domain is a ﬁeld. Proof. Let D be a ﬁnite integral domain and D∗ be the set of nonzero elements of D. We must show that every element in D∗ has an inverse. For each a ∈ D∗ we can deﬁne a map λa : D∗ → D∗ by λa(d) = ad. This map makes sense, because if a = 0 and d = 0, then ad = 0. The map λa is one-to-one, since for d1, d2 ∈ D∗, ad1 = λa(d1) = λa(d2) = ad2 implies d1 = d2 by left cancellation. Since D∗ is a ﬁnite set, the map λa must also be onto; hence, for some d ∈ D∗, λa(d) = ad = 1. Therefore, a has a left inverse. Since D is commutative, d must also be a right inverse for a. Consequently, D is a ﬁeld. For any nonnegative integer n and any element r in a ring R we write r + · · · + r (n times) as nr. We deﬁne the characteristic of a ring R to be the least positive integer n such that nr = 0 for all r ∈ R. If no such integer exists, then the characteristic of R is deﬁned to be 0. Example 12. For every prime p, Zp is a ﬁeld of characteristic p. By Proposition 3.1, every nonzero element in Zp has an inverse; hence, Zp is a ﬁeld. If a is
any nonzero element in the ﬁeld, then pa = 0, since the order of any nonzero element in the abelian group Zp is p. Lemma 16.5 Let R be a ring with identity. If 1 has order n, then the characteristic of R is n. Proof. If 1 has order n, then n is the least positive integer such that n1 = 0. Thus, for all r ∈ R, nr = n(1r) = (n1)r = 0r = 0. On the other hand, if no positive n exists such that n1 = 0, then the characteristic of R is zero. 250 CHAPTER 16 RINGS Theorem 16.6 The characteristic of an integral domain is either prime or zero. Proof. Let D be an integral domain and suppose that the characteristic of D is n with n = 0. If n is not prime, then n = ab, where 1 < a < n and 1 < b < n. By Lemma 16.5, we need only consider the case n1 = 0. Since 0 = n1 = (ab)1 = (a1)(b1) and there are no zero divisors in D, either a1 = 0 or b1 = 0. Hence, the characteristic of D must be less than n, which is a contradiction. Therefore, n must be prime. 16.3 Ring Homomorphisms and Ideals In the study of groups, a homomorphism is a map that preserves the operation of the group. Similarly, a homomorphism between rings preserves the operations of addition and multiplication in the ring. More speciﬁcally, if R and S are rings, then a ring homomorphism is a map φ : R → S satisfying φ(a + b) = φ(a) + φ(b) φ(ab) = φ(a)φ(b) for all a, b ∈ R. If φ : R → S is a one-to-one and onto homomorphism, then φ is called an isomorphism of rings. The set of elements that a ring homomorphism maps to 0 plays a fundamental role in the theory of rings. For any ring homomorphism φ : R → S, we deﬁne the kernel of a ring homomorphism to be the set ker φ = {r ∈ R
: φ(r) = 0}. Example 13. For any integer n we can deﬁne a ring homomorphism φ : Z → Zn by a → a (mod n). This is indeed a ring homomorphism, since and φ(a + b) = (a + b) (mod n) = a (mod n) + b (mod n) = φ(a) + φ(b) φ(ab) = ab (mod n) = a (mod n) · b (mod n) = φ(a)φ(b). 16.3 RING HOMOMORPHISMS AND IDEALS The kernel of the homomorphism φ is nZ. 251 Example 14. Let C[a, b] be the ring of continuous real-valued functions on an interval [a, b] as in Example 4. For a ﬁxed α ∈ [a, b], we can deﬁne a ring homomorphism φα : C[a, b] → R by φα(f ) = f (α). This is a ring homomorphism since φα(f + g) = (f + g)(α) = f (α) + g(α) = φα(f ) + φα(g) φα(f g) = (f g)(α) = f (α)g(α) = φα(f )φα(g). Ring homomorphisms of the type φα are called evaluation homomor- phisms. In the next proposition we will examine some fundamental properties of ring homomorphisms. The proof of the proposition is left as an exercise. Proposition 16.7 Let φ : R → S be a ring homomorphism. 1. If R is a commutative ring, then φ(R) is a commutative ring. 2. φ(0) = 0. 3. Let 1R and 1S be the identities for R and S, respectively. If φ is onto, then φ(1R) = 1S. 4. If R is a ﬁeld and φ(R) = 0, then φ(R) is a ﬁeld. In group theory we found that normal subgroups play a special role. These subgroups have
nice characteristics that make them more interesting to study than arbitrary subgroups. In ring theory the objects corresponding to normal subgroups are a special class of subrings called ideals. An ideal in a ring R is a subring I of R such that if a is in I and r is in R, then both ar and ra are in I; that is, rI ⊂ I and Ir ⊂ I for all r ∈ R. Example 15. Every ring R has at least two ideals, {0} and R. These ideals are called the trivial ideals. Let R be a ring with identity and suppose that I is an ideal in R such that 1 is in I. Since for any r ∈ R, r1 = r ∈ I by the deﬁnition of an ideal, I = R. Example 16. If a is any element in a commutative ring R with identity, then the set a = {ar : r ∈ R} 252 CHAPTER 16 RINGS is an ideal in R. Certainly, a is nonempty since both 0 = a0 and a = a1 are in a. The sum of two elements in a is again in a since ar +ar = a(r +r). The inverse of ar is −ar = a(−r) ∈ a. Finally, if we multiply an element ar ∈ a by an arbitrary element s ∈ R, we have s(ar) = a(sr). Therefore, a satisﬁes the deﬁnition of an ideal. If R is a commutative ring with identity, then an ideal of the form a = {ar : r ∈ R} is called a principal ideal. Theorem 16.8 Every ideal in the ring of integers Z is a principal ideal. Proof. The zero ideal {0} is a principal ideal since 0 = {0}. If I is any nonzero ideal in Z, then I must contain some positive integer m. There exists at least one such positive integer n in I by the Principle of Well-Ordering. Now let a be any element in I. Using the division algorithm, we know that there exist integers q and r such that a = nq + r where 0 ≤ r < n. This equation tells us that r = a − nq ∈ I, but r must be 0 since n is the least positive element in I. Therefore, a = nq and I = n
. Example 17. The set nZ is ideal in the ring of integers. If na is in nZ and b is in Z, then nab is in nZ as required. In fact, by Theorem 16.8, these are the only ideals of Z. Proposition 16.9 The kernel of any ring homomorphism φ : R → S is an ideal in R. Proof. We know from group theory that ker φ is an additive subgroup of R. Suppose that r ∈ R and a ∈ ker φ. Then we must show that ar and ra are in ker φ. However, and φ(ar) = φ(a)φ(r) = 0φ(r) = 0 φ(ra) = φ(r)φ(a) = φ(r)0 = 0. Remark. In our deﬁnition of an ideal we have required that rI ⊂ I and Ir ⊂ I for all r ∈ R. Such ideals are sometimes referred to as two-sided 16.3 RING HOMOMORPHISMS AND IDEALS 253 ideals. We can also consider one-sided ideals; that is, we may require only that either rI ⊂ I or Ir ⊂ I for r ∈ R hold but not both. Such ideals are called left ideals and right ideals, respectively. Of course, in a commutative ring any ideal must be two-sided. In this text we will concentrate on two-sided ideals. Theorem 16.10 Let I be an ideal of R. The factor group R/I is a ring with multiplication deﬁned by (r + I)(s + I) = rs + I. Proof. We already know that R/I is an abelian group under addition. Let r+I and s+I be in R/I. We must show that the product (r+I)(s+I) = rs+I is independent of the choice of coset; that is, if r ∈ r + I and s ∈ s + I, then rs must be in rs + I. Since r ∈ r + I, there exists an element a in I such that r = r + a. Similarly, there exists a b ∈ I such that s = s + b. Notice that rs = (r + a)(s + b)
= rs + as + rb + ab and as + rb + ab ∈ I since I is an ideal; consequently, rs ∈ rs + I. We will leave as an exercise the veriﬁcation of the associative law for multiplication and the distributive laws. The ring R/I in Theorem 16.10 is called the factor or quotient ring. Just as with group homomorphisms and normal subgroups, there is a relationship between ring homomorphisms and ideals. Theorem 16.11 Let I be an ideal of R. The map ψ : R → R/I deﬁned by ψ(r) = r + I is a ring homomorphism of R onto R/I with kernel I. Proof. Certainly ψ : R → R/I is a surjective abelian group homomorphism. It remains to show that ψ works correctly under ring multiplication. Let r and s be in R. Then ψ(r)ψ(s) = (r + I)(s + I) = rs + I = ψ(rs), which completes the proof of the theorem. The map ψ : R → R/I is often called the natural or canonical homomorphism. In ring theory we have isomorphism theorems relating ideals and ring homomorphisms similar to the isomorphism theorems for groups that relate normal subgroups and homomorphisms in Chapter 11. We will prove only the First Isomorphism Theorem for rings in this chapter and leave the proofs of the other two theorems as exercises. All of the proofs are similar to the proofs of the isomorphism theorems for groups. 254 CHAPTER 16 RINGS Theorem 16.12 (First Isomorphism Theorem) Let φ : R → S be a ring homomorphism. Then ker φ is an ideal of R. If ψ : R → R/ ker φ is the canonical homomorphism, then there exists a unique isomorphism η : R/ ker φ → φ(R) such that φ = ηψ. Proof. Let K = ker φ. By the First Isomorphism Theorem for groups, there exists a well-deﬁned group homomorphism η : R/K → ψ(R) deﬁned by η(r + K) = ψ
(r) for the additive abelian groups R and R/K. To show that this is a ring homomorphism, we need only show that η((r + K)(s + K)) = η(r + K)η(s + K); but η((r + K)(s + K)) = η(rs + K) = ψ(rs) = ψ(r)ψ(s) = η(r + K)η(s + K). Theorem 16.13 (Second Isomorphism Theorem) Let I be a subring of a ring R and J an ideal of R. Then I ∩ J is an ideal of I and I/I ∩ J ∼= (I + J)/J. Theorem 16.14 (Third Isomorphism Theorem) Let R be a ring and I and J be ideals of R where J ⊂ I. Then R/I ∼= R/J I/J. Theorem 16.15 (Correspondence Theorem) Let I be an ideal of a ring R. Then S → S/I is a one-to-one correspondence between the set of subrings S containing I and the set of subrings of R/I. Furthermore, the ideals of R containing I correspond to ideals of R/I. 16.4 Maximal and Prime Ideals In this particular section we are especially interested in certain ideals of commutative rings. These ideals give us special types of factor rings. More speciﬁcally, we would like to characterize those ideals I of a commutative ring R such that R/I is an integral domain or a ﬁeld. 16.4 MAXIMAL AND PRIME IDEALS 255 A proper ideal M of a ring R is a maximal ideal of R if the ideal M is not a proper subset of any ideal of R except R itself. That is, M is a maximal ideal if for any ideal I properly containing M, I = R. The following theorem completely characterizes maximal ideals for commutative rings with identity in terms of their corresponding factor rings. Theorem 16.16 Let R be a commutative ring with identity and M an ideal in R. Then M is a maximal ideal of R if and only if R/M is a ﬁeld. Proof. Let M be a maximal ideal in R. If R is a commut
ative ring, then R/M must also be a commutative ring. Clearly, 1 + M acts as an identity for R/M. We must also show that every nonzero element in R/M has an inverse. If a + M is a nonzero element in R/M, then a /∈ M. Deﬁne I to be the set {ra + m : r ∈ R and m ∈ M }. We will show that I is an ideal in R. The set I is nonempty since 0a + 0 = 0 is in I. If r1a + m1 and r2a + m2 are two elements in I, then (r1a + m1) − (r2a + m2) = (r1 − r2)a + (m1 − m2) is in I. Also, for any r ∈ R it is true that rI ⊂ I; hence, I is closed under multiplication and satisﬁes the necessary conditions to be an ideal. Therefore, by Proposition 16.2 and the deﬁnition of an ideal, I is an ideal properly containing M. Since M is a maximal ideal, I = R; consequently, by the deﬁnition of I there must be an m in M and a b in R such that 1 = ab + m. Therefore, 1 + M = ab + M = ba + M = (a + M )(b + M ). Conversely, suppose that M is an ideal and R/M is a ﬁeld. Since R/M is a ﬁeld, it must contain at least two elements: 0 + M = M and 1 + M. Hence, M is a proper ideal of R. Let I be any ideal properly containing M. We need to show that I = R. Choose a in I but not in M. Since a + M is a nonzero element in a ﬁeld, there exists an element b + M in R/M such that (a + M )(b + M ) = ab + M = 1 + M. Consequently, there exists an element m ∈ M such that ab + m = 1 and 1 is in I. Therefore, r1 = r ∈ I for all r ∈ R. Consequently, I = R. Example 18. Let pZ be an ideal in Z, where p
is prime. Then pZ is a maximal ideal since Z/pZ ∼= Zp is a ﬁeld. An ideal P in a commutative ring R is called a prime ideal if whenever ab ∈ P, then either a ∈ P or b ∈ P. Example 19. It is easy to check that the set P = {0, 2, 4, 6, 8, 10} is an ideal in Z12. This ideal is prime. In fact, it is a maximal ideal. 256 CHAPTER 16 RINGS Proposition 16.17 Let R be a commutative ring with identity. Then P is a prime ideal in R if and only if R/P is an integral domain. Proof. First let us assume that P is an ideal in R and R/P is an integral domain. Suppose that ab ∈ P. If a + P and b + P are two elements of R/P such that (a + P )(, then either a + P = P or b + P = P. This means that either a is in P or b is in P, which shows that P must be prime. Conversely, suppose that P is prime and (a + P )(b + P ) = ab + P = 0 + P = P. Then ab ∈ P. If a /∈ P, then b must be in P by the deﬁnition of a prime ideal; hence, b + P = 0 + P and R/P is an integral domain. Example 20. Every ideal in Z is of the form nZ. The factor ring Z/nZ ∼= Zn is an integral domain only when n is prime. It is actually a ﬁeld. Hence, the nonzero prime ideals in Z are the ideals pZ, where p is prime. This example really justiﬁes the use of the word “prime” in our deﬁnition of prime ideals. Since every ﬁeld is an integral domain, we have the following corollary. Corollary 16.18 Every maximal ideal in a commutative ring with identity is also a prime ideal. Historical Note Amalie Emmy Noether, one of the outstanding mathematicians of this century, was born in Erlangen, Germany in 1882. She was the daughter of Max Noether (1844–1921), a distinguished mathematician at the University of Erlang
en. Together with Paul Gordon (1837–1912), Emmy Noether’s father strongly inﬂuenced her early education. She entered the University of Erlangen at the age of 18. Although women had been admitted to universities in England, France, and Italy for decades, there was great resistance to their presence at universities in Germany. Noether was one of only two women among the university’s 986 students. After completing her doctorate under Gordon in 1907, she continued to do research at Erlangen, occasionally lecturing when her father was ill. Noether went to G¨ottingen to study in 1916. David Hilbert and Felix Klein tried unsuccessfully to secure her an appointment at G¨ottingen. Some of the faculty objected to women lecturers, saying, “What will our soldiers think when they return to the university and are expected to learn at the feet of a woman?” Hilbert, annoyed 16.5 AN APPLICATION TO SOFTWARE DESIGN 257 at the question, responded, “Meine Herren, I do not see that the sex of a candidate is an argument against her admission as a Privatdozent. After all, the Senate is not a bathhouse.” At the end of World War I, attitudes changed and conditions greatly improved for women. After Noether passed her habilitation examination in 1919, she was given a title and was paid a small sum for her lectures. In 1922, Noether became a Privatdozent at G¨ottingen. Over the next 11 years she used axiomatic methods to develop an abstract theory of rings and ideals. Though she was not good at lecturing, Noether was an inspiring teacher. One of her many students was B. L. van der Waerden, author of the ﬁrst text treating abstract algebra from a modern point of view. Some of the other mathematicians Noether inﬂuenced or closely worked with were Alexandroﬀ, Artin, Brauer, Courant, Hasse, Hopf, Pontryagin, von Neumann, and Weyl. One of the high points of her career was an invitation to address the International Congress of Mathematicians in Zurich in 1932. In spite of all the recognition she received from her colleagues, Noether’s abilities were never recognized as they should have been during her lifetime. She was never promoted to full professor by the Prussian academic
bureaucracy. In 1933, Noether, a Jew, was banned from participation in all academic activities in Germany. She emigrated to the United States, took a position at Bryn Mawr College, and became a member of the Institute for Advanced Study at Princeton. Noether died suddenly on April 14, 1935. After her death she was eulogized by such notable scientists as Albert Einstein. 16.5 An Application to Software Design The Chinese Remainder Theorem is a result from elementary number theory about the solution of systems of simultaneous congruences. The Chinese mathematician Sun-ts¨ı wrote about the theorem in the ﬁrst century A.D. This theorem has some interesting consequences in the design of software for parallel processors. Lemma 16.19 Let m and n be positive integers such that gcd(m, n) = 1. Then for a, b ∈ Z the system x ≡ a (mod m) x ≡ b (mod n) has a solution. If x1 and x2 are two solutions of the system, then x1 ≡ x2 (mod mn). Proof. The equation x ≡ a (mod m) has a solution since a + km satisﬁes the equation for all k ∈ Z. We must show that there exists an integer k1 such that a + k1m ≡ b (mod n). 258 CHAPTER 16 RINGS This is equivalent to showing that k1m ≡ (b − a) (mod n) has a solution for k1. Since m and n are relatively prime, there exist integers s and t such that ms + nt = 1. Consequently, or (b − a)ms = (b − a) − (b − a)nt, [(b − a)s]m ≡ (b − a) (mod n). Now let k1 = (b − a)s. To show that any two solutions are congruent modulo mn, let c1 and c2 be two solutions of the system. That is, for i = 1, 2. Then ci ≡ a (mod m) (mod n) ci ≡ b c2 ≡ c1 c2 ≡ c1 (mod m) (mod n). Therefore, both m and n divide c1 − c2. Consequently, c2 ≡ c1 (mod mn). Example 21. Let us solve the system x ≡ 3 (mod 4) x ≡ 4 (mod 5
). Using the Euclidean algorithm, we can ﬁnd integers s and t such that 4s + 5t = 1. Two such integers are s = −1 and t = 1. Consequently, x = a + k1m = 3 + 4k1 = 3 + 4[(5 − 4)4] = 19. 16.5 AN APPLICATION TO SOFTWARE DESIGN 259 Theorem 16.20 (Chinese Remainder Theorem) Let n1, n2,..., nk be positive integers such that gcd(ni, nj) = 1 for i = j. Then for any integers a1,..., ak, the system x ≡ a1 x ≡ a2... x ≡ ak (mod n1) (mod n2) (mod nk) has a solution. Furthermore, any two solutions of the system are congruent modulo n1n2 · · · nk. Proof. We will use mathematical induction on the number of equations in the system. If there are k = 2 equations, then the theorem is true by Lemma 16.19. Now suppose that the result is true for a system of k equations or less and that we wish to ﬁnd a solution of x ≡ a1 x ≡ a2... x ≡ ak+1 (mod n1) (mod n2) (mod nk+1). Considering the ﬁrst k equations, there exists a solution that is unique modulo n1 · · · nk, say a. Since n1 · · · nk and nk+1 are relatively prime, the system x ≡ a (mod n1 · · · nk) (mod nk+1) x ≡ ak+1 has a solution that is unique modulo n1... nk+1 by the lemma. Example 22. Let us solve the system x ≡ 3 (mod 4) x ≡ 4 (mod 5) x ≡ 1 (mod 9) x ≡ 5 (mod 7). 260 CHAPTER 16 RINGS From Example 21 we know that 19 is a solution of the ﬁrst two congruences and any other solution of the system is congruent to 19 (mod 20). Hence, we can reduce the system to a system of three congruences: x ≡ 19 (mod 20) x ≡ 1 (mod 9) x ≡ 5 (mod 7). Solving the next two equations,
we can reduce the system to x ≡ 19 (mod 180) x ≡ 5 (mod 7). Solving this last system, we ﬁnd that 19 is a solution for the system that is unique up to modulo 1260. One interesting application of the Chinese Remainder Theorem in the design of computer software is that the theorem allows us to break up a calculation involving large integers into several less formidable calculations. Most computers will handle integer calculations only up to a certain size. For example, the largest integer available on many workstations is 231 − 1 = 2,147,483,647. Special software is required for calculations involving larger integers which cannot be added directly by the machine. However, by using the Chinese Remainder Theorem we can break down large integer additions and multiplications into calculations that the computer can handle directly. This is especially useful on parallel processing computers which have the ability to run several programs concurrently. Most computers have a single central processing unit (CPU), which can only add two numbers at a time. To add a list of ten numbers, the CPU must do nine additions in sequence. However, a parallel processing computer has more than one CPU. A computer with 10 CPUs, for example, can perform 10 diﬀerent additions at the same time. If we can take a large integer and break it down into parts, sending each part to a diﬀerent CPU, then by performing several additions or multiplications simultaneously on those parts, we can work with an integer that the computer would not be able to handle as a whole. Example 23. Suppose that we wish to multiply 2134 by 1531. We will use the integers 95, 97, 98, and 99 because they are relatively prime. We can EXERCISES 261 break down each integer into four parts: and 2134 ≡ 44 (mod 95) 2134 ≡ 0 (mod 97) 2134 ≡ 76 (mod 98) 2134 ≡ 55 (mod 99) 1531 ≡ 11 (mod 95) 1531 ≡ 76 (mod 97) 1531 ≡ 61 (mod 98) 1531 ≡ 46 (mod 99). Multiplying the corresponding equations, we obtain 2134 · 1531 ≡ 44 · 11 ≡ 9 (mod 95) 2134 · 1531 ≡ 0 · 76 ≡ 0 (mod 97) 2134 · 1531 ≡ 76 · 61 ≡ 30 (mod 98) 2134 · 1531 ≡ 55 · 46 ≡ 55 (mod 99). Each of these four computations can be sent
to a diﬀerent processor if our computer has several CPUs. By the above calculation, we know that 2134 · 1531 is a solution of the system x ≡ 9 (mod 95) x ≡ 0 (mod 97) x ≡ 30 (mod 98) x ≡ 55 (mod 99). The Chinese Remainder Theorem tells us that solutions are unique up to modulo 95 · 97 · 98 · 99 = 89,403,930. Solving this system of congruences for x tells us that 2134 · 1531 = 3,267,154. The conversion of the computation into the four subcomputations will take some computing time. In addition, solving the system of congruences can also take considerable time. However, if we have many computations to be performed on a particular set of numbers, it makes sense to transform the problem as we have done above and to perform the necessary calculations simultaneously. 262 Exercises CHAPTER 16 RINGS 1. Which of the following sets are rings with respect to the usual operations of addition and multiplication? If the set is a ring, is it also a ﬁeld? 2, √ √ √ 2 ) = {a + b (a) 7Z (b) Z18 √ (c) Q( (d) Q( (e) Z[ √ (f) R = {a + b 3 (g) Z[i] = {a + bi : a, b ∈ Z and i2 = −1} √ √ (h) Q( 3 3 ) = {a + b 3 2 : a, b ∈ Q} √ 2 + c 3 : a, b ∈ Q} 3 : a, b ∈ Za + b 3 ] = {, b, c ∈ Q} 6 : a, b, c, d ∈ Q} 2. Let R be the ring of 2 × 2 matrices of the form a b 0 0, where a, b ∈ R. Show that although R is a ring that has no identity, we can ﬁnd a subring S of R with an identity. 3. List or characterize all of the units in each of the following rings. (a) Z10 (b) Z12 (c) Z7 (d) M2(Z), the 2 × 2 matrices with entries in Z (e) M2(Z2), the 2 × 2 mat
rices with entries in Z2 4. Find all of the ideals in each of the following rings. Which of these ideals are maximal and which are prime? (a) Z18 (b) Z25 (c) M2(R), the 2 × 2 matrices with entries in R (d) M2(Z), the 2 × 2 matrices with entries in Z (e) Q 5. For each of the following rings R with ideal I, give an addition table and a multiplication table for R/I. (a) R = Z and I = 6Z (b) R = Z12 and I = {0, 3, 6, 9} EXERCISES 263 6. Find all homomorphisms φ : Z/6Z → Z/15Z. 7. Prove that R is not isomorphic to C. 8. Prove or disprove: The ring Q( 2 ) = {a + b the ring Q( 3 ) = {a + b 3 : a, b ∈ Q}. √ √ √ √ 2 : a, b ∈ Q} is isomorphic to 9. What is the characteristic of the ﬁeld formed by the set of matrices with entries in Z2? 10. Deﬁne a map φ : C → M2(R) by φ(a + bi) = a b −b a. Show that φ is an isomorphism of C with its image in M2(R). 11. Prove that the Gaussian integers, Z[i], are an integral domain. √ √ 12. Prove that Z[ 3 i] = {a + b 3 i : a, b ∈ Z} is an integral domain. 13. Solve each of the following systems of congruences. (a) (b) x ≡ 2 (mod 5) x ≡ 6 (mod 11) x ≡ 3 (mod 7) x ≡ 0 (mod 8) x ≡ 5 (mod 15) (c) (d) x ≡ 2 (mod 4) x ≡ 4 (mod 7) x ≡ 7 (mod 9) x ≡ 5 (mod 11) x ≡ 3 (mod 5) x ≡ 0 (mod 8) x ≡ 1 (mod 11) x ≡ 5 (mod 13) 14. Use the method of parallel computation outlined in the text to calculate 2234 + 4121 by dividing the calculation into four separate
additions modulo 95, 97, 98, and 99. 15. Explain why the method of parallel computation outlined in the text fails for 2134 · 1531 if we attempt to break the calculation down into two smaller calculations modulo 98 and 99. 264 CHAPTER 16 RINGS 16. If R is a ﬁeld, show that the only two ideals of R are {0} and R itself. 17. Let a be any element in a ring R with identity. Show that (−1)a = −a. 18. Let φ : R → S be a ring homomorphism. Prove each of the following statements. (a) If R is a commutative ring, then φ(R) is a commutative ring. (b) φ(0) = 0. (c) Let 1R and 1S be the identities for R and S, respectively. If φ is onto, then φ(1R) = 1S. (d) If R is a ﬁeld and φ(R) = 0, then φ(R) is a ﬁeld. 19. Prove that the associative law for multiplication and the distributive laws hold in R/I. 20. Prove the Second Isomorphism Theorem for rings: Let I be a subring of a ring R and J an ideal in R. Then I ∩ J is an ideal in I and I/I ∩ J ∼= I + J/J. 21. Prove the Third Isomorphism Theorem for rings: Let R be a ring and I and J be ideals of R, where J ⊂ I. Then R/I ∼= R/J I/J. 22. Prove the Correspondence Theorem: Let I be an ideal of a ring R. Then S → S/I is a one-to-one correspondence between the set of subrings S containing I and the set of subrings of R/I. Furthermore, the ideals of R correspond to ideals of R/I. 23. Let R be a ring and S a subset of R. Show that S is a subring of R if and only if each of the following conditions is satisﬁed. (a) S = ∅. (b) rs ∈ S for all r, s ∈ S. (c) r − s ∈ S for all r, s
∈ S. 24. Let R be a ring with a collection of subrings {Rα}. Prove that Rα is a subring of R. Give an example to show that the union of two subrings cannot be a subring. 25. Let {Iα}α∈A be a collection of ideals in a ring R. Prove that α∈A Iα is also an ideal in R. Give an example to show that if I1 and I2 are ideals in R, then I1 ∪ I2 may not be an ideal. 26. Let R be an integral domain. Show that if the only ideals in R are {0} and R itself, R must be a ﬁeld. EXERCISES 265 27. Let R be a commutative ring. An element a in R is nilpotent if an = 0 for some positive integer n. Show that the set of all nilpotent elements forms an ideal in R. 28. A ring R is a Boolean ring if for every a ∈ R, a2 = a. Show that every Boolean ring is a commutative ring. 29. Let R be a ring, where a3 = a for all a ∈ R. Prove that R must be a commutative ring. 30. Let R be a ring with identity 1R and S a subring of R with identity 1S. Prove or disprove that 1R = 1S. 31. If we do not require the identity of a ring to be distinct from 0, we will not have a very interesting mathematical structure. Let R be a ring such that 1 = 0. Prove that R = {0}. 32. Let S be a subset of a ring R. Prove that there is a subring R of R that contains S. 33. Let R be a ring. Deﬁne the center of R to be Z(R) = {a ∈ R : ar = ra for all r ∈ R }. Prove that Z(R) is a commutative subring of R. 34. Let p be prime. Prove that Z(p) = {a/b : a, b ∈ Z and gcd(b, p) = 1} is a ring. The ring Z(p) is called the ring of integers localized at p. 35. Prove or disprove: Every ﬁnite integral domain
is isomorphic to Zp. 36. Let R be a ring with identity. (a) Let u be a unit in R. Deﬁne a map iu : R → R by r → uru−1. Prove that iu is an automorphism of R. Such an automorphism of R is called an inner automorphism of R. Denote the set of all inner automorphisms of R by Inn(R). (b) Denote the set of all automorphisms of R by Aut(R). Prove that Inn(R) is a normal subgroup of Aut(R). (c) Let U (R) be the group of units in R. Prove that the map φ : U (R) → Inn(R) deﬁned by u → iu is a homomorphism. Determine the kernel of φ. (d) Compute Aut(Z), Inn(Z), and U (Z). 37. Let R and S be arbitrary rings. Show that their Cartesian product is a ring if we deﬁne addition and multiplication in R × S by 266 CHAPTER 16 RINGS (a) (r, s) + (r, s) = (r + r, s + s) (b) (r, s)(r, s) = (rr, ss) 38. An element x in a ring is called an idempotent if x2 = x. Prove that the only idempotents in an integral domain are 0 and 1. Find a ring with a idempotent x not equal to 0 or 1. 39. Let gcd(a, n) = d and gcd(b, d) = 1. Prove that ax ≡ b (mod n) does not have a solution. 40. The Chinese Remainder Theorem for Rings. Let R be a ring and I and J be ideals in R such that I + J = R. (a) Show that for any r and s in R, the system of equations x ≡ r (mod I) x ≡ s (mod J) has a solution. (b) In addition, prove that any two solutions of the system are congruent modulo I ∩ J. (c) Let I and J be ideals in a ring R such that I + J = R. Show that there exists a ring isomorphism R/(I ∩
J) ∼= R/I × R/J. Programming Exercise Write a computer program implementing fast addition and multiplication using the Chinese Remainder Theorem and the method outlined in the text. References and Suggested Readings [1] Anderson, F. W. and Fuller, K. R. Rings and Categories of Modules. 2nd ed. Springer, New York, 1992. [2] Atiyah, M. F. and MacDonald, I. G. Introduction to Commutative Algebra. Westview Press, Boulder, CO, 1994. [3] Herstein, I. N. Noncommutative Rings. Mathematical Association of America, Washington, DC, 1994. [4] Kaplansky, I. Commutative Rings. Revised edition. University of Chicago Press, Chicago, 1974. [5] Knuth, D. E. The Art of Computer Programming: Semi-Numerical Algorithms, vol. 2. 3rd ed. Addison-Wesley Professional, Boston, 1997. EXERCISES 267 [6] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998. A good source for applications. [7] Mackiw, G. Applications of Abstract Algebra. Wiley, New York, 1985. [8] McCoy, N. H. Rings and Ideals. Carus Monograph Series, No. 8. Mathemati- cal Association of America, Washington, DC, 1968. [9] McCoy, N. H. The Theory of Rings. Chelsea, New York, 1972. [10] Zariski, O. and Samuel, P. Commutative Algebra, vols. I and II. Springer, New York, 1975, 1960. Sage Rings are at the heart of Sage’s design, so you will ﬁnd a wide range of possibilities for computing with rings and ﬁelds. Ideals, quotients, and homomorphisms are all available. 17 Polynomials Most people are fairly familiar with polynomials by the time they begin to study abstract algebra. When we examine polynomial expressions such as p(x) = x3 − 3x + 2 q(x) = 3x2 − 6x + 5, we have a pretty good idea of what p(x) + q(x) and p(x)q(x) mean. We just add
and multiply polynomials as functions; that is, and (p + q)(x) = p(x) + q(x) = (x3 − 3x + 2) + (3x2 − 6x + 5) = x3 + 3x2 − 9x + 7 (pq)(x) = p(x)q(x) = (x3 − 3x + 2)(3x2 − 6x + 5) = 3x5 − 6x4 − 4x3 + 24x2 − 27x + 10. It is probably no surprise that polynomials form a ring. In this chapter we shall emphasize the algebraic structure of polynomials by studying polynomial rings. We can prove many results for polynomial rings that are similar to the theorems we proved for the integers. Analogs of prime numbers, of the division algorithm, and of the Euclidean algorithm exist for polynomials. 268 17.1 POLYNOMIAL RINGS 269 17.1 Polynomial Rings Throughout this chapter we shall assume that R is a commutative ring with identity. Any expression of the form f (x) = n i=0 aixi = a0 + a1x + a2x2 + · · · + anxn, where ai ∈ R and an = 0, is called a polynomial over R with indeterminate x. The elements a0, a1,..., an are called the coeﬃcients of f. The coeﬃcient an is called the leading coeﬃcient. A polynomial is called monic if the leading coeﬃcient is 1. If n is the largest nonnegative number for which an = 0, we say that the degree of f is n and write deg f (x) = n. If no such n exists—that is, if f = 0 is the zero polynomial—then the degree of f is deﬁned to be −∞. We will denote the set of all polynomials with coeﬃcients in a ring R by R[x]. Two polynomials are equal exactly when their corresponding coeﬃcients are equal; that is, if we let p(x) = a0 + a1x + · · · + anxn q(x) =
b0 + b1x + · · · + bmxm, then p(x) = q(x) if and only if ai = bi for all i ≥ 0. To show that the set of all polynomials forms a ring, we must ﬁrst deﬁne addition and multiplication. We deﬁne the sum of two polynomials as follows. Let p(x) = a0 + a1x + · · · + anxn q(x) = b0 + b1x + · · · + bmxm. Then the sum of p(x) and q(x) is p(x) + q(x) = c0 + c1x + · · · + ckxk, where ci = ai + bi for each i. We deﬁne the product of p(x) and q(x) to be p(x)q(x) = c0 + c1x + · · · + cm+nxm+n, where ci = i k=0 akbi−k = a0bi + a1bi−1 + · · · + ai−1b1 + aib0 270 CHAPTER 17 POLYNOMIALS for each i. Notice that in each case some of the coeﬃcients may be zero. Example 1. Suppose that and p(x) = 3 + 0x + 0x2 + 2x3 + 0x4 q(x) = 2 + 0x − x2 + 0x3 + 4x4 are polynomials in Z[x]. If the coeﬃcient of some term in a polynomial is zero, then we usually just omit that term. In this case we would write p(x) = 3 + 2x3 and q(x) = 2 − x2 + 4x4. The sum of these two polynomials is p(x) + q(x) = 5 − x2 + 2x3 + 4x4. The product, p(x)q(x) = (3 + 2x3)(2 − x2 + 4x4) = 6 − 3x2 + 4x3 + 12x4 − 2x5 + 8x7, can be calculated either by determining the ci’s in the deﬁ
nition or by simply multiplying polynomials in the same way as we have always done. Example 2. Let p(x) = 3 + 3x3 and q(x) = 4 + 4x2 + 4x4 be polynomials in Z12[x]. The sum of p(x) and q(x) is 7 + 4x2 + 3x3 + 4x4. The product of the two polynomials is the zero polynomial. This example tells us that R[x] cannot be an integral domain if R is not an integral domain. Theorem 17.1 Let R be a commutative ring with identity. Then R[x] is a commutative ring with identity. Proof. Our ﬁrst task is to show that R[x] is an abelian group under polynomial addition. The zero polynomial, f (x) = 0, is the additive identity. Given a polynomial p(x) = n i=0 aixi, the inverse of p(x) is easily veriﬁed to be −p(x) = n i=0 aixi. Commutativity and associativity follow immediately from the deﬁnition of polynomial addition and from the fact that addition in R is both commutative and associative. i=0(−ai)xi = − n 17.1 POLYNOMIAL RINGS 271 To show that polynomial multiplication is associative, let p(x) = q(x) = r(x) = m i=0 n i=0 p i=0 aixi, bixi, cixi. Then [p(x)q(x)]r(x) = m n p cixi bixi aixi i=0 i=0   m+n i   ajbi−j i=0 p cixi     xi i=0 j=0 m+n+p   i j akbj−k i=0   xi cj i=0 j=0 k=0 m+n+p   ajbkcr   xi i=0 j+
k+l=i m+n+p   i i−j aj  bkci−j−k  xi i=0 m i=0 m i=0 j=0   aixi k=0   n+p i   xi   bjci−j i=0 n j=0 bixi aixi p cixi i=0 i=0 = = = = = = = p(x)[q(x)r(x)] The commutativity and distribution properties of polynomial multiplication are proved in a similar manner. We shall leave the proofs of these properties as an exercise. Proposition 17.2 Let p(x) and q(x) be polynomials in R[x], where R is an integral domain. Then deg p(x) + deg q(x) = deg(p(x)q(x)). Furthermore, R[x] is an integral domain. 272 CHAPTER 17 POLYNOMIALS Proof. Suppose that we have two nonzero polynomials p(x) = amxm + · · · + a1x + a0 and q(x) = bnxn + · · · + b1x + b0 with am = 0 and bn = 0. The degrees of p and q are m and n, respectively. The leading term of p(x)q(x) is ambnxm+n, which cannot be zero since R is an integral domain; hence, the degree of p(x)q(x) is m + n, and p(x)q(x) = 0. Since p(x) = 0 and q(x) = 0 imply that p(x)q(x) = 0, we know that R[x] must also be an integral domain. We also want to consider polynomials in two or more variables, such as x2 − 3xy + 2y3. Let R be a ring and suppose that we are given two indeterminates x and y. Certainly we can form the ring (R[x])[y]. It is straightforward but perhaps tedious to show that (R[x])[y] ∼= R([y])[x]. We shall identify these two rings by this
isomorphism and simply write R[x, y]. The ring R[x, y] is called the ring of polynomials in two indeterminates x and y with coeﬃcients in R. We can deﬁne the ring of polynomials in n indeterminates with coeﬃcients in R similarly. We shall denote this ring by R[x1, x2,..., xn]. Theorem 17.3 Let R be a commutative ring with identity and α ∈ R. Then we have a ring homomorphism φα : R[x] → R deﬁned by φα(p(x)) = p(α) = anαn + · · · + a1α + a0, where p(x) = anxn + · · · + a1x + a0. i=0 aixi and q(x) = m Proof. Let p(x) = n i=0 bixi. It is easy to show that φα(p(x) + q(x)) = φα(p(x)) + φα(q(x)). To show that multiplication is preserved under the map φα, observe that φα(p(x))φα(q(x)) = p(α)q(α) = = n m biαi aiαi i=0 m+n i i=0 k=0 i=0 akbi−k αi = φα(p(x)q(x)). The map φα : R[x] → R is called the evaluation homomorphism at α. 17.2 THE DIVISION ALGORITHM 273 17.2 The Division Algorithm Recall that the division algorithm for integers (Theorem 2.3) says that if a and b are integers with b > 0, then there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. The algorithm by which q and r are found is just long division. A similar theorem exists for polynomials. The division algorithm for polynomials has several important consequences. Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review Theorem 2.3 at this point. Theorem 17.4 (Division Al
gorithm) Let f (x) and g(x) be two nonzero polynomials in F [x], where F is a ﬁeld and g(x) is a nonconstant polynomial. Then there exist unique polynomials q(x), r(x) ∈ F [x] such that f (x) = g(x)q(x) + r(x), where either deg r(x) < deg g(x) or r(x) is the zero polynomial. Proof. We will ﬁrst consider the existence of q(x) and r(x). Let S = {f (x) − g(x)h(x) : h(x) ∈ F [x]} and assume that g(x) = a0 + a1x + · · · + anxn is a polynomial of degree n. This set is nonempty since f (x) ∈ S. If f (x) is the zero polynomial, then 0 = f (x) = 0 · g(x) + 0; hence, both q and r must also be the zero polynomial. Now suppose that the zero polynomial is not in S. In this case the degree of every polynomial in S is nonnegative. Choose a polynomial r(x) of smallest degree in S; hence, there must exist a q(x) ∈ F [x] such that or r(x) = f (x) − g(x)q(x), f (x) = g(x)q(x) + r(x). We need to show that the degree of r(x) is less than the degree of g(x). Assume that deg g(x) ≤ deg r(x). Say r(x) = b0 + b1x + · · · + bmxm and m ≥ n. Then f (x) − g(x)[q(x) + (bm/an)xm−n] = f (x) − g(x)q(x) − (bm/an)xm−ng(x) = r(x) − (bm/an)xm−ng(x) = r(x) − bmxm + terms of lower degree 274 CHAPTER 17 POLYNOMIALS is in S. This
is a polynomial of lower degree than r(x), which contradicts the fact that r(x) is a polynomial of smallest degree in S; hence, deg r(x) < deg g(x). To show that q(x) and r(x) are unique, suppose that there exist two other polynomials q(x) and r(x) such that f (x) = g(x)q(x) + r(x) and deg r(x) < deg g(x) or r(x) = 0, so that f (x) = g(x)q(x) + r(x) = g(x)q(x) + r(x), and g(x)[q(x) − q(x)] = r(x) − r(x). If g is not the zero polynomial, then deg(g(x)[q(x) − q(x)]) = deg(r(x) − r(x)) ≥ deg g(x). However, the degrees of both r(x) and r(x) are strictly less than the degree of g(x); therefore, r(x) = r(x) and q(x) = q(x). Example 3. The division algorithm merely formalizes long division of polynomials, a task we have been familiar with since high school. For example, suppose that we divide x3 − x2 + 2x − 3 by x − 2. x2 + x − 2 x3 − x + 4 x2 + 2x − 3 x3 − 2x2 x2 + 2x − 3 x2 − 2x 4x − 3 4x − 8 5 Hence, x3 − x2 + 2x − 3 = (x − 2)(x2 + x + 4) + 5. Let p(x) be a polynomial in F [x] and α ∈ F. We say that α is a zero or root of p(x) if p(x) is in the kernel of the evaluation homomorphism φα. All we are really saying here is that α is a zero of p(x) if p(α) = 0. Corollary 17.5 Let F be a ﬁeld. An element α ∈ F is a zero of p(x) ∈ F [x] if and
only if x − α is a factor of p(x) in F [x]. Proof. Suppose that α ∈ F and p(α) = 0. By the division algorithm, there exist polynomials q(x) and r(x) such that p(x) = (x − α)q(x) + r(x) 17.2 THE DIVISION ALGORITHM 275 and the degree of r(x) must be less than the degree of x − α. Since the degree of r(x) is less than 1, r(x) = a for a ∈ F ; therefore, But p(x) = (x − α)q(x) + a. 0 = p(α) = 0 · q(x) + a = a; consequently, p(x) = (x − α)q(x), and x − α is a factor of p(x). Conversely, suppose that x − α is a factor of p(x); say p(x) = (x − α)q(x). Then p(α) = 0 · q(x) = 0. Corollary 17.6 Let F be a ﬁeld. A nonzero polynomial p(x) of degree n in F [x] can have at most n distinct zeros in F. Proof. We will use induction on the degree of p(x). If deg p(x) = 0, then p(x) is a constant polynomial and has no zeros. Let deg p(x) = 1. Then p(x) = ax + b for some a and b in F. If α1 and α2 are zeros of p(x), then aα1 + b = aα2 + b or α1 = α2. Now assume that deg p(x) > 1. If p(x) does not have a zero in F, then we are done. On the other hand, if α is a zero of p(x), then p(x) = (x − α)q(x) for some q(x) ∈ F [x] by Corollary 17.5. The degree of q(x) is n − 1 by Proposition 17.2. Let β be some other zero of p(x) that is distinct from α. Then p(β) = (β − α)
q(β) = 0. Since α = β and F is a ﬁeld, q(β) = 0. By our induction hypothesis, p(x) can have at most n − 1 zeros in F that are distinct from α. Therefore, p(x) has at most n distinct zeros in F. Let F be a ﬁeld. A monic polynomial d(x) is a greatest common divisor of polynomials p(x), q(x) ∈ F [x] if d(x) evenly divides both p(x) and q(x); and, if for any other polynomial d(x) dividing both p(x) and q(x), d(x) \| d(x). We write d(x) = gcd(p(x), q(x)). Two polynomials p(x) and q(x) are relatively prime if gcd(p(x), q(x)) = 1. Proposition 17.7 Let F be a ﬁeld and suppose that d(x) is the greatest common divisor of two polynomials p(x) and q(x) in F [x]. Then there exist polynomials r(x) and s(x) such that d(x) = r(x)p(x) + s(x)q(x). Furthermore, the greatest common divisor of two polynomials is unique. 276 CHAPTER 17 POLYNOMIALS Proof. Let d(x) be the monic polynomial of smallest degree in the set S = {f (x)p(x) + g(x)q(x) : f (x), g(x) ∈ F [x]}. We can write d(x) = r(x)p(x) + s(x)q(x) for two polynomials r(x) and s(x) in F [x]. We need to show that d(x) divides both p(x) and q(x). We shall ﬁrst show that d(x) divides p(x). By the division algorithm, there exist polynomials a(x) and b(x) such that p(x) = a(x)d(x) + b(x), where b(
x) is either the zero polynomial or deg b(x) < deg d(x). Therefore, b(x) = p(x) − a(x)d(x) = p(x) − a(x)(r(x)p(x) + s(x)q(x)) = p(x) − a(x)r(x)p(x) − a(x)s(x)q(x) = p(x)(1 − a(x)r(x)) + q(x)(−a(x)s(x)) is a linear combination of p(x) and q(x) and therefore must be in S. However, b(x) must be the zero polynomial since d(x) was chosen to be of smallest degree; consequently, d(x) divides p(x). A symmetric argument shows that d(x) must also divide q(x); hence, d(x) is a common divisor of p(x) and q(x). To show that d(x) is a greatest common divisor of p(x) and q(x), suppose that d(x) is another common divisor of p(x) and q(x). We will show that d(x) \| d(x). Since d(x) is a common divisor of p(x) and q(x), there exist polynomials u(x) and v(x) such that p(x) = u(x)d(x) and q(x) = v(x)d(x). Therefore, d(x) = r(x)p(x) + s(x)q(x) = r(x)u(x)d(x) + s(x)v(x)d(x) = d(x)[r(x)u(x) + s(x)v(x)]. Since d(x) \| d(x), d(x) is a greatest common divisor of p(x) and q(x). Finally, we must show that the greatest common divisor of p(x) and q(x)) is unique. Suppose that d(x) is another greatest common divisor of p(x) and q(x). We have just shown that there exist polynomials
u(x) and v(x) in F [x] such that d(x) = d(x)[r(x)u(x) + s(x)v(x)]. Since deg d(x) = deg d(x) + deg[r(x)u(x) + s(x)v(x)] and d(x) and d(x) are both greatest common divisors, deg d(x) = deg d(x). Since d(x) and d(x) are both monic polynomials of the same degree, it must be the case that d(x) = d(x). Notice the similarity between the proof of Proposition 17.7 and the proof of Theorem 2.4. 17.3 IRREDUCIBLE POLYNOMIALS 277 17.3 Irreducible Polynomials A nonconstant polynomial f (x) ∈ F [x] is irreducible over a ﬁeld F if f (x) cannot be expressed as a product of two polynomials g(x) and h(x) in F [x], where the degrees of g(x) and h(x) are both smaller than the degree of f (x). Irreducible polynomials function as the “prime numbers” of polynomial rings. Example 4. The polynomial x2 − 2 ∈ Q[x] is irreducible since it cannot be factored any further over the rational numbers. Similarly, x2 + 1 is irreducible over the real numbers. Example 5. The polynomial p(x) = x3 + x2 + 2 is irreducible over Z3[x]. Suppose that this polynomial was reducible over Z3[x]. By the division algorithm there would have to be a factor of the form x − a, where a is some element in Z3[x]. Hence, it would have to be true that p(a) = 0. However, p(0) = 2 p(1) = 1 p(2) = 2. Therefore, p(x) has no zeros in Z3 and must be irreducible. Lemma 17.8 Let p(x) ∈ Q[x]. Then p(x) = r s (a0 + a1x + · · ·
+ anxn), where r, s, a0,..., an are integers, the ai’s are relatively prime, and r and s are relatively prime. Proof. Suppose that p(x) = b0 c0 + b1 c1 x + · · · + bn cn xn, where the bi’s and the ci’s are integers. We can rewrite p(x) as p(x) = 1 c0 · · · cn (d0 + d1x + · · · + dnxn), where d0,..., dn are integers. Let d be the greatest common divisor of d0,..., dn. Then p(x) = d c0 · · · cn (a0 + a1x + · · · + anxn), 278 CHAPTER 17 POLYNOMIALS where di = dai and the ai’s are relatively prime. Reducing d/(c0 · · · cn) to its lowest terms, we can write p(x) = r s (a0 + a1x + · · · + anxn), where gcd(r, s) = 1. Theorem 17.9 (Gauss’s Lemma) Let p(x) ∈ Z[x] be a monic polynomial such that p(x) factors into a product of two polynomials α(x) and β(x) in Q[x], where the degrees of both α(x) and β(x) are less than the degree of p(x). Then p(x) = a(x)b(x), where a(x) and b(x) are monic polynomials in Z[x] with deg α(x) = deg a(x) and deg β(x) = deg b(x). Proof. By Lemma 17.8, we can assume that α(x) = β(x) = c1 d1 c2 d2 (a0 + a1x + · · · + amxm) = c2 d2 (b0 + b1x + · · · + bnxn) = α1(x) c1 d1 β1(x), where the ai’s are relatively prime and the bi’s are relatively
prime. Consequently, p(x) = α(x)β(x) = c1c2 d1d2 α1(x)β1(x) = c d α1(x)β1(x), where c/d is the product of c1/d1 and c2/d2 expressed in lowest terms. Hence, dp(x) = cα1(x)β1(x). If d = 1, then cambn = 1 since p(x) is a monic polynomial. Hence, either c = 1 or c = −1. If c = 1, then either am = bn = 1 or am = bn = −1. In the ﬁrst case p(x) = α1(x)β1(x), where α1(x) and β1(x) are monic polynomials with deg α(x) = deg α1(x) and deg β(x) = deg β1(x). In the second case a(x) = −α1(x) and b(x) = −β1(x) are the correct monic polynomials since p(x) = (−α1(x))(−β1(x)) = a(x)b(x). The case in which c = −1 can be handled similarly. Now suppose that d = 1. Since gcd(c, d) = 1, there exists a prime p such that p \| d and p \|c. Also, since the coeﬃcients of α1(x) are relatively prime, there exists a coeﬃcient ai such that p \|ai. Similarly, there exists a coeﬃcient 1(x) be the polynomials in Zp[x] bj of β1(x) such that p \|bj. Let α 1(x) and β obtained by reducing the coeﬃcients of α1(x) and β1(x) modulo p. Since p \| d, α 1(x) 1(x) is the zero polynomial and Zp[x] is an integral domain. Therefore, nor β d = 1 and the theorem is proven. 1(x) = 0 in Zp[x]. However, this is impossible since neither α 1(x)β 17.
3 IRREDUCIBLE POLYNOMIALS 279 Corollary 17.10 Let p(x) = xn + an−1xn−1 + · · · + a0 be a polynomial with coeﬃcients in Z and a0 = 0. If p(x) has a zero in Q, then p(x) also has a zero α in Z. Furthermore, α divides a0. Proof. Let p(x) have a zero a ∈ Q. Then p(x) must have a linear factor x − a. By Gauss’s Lemma, p(x) has a factorization with a linear factor in Z[x]. Hence, for some α ∈ Z p(x) = (x − α)(xn−1 + · · · − a0/α). Thus a0/α ∈ Z and so α \| a0. Example 6. Let p(x) = x4 − 2x3 + x + 1. We shall show that p(x) is irreducible over Q[x]. Assume that p(x) is reducible. Then either p(x) has a linear factor, say p(x) = (x − α)q(x), where q(x) is a polynomial of degree three, or p(x) has two quadratic factors. If p(x) has a linear factor in Q[x], then it has a zero in Z. By Corollary 17.10, any zero must divide 1 and therefore must be ±1; however, p(1) = 1 and p(−1) = 3. Consequently, we have eliminated the possibility that p(x) has any linear factors. Therefore, if p(x) is reducible it must factor into two quadratic polyno- mials, say p(x) = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd, where each factor is in Z[x] by Gauss’s Lemma. Hence, a + c = −2 ac + b + d = 0 ad + bc = 1 bd = 1. Since bd = 1, either b = d = 1 or b = d = −1. In
either case b = d and so ad + bc = b(a + c) = 1. Since a + c = −2, we know that −2b = 1. This is impossible since b is an integer. Therefore, p(x) must be irreducible over Q. 280 CHAPTER 17 POLYNOMIALS Theorem 17.11 (Eisenstein’s Criterion) Let p be a prime and suppose that f (x) = anxn + · · · + a0 ∈ Z[x]. If p \| ai for i = 0, 1,..., n − 1, but p \|an and p2 \|a0, then f (x) is irreducible over Q. Proof. By Gauss’s Lemma, we need only show that f (x) does not factor into polynomials of lower degree in Z[x]. Let f (x) = (brxr + · · · + b0)(csxs + · · · + c0) be a factorization in Z[x], with br and cs not equal to zero and r, s < n. Since p2 does not divide a0 = b0c0, either b0 or c0 is not divisible by p. Suppose that p \|b0 and p \| c0. Since p \|an and an = brcs, neither br nor cs is divisible by p. Let m be the smallest value of k such that p \|ck. Then am = b0cm + b1cm−1 + · · · + bmc0 is not divisible by p, since each term on the right-hand side of the equation is divisible by p except for b0cm. Therefore, m = n since ai is divisible by p for m < n. Hence, f (x) cannot be factored into polynomials of lower degree and therefore must be irreducible. Example 7. The polynomial p(x) = 16x5 − 9x4 + 3x2 + 6x − 21 is easily seen to be irreducible over Q by Eisenstein’s Criterion if we let p = 3. Eisenstein’s Criterion is more useful in constructing irreducible polynomials of a certain degree over Q than in determining the irreducibility of an arbitrary polynomial in Q[x
]: given an arbitrary polynomial, it is not very likely that we can apply Eisenstein’s Criterion. The real value of Theorem 17.11 is that we now have an easy method of generating irreducible polynomials of any degree. Ideals in F [x] Let F be a ﬁeld. Recall that a principal ideal in F [x] is an ideal p(x) generated by some polynomial p(x); that is, p(x) = {p(x)q(x) : q(x) ∈ F [x]}. 17.3 IRREDUCIBLE POLYNOMIALS 281 Example 8. The polynomial x2 in F [x] generates the ideal x2 consisting of all polynomials with no constant term or term of degree 1. Theorem 17.12 If F is a ﬁeld, then every ideal in F [x] is a principal ideal. Proof. Let I be an ideal of F [x]. If I is the zero ideal, the theorem is easily true. Suppose that I is a nontrivial ideal in F [x], and let p(x) ∈ I be a nonzero element of minimal degree. If deg p(x) = 0, then p(x) is a nonzero constant and 1 must be in I. Since 1 generates all of F [x], 1 = I = F [x] and I is again a principal ideal. Now assume that deg p(x) ≥ 1 and let f (x) be any element in I. By the division algorithm there exist q(x) and r(x) in F [x] such that f (x) = p(x)q(x) + r(x) and deg r(x) < deg p(x). Since f (x), p(x) ∈ I and I is an ideal, r(x) = f (x) − p(x)q(x) is also in I. However, since we chose p(x) to be of minimal degree, r(x) must be the zero polynomial. Since we can write any element f (x) in I as p(x)q(x) for some q(x) ∈ F [x], it must be the case that I = p(x). It is not the case that every ideal in the
ring F [x, y] is a Example 9. principal ideal. Consider the ideal of F [x, y] generated by the polynomials x and y. This is the ideal of F [x, y] consisting of all polynomials with no constant term. Since both x and y are in the ideal, no single polynomial can generate the entire ideal. Theorem 17.13 Let F be a ﬁeld and suppose that p(x) ∈ F [x]. Then the ideal generated by p(x) is maximal if and only if p(x) is irreducible. Proof. Suppose that p(x) generates a maximal ideal of F [x]. Then p(x) is also a prime ideal of F [x]. Since a maximal ideal must be properly contained inside F [x], p(x) cannot be a constant polynomial. Let us assume that p(x) factors into two polynomials of lesser degree, say p(x) = f (x)g(x). Since p(x) is a prime ideal one of these factors, say f (x), is in p(x) and therefore be a multiple of p(x). But this would imply that p(x) ⊂ f (x), which is impossible since p(x) is maximal. Conversely, suppose that p(x) is irreducible over F [x]. Let I be an ideal in F [x] containing p(x). By Theorem 17.12, I is a principal ideal; hence, I = f (x) for some f (x) ∈ F [x]. Since p(x) ∈ I, it must be the case that p(x) = f (x)g(x) for some g(x) ∈ F [x]. However, p(x) is irreducible; hence, either f (x) or g(x) is a constant polynomial. If f (x) is constant, then I = F [x] and we are done. If g(x) is constant, then f (x) is a constant 282 CHAPTER 17 POLYNOMIALS multiple of I and I = p(x). Thus, there are no proper ideals of F [x] that properly contain p(x). Historical Note Throughout history, the solution of polynomial equations has been a challenging
problem. The Babylonians knew how to solve the equation ax2 + bx + c = 0. Omar Khayyam (1048–1131) devised methods of solving cubic equations through the use of geometric constructions and conic sections. The algebraic solution of the general cubic equation ax3 + bx2 + cx + d = 0 was not discovered until the sixteenth century. An Italian mathematician, Luca Pacioli (ca. 1445–1509), wrote in Summa de Arithmetica that the solution of the cubic was impossible. This was taken as a challenge by the rest of the mathematical community. Scipione del Ferro (1465–1526), of the University of Bologna, solved the “de- pressed cubic,” ax3 + cx + d = 0. He kept his solution an absolute secret. This may seem surprising today, when mathematicians are usually very eager to publish their results, but in the days of the Italian Renaissance secrecy was customary. Academic appointments were not easy to secure and depended on the ability to prevail in public contests. Such challenges could be issued at any time. Consequently, any major new discovery was a valuable weapon in such a contest. If an opponent presented a list of problems to be solved, del Ferro could in turn present a list of depressed cubics. He kept the secret of his discovery throughout his life, passing it on only on his deathbed to his student Antonio Fior (ca. 1506–?). Although Fior was not the equal of his teacher, he immediately issued a challenge to Niccolo Fontana (1499–1557). Fontana was known as Tartaglia (the Stammerer). As a youth he had suﬀered a blow from the sword of a French soldier during an attack on his village. He survived the savage wound, but his speech was permanently impaired. Tartaglia sent Fior a list of 30 various mathematical problems; Fior countered by sending Tartaglia a list of 30 depressed cubics. Tartaglia would either solve all 30 of the problems or absolutely fail. After much eﬀort Tartaglia ﬁnally succeeded in solving the depressed cubic and defeated Fior, who faded into obscurity. At this point another mathematician, Gerolamo Cardano (1501–1576), entered the story. Cardano wrote to Tartaglia, begging him for the solution to the depressed cubic. Tartaglia refused several of his requests,
then ﬁnally revealed the solution to Cardano after the latter swore an oath not to publish the secret or to pass it on to anyone else. Using the knowledge that he had obtained from Tartaglia, Cardano eventually solved the general cubic ax3 + bx2 + cx + d = 0. Cardano shared the secret with his student, Ludovico Ferrari (1522–1565), who solved the general quartic equation, ax4 + bx3 + cx2 + dx + e = 0. EXERCISES 283 In 1543, Cardano and Ferrari examined del Ferro’s papers and discovered that he had also solved the depressed cubic. Cardano felt that this relieved him of his obligation to Tartaglia, so he proceeded to publish the solutions in Ars Magna (1545), in which he gave credit to del Ferro for solving the special case of the cubic. This resulted in a bitter dispute between Cardano and Tartaglia, who published the story of the oath a year later. Exercises 1. List all of the polynomials of degree 3 or less in Z2[x]. 2. Compute each of the following. (a) (5x2 + 3x − 4) + (4x2 − x + 9) in Z12 (b) (5x2 + 3x − 4)(4x2 − x + 9) in Z12 (c) (7x3 + 3x2 − x) + (6x2 − 8x + 4) in Z9 (d) (3x2 + 2x − 4) + (4x2 + 2) in Z5 (e) (3x2 + 2x − 4)(4x2 + 2) in Z5 (f) (5x2 + 3x − 2)2 in Z12 3. Use the division algorithm to ﬁnd q(x) and r(x) such that a(x) = q(x)b(x) + r(x) with deg r(x) < deg b(x) for each of the following pairs of polynomials. (a) a(x) = 5x3 + 6x2 − 3x + 4 and b(x) = x − 2 in Z7[x] (b) a(x) = 6x4 − 2x3 + x2 − 3x + 1 and b
(x) = x2 + x − 2 in Z7[x] (c) a(x) = 4x5 − x3 + x2 + 4 and b(x) = x3 − 2 in Z5[x] (d) a(x) = x5 + x3 − x2 − x and b(x) = x3 + x in Z2[x] 4. Find the greatest common divisor of each of the following pairs p(x) and q(x) of polynomials. If d(x) = gcd(p(x), q(x)), ﬁnd two polynomials a(x) and b(x) such that a(x)p(x) + b(x)q(x) = d(x). (a) p(x) = 7x3 + 6x2 − 8x + 4 and q(x) = x3 + x − 2, where p(x), q(x) ∈ Q[x] (b) p(x) = x3 + x2 − x + 1 and q(x) = x3 + x − 1, where p(x), q(x) ∈ Z2[x] (c) p(x) = x3 + x2 − 4x + 4 and q(x) = x3 + 3x − 2, where p(x), q(x) ∈ Z5[x] (d) p(x) = x3 − 2x + 4 and q(x) = 4x3 + x + 3, where p(x), q(x) ∈ Q[x] 5. Find all of the zeros for each of the following polynomials. (a) 5x3 + 4x2 − x + 9 in Z12 (b) 3x3 − 4x2 − x + 4 in Z5 (c) 5x4 + 2x2 − 3 in Z7 (d) x3 + x + 1 in Z2 284 CHAPTER 17 POLYNOMIALS 6. Find all of the units in Z[x]. 7. Find a unit p(x) in Z4[x] such that deg p(x) > 1. 8. Which of the following polynomials are irreducible over Q[x]? (a) x4
− 2x3 + 2x2 + x + 4 (b) x4 − 5x3 + 3x − 2 (c) 3x5 − 4x3 − 6x2 + 6 (d) 5x5 − 6x4 − 3x2 + 9x − 15 9. Find all of the irreducible polynomials of degrees 2 and 3 in Z2[x]. 10. Give two diﬀerent factorizations of x2 + x + 8 in Z10[x]. 11. Prove or disprove: There exists a polynomial p(x) in Z6[x] of degree n with more than n distinct zeros. 12. If F is a ﬁeld, show that F [x1,..., xn] is an integral domain. 13. Show that the division algorithm does not hold for Z[x]. Why does it fail? 14. Prove or disprove: xp + a is irreducible for any a ∈ Zp, where p is prime. 15. Let f (x) be irreducible. If f (x) \| p(x)q(x), prove that either f (x) \| p(x) or f (x) \| q(x). 16. Suppose that R and S are isomorphic rings. Prove that R[x] ∼= S[x]. 17. Let F be a ﬁeld and a ∈ F. If p(x) ∈ F [x], show that p(a) is the remainder obtained when p(x) is divided by x − a. 18. Let Q∗ be the multiplicative group of positive rational numbers. Prove that Q∗ is isomorphic to (Z[x], +). 19. Cyclotomic Polynomials. The polynomial Φn(x) = xn − 1 x − 1 = xn−1 + xn−2 + · · · + x + 1 is called the cyclotomic polynomial. Show that Φp(x) is irreducible over Q for any prime p. 20. If F is a ﬁeld, show that there are inﬁnitely many irreducible polynomials in F [x]. 21. Let R be a commutative ring with identity. Prove that multiplication is
commutative in R[x]. 22. Let R be a commutative ring with identity. Prove that multiplication is distributive in R[x]. 23. Show that xp − x has p distinct zeros in Zp, for any prime p. Conclude that therefore xp − x = x(x − 1)(x − 2) · · · (x − (p − 1)). EXERCISES 285 24. Let F be a ring and f (x) = a0 + a1x + · · · + anxn be in F [x]. Deﬁne f (x) = a1 + 2a2x + · · · + nanxn−1 to be the derivative of f (x). (a) Prove that (f + g)(x) = f (x) + g(x). Conclude that we can deﬁne a homomorphism of abelian groups D : F [x] → F [x] by (D(f (x)) = f (x). (b) Calculate the kernel of D if charF = 0. (c) Calculate the kernel of D if charF = p. (d) Prove that (f g)(x) = f (x)g(x) + f (x)g(x). (e) Suppose that we can factor a polynomial f (x) ∈ F [x] into linear factors, say f (x) = a(x − a1)(x − a2) · · · (x − an). Prove that f (x) has no repeated factors if and only if f (x) and f (x) are relatively prime. 25. Let F be a ﬁeld. Show that F [x] is never a ﬁeld. 26. Let R be an integral domain. Prove that R[x1,..., xn] is an integral domain. 27. Let R be a commutative ring with identity. Show that R[x] has a subring R isomorphic to R. 28. Let p(x) and q(x) be polynomials in R[x], where R is a commutative ring with identity. Prove that deg(p(x) + q(x)) ≤ max(deg p(x), deg q(x)). Additional Exercises
: Solving the Cubic and Quartic Equations 1. Solve the general quadratic equation to obtain ax2 + bx + c = 0 −b ± x = √ b2 − 4ac 2a. The discriminant of the quadratic equation ∆ = b2 − 4ac determines the nature of the solutions of the equation. If ∆ > 0, the equation has two distinct real solutions. If ∆ = 0, the equation has a single repeated real root. If ∆ < 0, there are two distinct imaginary solutions. 2. Show that any cubic equation of the form x3 + bx2 + cx + d = 0 can be reduced to the form y3 + py + q = 0 by making the substitution x = y − b/3. 286 CHAPTER 17 POLYNOMIALS 3. Prove that the cube roots of 1 are given by √ −1 + i 2 √ −1 − i 3 3 ω = ω2 = 2 p 3z 4. Make the substitution ω3 = 1. y = z − for y in the equation y3 + py + q = 0 and obtain two solutions A and B for z3. 5. Show that the product of the solutions obtained in (4) is −p3/27, deducing √ that 3 AB = −p/3. 6. Prove that the possible solutions for z in (4) are given by √ 3 √ A, ω 3 √ A, ω2 3 A, √ 3 √ B, ω 3 √ B, ω2 3 B and use this result to show that the three possible solutions for y are ωi 3 − q 2 + p3 27 + q2 4 + ω2i 3 − q 2 − p3 27 + q2 4, where i = 0, 1, 2. 7. The discriminant of the cubic equation is ∆ = p3 27 + q2 4. Show that y3 + py + q = 0 (a) has three real roots, at least two of which are equal, if ∆ = 0. (b) has one real root and two conjugate imaginary roots if ∆ > 0. (c) has three distinct real roots if ∆ < 0. 8. Solve the following cubic equations. (a) x3 − 4x2 + 11x + 30 = 0 (b
) x3 − 3x + 5 = 0 (c) x3 − 3x + 2 = 0 (d) x3 + x + 3 = 0 9. Show that the general quartic equation can be reduced to x4 + ax3 + bx2 + cx + d = 0 y4 + py2 + qy + r = 0 by using the substitution x = y − a/4. EXERCISES 10. Show that y2 + 2 1 2 z = (z − p)y2 − qy + z2 − r. 1 4 287 11. Show that the right-hand side of (10) can be put in the form (my + k)2 if and only if q2 − 4(z − p) z2 − r = 0. 1 4 12. From (11) obtain the resolvent cubic equation z3 − pz2 − 4rz + (4pr − q2) = 0. Solving the resolvent cubic equation, put the equation found in (10) in the form y2 + 2 1 2 z = (my + k)2 to obtain the solution of the quartic equation. 13. Use this method to solve the following quartic equations. (a) x4 − x2 − 3x + 2 = 0 (b) x4 + x3 − 7x2 − x + 6 = 0 (c) x4 − 2x2 + 4x − 3 = 0 (d) x4 − 4x3 + 3x2 − 5x + 2 = 0 Sage Polynomial rings are very important for computational approaches to algebra, and so Sage makes it very easy to compute with polynomials, over rings, or over ﬁelds. And it is trivial to check if a polynomial is irreducible. 18 Integral Domains One of the most important rings we study is the ring of integers. It was our ﬁrst example of an algebraic structure: the ﬁrst polynomial ring that we examined was Z[x]. We also know that the integers sit naturally inside the ﬁeld of rational numbers, Q. The ring of integers is the model for all integral domains. In this chapter we will examine integral domains in general, answering questions about the ideal structure of integral domains, polynomial rings over integral domains, and whether or not an integral domain can be embedded in a
ﬁeld. 18.1 Fields of Fractions Every ﬁeld is also an integral domain; however, there are many integral domains that are not ﬁelds. For example, the integers Z are an integral domain but not a ﬁeld. A question that naturally arises is how we might associate an integral domain with a ﬁeld. There is a natural way to construct the rationals Q from the integers: the rationals can be represented as formal quotients of two integers. The rational numbers are certainly a ﬁeld. In fact, it can be shown that the rationals are the smallest ﬁeld that contains the integers. Given an integral domain D, our question now becomes how to construct a smallest ﬁeld F containing D. We will do this in the same way as we constructed the rationals from the integers. An element p/q ∈ Q is the quotient of two integers p and q; however, diﬀerent pairs of integers can represent the same rational number. For instance, 1/2 = 2/4 = 3/6. We know that a b = c d if and only if ad = bc. A more formal way of considering this problem is to examine fractions in terms of equivalence relations. We can think of 288 18.1 FIELDS OF FRACTIONS 289 elements in Q as ordered pairs in Z × Z. A quotient p/q can be written as (p, q). For instance, (3, 7) would represent the fraction 3/7. However, there are problems if we consider all possible pairs in Z × Z. There is no fraction 5/0 corresponding to the pair (5, 0). Also, the pairs (3, 6) and (2, 4) both represent the fraction 1/2. The ﬁrst problem is easily solved if we require the second coordinate to be nonzero. The second problem is solved by considering two pairs (a, b) and (c, d) to be equivalent if ad = bc. If we use the approach of ordered pairs instead of fractions, then we can study integral domains in general. Let D be any integral domain and let S = {(a, b) : a, b ∈ D and b = 0}. Deﬁne a relation on S by (a, b) ∼ (c, d) if ad = bc
. Lemma 18.1 The relation ∼ between elements of S is an equivalence relation. Proof. Since D is commutative, ab = ba; hence, ∼ is reﬂexive on D. Now suppose that (a, b) ∼ (c, d). Then ad = bc or cb = da. Therefore, (c, d) ∼ (a, b) and the relation is symmetric. Finally, to show that the relation is transitive, let (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). In this case ad = bc and cf = de. Multiplying both sides of ad = bc by f yields af d = adf = bcf = bde = bed. Since D is an integral domain, we can deduce that af = be or (a, b) ∼ (e, f ). We will denote the set of equivalence classes on S by FD. We now need to deﬁne the operations of addition and multiplication on FD. Recall how fractions are added and multiplied in Q = = ; ad + bc bd. ac bd It seems reasonable to deﬁne the operations of addition and multiplication on FD in a similar manner. If we denote the equivalence class of (a, b) ∈ S by [a, b], then we are led to deﬁne the operations of addition and multiplication on FD by [a, b] + [c, d] = [ad + bc, bd] 290 and CHAPTER 18 INTEGRAL DOMAINS [a, b] · [c, d] = [ac, bd], respectively. The next lemma demonstrates that these operations are independent of the choice of representatives from each equivalence class. Lemma 18.2 The operations of addition and multiplication on FD are welldeﬁned. Proof. We will prove that the operation of addition is well-deﬁned. The proof that multiplication is well-deﬁned is left as an exercise. Let [a1, b1] = [a2, b2] and [c1, d1] = [c2, d2]. We must show that [a1d1 + b1c1, b1d1] = [a2d2 + b2c2, b2d2] or, equivalently, that (a
1d1 + b1c1)(b2d2) = (b1d1)(a2d2 + b2c2). Since [a1, b1] = [a2, b2] and [c1, d1] = [c2, d2], we know that a1b2 = b1a2 and c1d2 = d1c2. Therefore, (a1d1 + b1c1)(b2d2) = a1d1b2d2 + b1c1b2d2 = a1b2d1d2 + b1b2c1d2 = b1a2d1d2 + b1b2d1c2 = (b1d1)(a2d2 + b2c2). Lemma 18.3 The set of equivalence classes of S, FD, under the equivalence relation ∼, together with the operations of addition and multiplication deﬁned by [a, b] + [c, d] = [ad + bc, bd] [a, b] · [c, d] = [ac, bd], is a ﬁeld. Proof. The additive and multiplicative identities are [0, 1] and [1, 1], respectively. To show that [0, 1] is the additive identity, observe that [a, b] + [0, 1] = [a1 + b0, b1] = [a, b]. 18.1 FIELDS OF FRACTIONS 291 It is easy to show that [1, 1] is the multiplicative identity. Let [a, b] ∈ FD such that a = 0. Then [b, a] is also in FD and [a, b] · [b, a] = [1, 1]; hence, [b, a] is the multiplicative inverse for [a, b]. Similarly, [−a, b] is the additive inverse of [a, b]. We leave as exercises the veriﬁcation of the associative and commutative properties of multiplication in FD. We also leave it to the reader to show that FD is an abelian group under addition. It remains to show that the distributive property holds in FD; however, [a, b][e, f ] + [c, d][e, f ] =
[ae, bf ] + [ce, df ] = [aedf + bf ce, bdf 2] = [aed + bce, bdf ] = [ade + bce, bdf ] = ([a, b] + [c, d])[e, f ] and the lemma is proved. The ﬁeld FD in Lemma 18.3 is called the ﬁeld of fractions or ﬁeld of quotients of the integral domain D. Theorem 18.4 Let D be an integral domain. Then D can be embedded in a ﬁeld of fractions FD, where any element in FD can be expressed as the quotient of two elements in D. Furthermore, the ﬁeld of fractions FD is unique in the sense that if E is any ﬁeld containing D, then there exists a map ψ : FD → E giving an isomorphism with a subﬁeld of E such that ψ(a) = a for all elements a ∈ D. Proof. We will ﬁrst demonstrate that D can be embedded in the ﬁeld FD. Deﬁne a map φ : D → FD by φ(a) = [a, 1]. Then for a and b in D, φ(a + b) = [a + b, 1] = [a, 1] + [b, 1] = φ(a) + φ(b) and φ(ab) = [ab, 1] = [a, 1][b, 1] = φ(a)φ(b); hence, φ is a homomorphism. To show that φ is one-to-one, suppose that φ(a) = φ(b). Then [a, 1] = [b, 1], or a = a1 = 1b = b. Finally, any element of FD can be expressed as the quotient of two elements in D, since φ(a)[φ(b)]−1 = [a, 1][b, 1]−1 = [a, 1] · [1, b] = [a, b]. Now let E be a ﬁeld containing D and deﬁne a map ψ : FD → E by ψ([a, b]) = ab−
1. To show that ψ is well-deﬁned, let [a1, b1] = [a2, b2]. Then a1b2 = b1a2. Therefore, a1b−1 2 and ψ([a1, b1]) = ψ([a2, b2]). 1 = a2b−1 292 CHAPTER 18 INTEGRAL DOMAINS If [a, b] and [c, d] are in FD, then and ψ([a, b] + [c, d]) = ψ([ad + bc, bd]) = (ad + bc)(bd)−1 = ab−1 + cd−1 = ψ([a, b]) + ψ([c, d]) ψ([a, b] · [c, d]) = ψ([ac, bd]) = (ac)(bd)−1 = ab−1cd−1 = ψ([a, b])ψ([c, d]). Therefore, ψ is a homomorphism. To complete the proof of the theorem, we need to show that ψ is one-toone. Suppose that ψ([a, b]) = ab−1 = 0. Then a = 0b = 0 and [a, b] = [0, b]. Therefore, the kernel of ψ is the zero element [0, b] in FD, and ψ is injective. Example 1. Since Q is a ﬁeld, Q[x] is an integral domain. The ﬁeld of fractions of Q[x] is the set of all rational expressions p(x)/q(x), where p(x) and q(x) are polynomials over the rationals and q(x) is not the zero polynomial. We will denote this ﬁeld by Q(x). We will leave the proofs of the following corollaries of Theorem 18.4 as exercises. Corollary 18.5 Let F be a ﬁeld of characteristic zero. Then F contains a subﬁeld isomorphic to Q. Corollary 18.6 Let F be a ﬁeld of characteristic p. Then F contains a subﬁeld isomorphic to Zp. 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. 18.2 FACTORIZATION IN INTEGRAL DOMAINS 293 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 ∈ R such that b = ac. A unit in R is an element that has a multiplicative inverse. Two elements a and b in R are said to be associates if there exists a unit u in R such that a = ub. Let D be an integral domain. A nonzero element p ∈ D that is not a unit is said to be irreducible provided that whenever p = ab, either a or b is a unit. Furthermore, p is prime if whenever p \| ab either p \| a or p \| b. Example 2. 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 x2, y2, and xy. Each of these elements is irreducible in R; however, xy is not prime, since xy divides x2y2 but does not divide either x2 or y2. The Fundamental Theorem of Arithmetic states that every positive integer n > 1 can be factored into a product of prime numbers p1 · · · pk, where the pi’s are not necessarily distinct. We also know that such factorizations are unique up to the order of the pi’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. 1. Let a ∈ D such that a = 0 and a is not a unit. Then a can be written as the product of irreducible elements in D. 2. Let a = p1 · · · pr = q1 · · · qs, where the pi
’s and the qi’s are irreducible. Then r = s and there is a π ∈ Sr such that pi and qπ(j) are associates for j = 1,..., r. Example 3. The integers are a unique factorization domain by the Funda- mental Theorem of Arithmetic. √ √ Example 4. Not every integral domain is a unique factorization domain. The subring Z[ 3 i} of the complex numbers is an integral domain 3 i] = {a + b 3 i] → N ∪ {0} 3 i and deﬁne ν : Z[ (Exercise 12, Chapter 16). Let z = a+b by ν(z) = \|z\|2 = a2 + 3b2. It is clear that ν(z) ≥ 0 with equality when z = 0. Also, from our knowledge of complex numbers we know that ν(zw) = 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[ 3 i] are 1 and −1. √ √ √ 294 CHAPTER 18 INTEGRAL DOMAINS We claim that 4 has two distinct factorizations into irreducible elements1 − 3 i)(1 + 3 i). We must show that each of these factors is an irreducible element in Z[ 3 i]. If 2 is not irreducible, then 2 = zw for elements z, w in Z[ 3 i] where √ ν(z) = ν(w) = 2. However, there does not exist an element in z in Z[ 3 i] such that ν(z) = 2 because the equation a2 + 3b2 = 2 has no integer solutions. 3 i Therefore, 2 must be irreducible. A similar argument shows that both 1− and 1 + 3 i 3 i are irreducible. Since 2 is not a unit multiple of either 1 − √ 3 i, 4 has at least two distinct factorizations into irreducible elements. or 1 + √ √ √ √ √ Principal Ideal Domains Let R be a commutative ring with identity. Recall that a principal ideal generated by a ∈ R is an ideal of the form a =
{ra : r ∈ R}. An integral domain in which every ideal is principal is called a principal ideal domain, or PID. Lemma 18.7 Let D be an integral domain and let a, b ∈ D. Then 1. a \| b ⇔ b ⊂ a. 2. a and b are associates ⇔ b = a. 3. a is a unit in D ⇔ a = D. Proof. (1) Suppose that a \| b. Then b = ax for some x ∈ D. Hence, for every r in D, br = (ax)r = a(xr) and b ⊂ a. Conversely, suppose that b ⊂ a. Then b ∈ a. Consequently, b = ax for some x ∈ D. Thus, a \| b. (2) Since a and b are associates, there exists a unit u such that a = ub. Therefore, b \| a and a ⊂ b. Similarly, b ⊂ a. It follows that a = b. Conversely, suppose that a = b. By part (1), a \| b and b \| a. Then a = bx and b = ay for some x, y ∈ D. Therefore, a = bx = ayx. Since D is an integral domain, xy = 1; that is, x and y are units and a and b are associates. (3) An element a ∈ 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 18.8 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. 18.2 FACTORIZATION IN INTEGRAL DOMAINS 295 Proof. Suppose that p is a maximal ideal. If some element a in D divides p, then p ⊂ a. Since p is maximal, either D = a or 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 ⊂ a ⊂ D, then a \| p. Since p is irreducible, either a must be a unit or a and p are associates. Therefore, either D
= a or p = a. Thus, p is a maximal ideal. Corollary 18.9 Let D be a PID. If p is irreducible, then p is prime. Proof. Let p be irreducible and suppose that p \| ab. Then ab ⊂ p. By Corollary 16.18, since p is a maximal ideal, p must also be a prime ideal. Thus, either a ∈ p or b ∈ p. Hence, either p \| a or p \| b. Lemma 18.10 Let D be a PID. Let I1, I2,... be a set of ideals such that I1 ⊂ I2 ⊂ · · ·. Then there exists an integer N such that In = IN for all n ≥ N. Proof. We claim that I = ∞ i=1 Ii is an ideal of D. Certainly I is not empty, since I1 ⊂ I and 0 ∈ I. If a, b ∈ I, then a ∈ Ii and b ∈ Ij for some i and j in N. Without loss of generality we can assume that i ≤ j. Hence, a and b are both in Ij and so a − b is also in Ij. Now let r ∈ D and a ∈ I. Again, we note that a ∈ Ii for some positive integer i. Since Ii is an ideal, ra ∈ Ii 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 a ∈ D that generates I. Since a is in IN for some N ∈ N, we know that IN = I = a. Consequently, In = IN for n ≥ N. Any commutative ring satisfying the condition in Lemma 18.10 is said to satisfy the ascending chain condition, or ACC. Such rings are called Noetherian rings, after Emmy Noether. Theorem 18.11 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 = a1b1, where neither a1 nor b1 is a unit. Hence, a ⊂ a
1. By Lemma 18.7, we know that a = a1; otherwise, a and a1 would be associates and b1 would be a unit, which would contradict our assumption. Now suppose that a1 = a2b2, where neither a2 nor b2 is a 296 CHAPTER 18 INTEGRAL DOMAINS unit. By the same argument as before, a1 ⊂ a2. We can continue with this construction to obtain an ascending chain of ideals a ⊂ a1 ⊂ a2 ⊂ · · ·. By Lemma 18.10, there exists a positive integer N such that an = aN for all n ≥ N. Consequently, aN 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 = c1p1, where p1 is irreducible. If c1 is not a unit, we can repeat the preceding argument to conclude that a ⊂ c1. Either c1 is irreducible or c1 = c2p2, where p2 is irreducible and c2 is not a unit. Continuing in this manner, we obtain another chain of ideals a ⊂ c1 ⊂ c2 ⊂ · · ·. This chain must satisfy the ascending chain condition; therefore, a = p1p2 · · · pr for irreducible elements p1,..., pr. Uniqueness of the factorization. To show uniqueness, let a = p1p2 · · · pr = q1q2 · · · qs, where each pi and each qi is irreducible. Without loss of generality, we can assume that r < s. Since p1 divides q1q2 · · · qs, by Corollary 18.9 it must divide some qi. By rearranging the qi’s, we can assume that p1 \| q1; hence, q1 = u1p1 for some unit u1 in D. Therefore, or a = p1p2 · · · pr = u1p1q2 · · · qs p2 · · · pr = u1q2 · · · qs. Continuing in this manner, we can arrange the qi’s such that p2 = q2, p3 = q
3,..., pr = qr, to obtain u1u2 · · · urqr+1 · · · qs = 1. In this case qr+1 · · · qs is a unit, which contradicts the fact that qr+1,..., qs are irreducibles. Therefore, r = s and the factorization of a is unique. Corollary 18.12 Let F be a ﬁeld. Then F [x] is a UFD. 18.2 FACTORIZATION IN INTEGRAL DOMAINS 297 Example 5. Every PID is a UFD, but it is not the case that every UFD is a PID. In Corollary 18.22, we will prove that Z[x] is a UFD. However, Z[x] is not a PID. Let I = {5f (x) + xg(x) : f (x), g(x) ∈ Z[x]}. We can easily show that I is an ideal of Z[x]. Suppose that I = p(x). Since 5 ∈ I, 5 = f (x)p(x). In this case p(x) = p must be a constant. Since x ∈ I, x = pg(x); consequently, p = ±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 = 5f (x) + xg(x) for some f (x) and g(x) in Z[x]. Examining the constant term of this polynomial, we see that 3 = 5f (x), which is impossible. Euclidean Domains We have repeatedly used the division algorithm when proving results about either Z or 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 for each a ∈ D there is a nonneg- ative integer ν(a) satisfying the following conditions. 1. If a and b are nonzero elements in D, then ν(a) ≤ ν(ab). 2. Let a, b ∈ D and suppose that b = 0. Then there exist elements q, r ∈ D such that a = bq + 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 6. Absolute value on Z is a Euclidean valuation. Example 7. Let F be a ﬁeld. Then the degree of a polynomial in F [x] is a Euclidean valuation. Example 8. Recall that the Gaussian integers in Example 9 of Chapter 16 are deﬁned by Z[i] = {a + bi : a, b ∈ Z}. √ √ a2 + b2; however, We usually measure the size of a complex number a + bi by its absolute a2 + b2 may not be an integer. For our value, \|a + bi\| = valuation we will let ν(a + bi) = a2 + b2 to ensure that we have an integer. We claim that ν(a + bi) = a2 + b2 is a Euclidean valuation on Z[i]. Let z, w ∈ Z[i]. Then ν(zw) = \|zw\|2 = \|z\|2\|w\|2 = ν(z)ν(w). Since ν(z) ≥ 1 for every nonzero z ∈ Z[i], ν(z) = ν(z)ν(w). 298 CHAPTER 18 INTEGRAL DOMAINS Next, we must show that for any z = a + bi and w = c + di in Z[i] with w = 0, there exist elements q and r in Z[i] such that z = qw + r with either r = 0 or ν(r) < ν(w). We can view z and w as elements in Q(i) = {p + qi : p, q ∈ Q}, the ﬁeld of fractions of Z[i]. Observe that zw−1 = (a + bi) c − di c2 + d2 = ac + bd c2 + d2 + bc − ad c2 + d2 i n1 c2 + d2 + = m1 + m2 + = (m1 + m2i) + n1 c2 + d2 + = (m1 + m2i) + (s + ti) i n2 c2 + d2
n2 c2 + d2 i in 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 mi such that the fractional part satisﬁes \|ni/(a2 + b2)\| ≤ 1/2. For example, we write 9 8 15. Thus, s and t are the “fractional parts” of zw−1 = (m1 + m2i) + (s + ti). We also know that s2 + t2 ≤ 1/4 + 1/4 = 1/2. Multiplying by w, we have z = zw−1w = w(m1 + m2i) + w(s + ti) = qw + r, where q = m1 + m2i and r = w(s + ti). Since z and qw are in Z[i], r must be in Z[i]. Finally, we need to show that either r = 0 or ν(r) < ν(w). However, ν(r) = ν(w)ν(s + ti) ≤ 1 2 ν(w) < ν(w). Theorem 18.13 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 ∈ I such that ν(b) is minimal for all a ∈ I. Since D is a Euclidean domain, there exist elements q and r in D such that a = bq + r and either r = 0 or ν(r) < ν(b). But r = a − bq is in I since I is an ideal; therefore, r = 0 by the minimality of b. It follows that a = bq and I = b. 18.2 FACTORIZATION IN INTEGRAL DOMAINS 299 Corollary 18.14 Every Euclidean domain is a unique factorization domain. 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 17.9). Let D be a unique factorization domain and suppose that p(x) = anxn + · · · + a1x + a0 in D[x]. Then the content of p(x) is the greatest common divisor of a0,..., an. We say that p(x) is primitive if gcd(a0,..., an) = 1. Example 9. In Z[x] the polynomial p(x) = 5x4 − 3x3 + x − 4 is a primitive polynomial since the greatest common divisor of the coeﬃcients is 1; however, the polynomial q(x) = 4x2 − 6x + 8 is not primitive since the content of q(x) is 2. Theorem 18.15 (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. i=0 aixi and g(x) = n Proof. Let f (x) = m i=0 bixi. Suppose that p is a prime dividing the coeﬃcients of f (x)g(x). Let r be the smallest integer such that p \|ar and s be the smallest integer such that p \|bs. The coeﬃcient of xr+s in f (x)g(x) is cr+s = a0br+s + a1br+s−1 + · · · + ar+s−1b1 + ar+sb0. Since p divides a0,..., ar−1 and b0,..., bs−1, p divides every term of cr+s except for the term arbs. However, since p \| cr+s, either p divides ar or p divides bs. But this is impossible. Lemma 18.16 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) = cp1(x) and q(x) = dq1(x), where c and d are the contents of p(x) and q(x), respectively. Then p1(x) and q1(x) are primitive. We can now write p(x)q(x) = cdp1(x)q1(x). Since p1(x)q1(x) is primitive, the content of p(x)q(x) must be cd. 300 CHAPTER 18 INTEGRAL DOMAINS Lemma 18.17 Let D be a UFD and F its ﬁeld of fractions. Suppose that p(x) ∈ D[x] and p(x) = f (x)g(x), where f (x) and g(x) are in F [x]. Then p(x) = f1(x)g1(x), where f1(x) and g1(x) are in D[x]. Furthermore, deg f (x) = deg f1(x) and deg g(x) = deg g1(x). Proof. Let a and b be nonzero elements of D such that af (x), bg(x) are in D[x]. We can ﬁnd a1, b2 ∈ D such that af (x) = a1f1(x) and bg(x) = b1g1(x), where f1(x) and g1(x) are primitive polynomials in D[x]. Therefore, abp(x) = (a1f1(x))(b1g1(x)). Since f1(x) and g1(x) are primitive polynomials, it must be the case that ab \| a1b1 by Gauss’s Lemma. Thus there exists a c ∈ D such that p(x) = cf1(x)g1(x). Clearly, deg f (x) = deg f1(x) and deg g(x) = deg g1(x). The following corollaries are direct consequences of Lemma 18.17. Corollary 18.18 Let D be a UFD and F its ﬁeld of fractions. A primitive polynomial p(x) in D[x] is ir
reducible in F [x] if and only if it is irreducible in D[x]. Corollary 18.19 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) in F [x], then p(x) = f1(x)g1(x), where f1(x) and g1(x) are in D[x]. Furthermore, deg f (x) = deg f1(x) and deg g(x) = deg g1(x). Theorem 18.20 If D is a UFD, then D[x] is a UFD. Proof. Let p(x) be a nonzero polynomial in D[x]. If 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) = f1(x)f2(x) · · · fn(x) by a factorization of p(x), where each fi(x) is irreducible. Choose ai ∈ D such that aifi(x) is in D[x]. There exist b1,..., bn ∈ D such that aifi(x) = bigi(x), where gi(x) is a primitive polynomial in D[x]. By Corollary 18.18, each gi(x) is irreducible in D[x]. Consequently, we can write a1 · · · anp(x) = b1 · · · bng1(x) · · · gn(x). Let b = b1 · · · bn. Since g1(x) · · · gn(x) is primitive, a1 · · · an divides b. Therefore, p(x) = ag1(x) · · · gn(x), where a ∈ D. Since D is a UFD, we can factor a as uc1 · · · ck, where u is a unit and each of the ci’s is irreduc
ible in D. We will now show the uniqueness of this factorization. Let p(x) = a1 · · · amf1(x) · · · fn(x) = b1 · · · brg1(x) · · · gs(x) 18.2 FACTORIZATION IN INTEGRAL DOMAINS 301 be two factorizations of p(x), where all of the factors are irreducible in D[x]. By Corollary 18.18, each of the fi’s and gi’s is irreducible in F [x]. The ai’s and the bi’s are units in F. Since F [x] is a PID, it is a UFD; therefore, n = s. Now rearrange the gi(x)’s so that fi(x) and gi(x) are associates for i = 1,..., n. Then there exist c1,..., cn and d1,..., dn in D such that (ci/di)fi(x) = gi(x) or cifi(x) = digi(x). The polynomials fi(x) and gi(x) are primitive; hence, ci and di are associates in D. Thus, a1 · · · am = ub1 · · · br in D, where u is a unit in D. Since D is a unique factorization domain, m = s. Finally, we can reorder the bi’s so that ai and bi 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 18.21 Let F be a ﬁeld. Then F [x] is a UFD. Corollary 18.22 Z[x] is a UFD. Corollary 18.23 Let D be a UFD. Then D[x1,..., xn] is a UFD. Remark. It is important to notice that every Euclidean domain is a PID and every PID is a UFD. However, as demonstrated by our examples, 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]). 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 = 22n + 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 302 CHAPTER 18 INTEGRAL DOMAINS numbers. The acceptance of complex numbers was brought about by Gauss, who was the ﬁrst person to use the notation of i for −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 G¨ottingen, 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 G¨ottingen on February 23, 1855. Exercises 1. Let z = a + b √ 3 i be in Z[ √ Show that the only units of Z[ 3 i] are 1 and −1. 3 i]. If a2 + 3b2 = 1, show that z must be a unit. √ 2. The Gaussian integers, Z[i], are a UFD. Factor each of the following elements in Z[i] into a product of irreducibles. (a) 5 (b) 1 + 3i (c) 6 + 8i (d) 2 3. Let D be an integral domain. (a) Prove that FD is an abelian group under the operation of addition. EXERCISES 303 (b) Show that the operation of multiplication is well-deﬁned in the ﬁeld of fractions, FD. (c) Verify the associative and commutative properties for multiplication in FD. 4. Prove or disprove: Any subring of a ﬁeld F containing 1 is an integral domain. 5. Prove or disprove: If D is an integral domain, then every prime element in D is also irreducible in D. 6. Let F be a ﬁeld of characteristic zero. Prove that F contains a subﬁeld isomorphic to Q. 7. Let F be a ﬁeld. (a) Prove that the ﬁeld of fractions of F [x], denoted by F (x), is isomorphic to the set all rational expressions p(x)/q(x), where q(x) is not the zero polynomial. (b) Let p(x1,..., xn) and q(x1,..., xn) be polynomials
in F [x1,..., xn]. Show that the set of all rational expressions p(x1,..., xn)/q(x1,..., xn) is isomorphic to the ﬁeld of fractions of F [x1,..., xn]. We denote the ﬁeld of fractions of F [x1,..., xn] by F (x1,..., xn). 8. Let p be prime and denote the ﬁeld of fractions of Zp[x] by Zp(x). Prove that Zp(x) is an inﬁnite ﬁeld of characteristic p. 9. Prove that the ﬁeld of fractions of the Gaussian integers, Z[i], is Q(i) = {p + qi : p, q ∈ Q}. 10. A ﬁeld F is called a prime ﬁeld if it has no proper subﬁelds. If E is a subﬁeld of F and E is a prime ﬁeld, then E is a prime subﬁeld of F. (a) Prove that every ﬁeld contains a unique prime subﬁeld. (b) If F is a ﬁeld of characteristic 0, prove that the prime subﬁeld of F is isomorphic to the ﬁeld of rational numbers, Q. (c) If F is a ﬁeld of characteristic p, prove that the prime subﬁeld of F is isomorphic to Zp. √ √ 2 ] = {a + b 11. Let Z[ 2 : a, b ∈ Z}. √ 2 ] is an integral domain. (a) Prove that Z[ (b) Find all of the units in Z[ (c) Determine the ﬁeld of fractions of Z[ (d) Prove that Z[ ν(a + b 2 i) = a2 + 2b2. 2 ]. √ √ √ √ 2 ]. 2i] is a Euclidean domain under the Euclidean valuation 304 CHAPTER 18 INTEGRAL DOMAINS 12. Let D be a UFD.
An element d ∈ D is a greatest common divisor of a and b in D if d \| a and d \| b and d is divisible by any other element dividing both a and b. (a) If D is a PID and a and b are both nonzero elements of D, prove there exists a unique greatest common divisor of a and b up to associates. That is, if d and d are both greatest common divisors of a and b, then d and d are associates. We write gcd(a, b) for the greatest common divisor of a and b. (b) Let D be a PID and a and b be nonzero elements of D. Prove that there exist elements s and t in D such that gcd(a, b) = as + bt. 13. Let D be an integral domain. Deﬁne a relation on D by a ∼ b if a and b are associates in D. Prove that ∼ is an equivalence relation on D. 14. Let D be a Euclidean domain with Euclidean valuation ν. If u is a unit in D, show that ν(u) = ν(1). 15. Let D be a Euclidean domain with Euclidean valuation ν. If a and b are associates in D, prove that ν(a) = ν(b). √ 16. Show that Z[ 5 i] is not a unique factorization domain. 17. Prove or disprove: Every subdomain of a UFD is also a UFD. 18. An ideal of a commutative ring R is said to be ﬁnitely generated if there exist elements a1,..., an in R such that every element r ∈ R can be written as a1r1 + · · · + anrn for some r1,..., rn in R. Prove that R satisﬁes the ascending chain condition if and only if every ideal of R is ﬁnitely generated. 19. Let D be an integral domain with a descending chain of ideals I1 ⊃ I2 ⊃ I3 ⊃ · · ·. Suppose that there exists an N such that Ik = IN for all k ≥ N. A ring satisfying this condition is said to satisfy the descending chain condition, or DCC. Rings satisfying the DCC are called
Artinian rings, after Emil Artin. Show that if D satisﬁes the descending chain condition, it must satisfy the ascending chain condition. 20. Let R be a commutative ring with identity. We deﬁne a multiplicative subset of R to be a subset S such that 1 ∈ S and ab ∈ S if a, b ∈ S. (a) Deﬁne a relation ∼ on R × S by (a, s) ∼ (a, s) if there exists an s∗ ∈ S such that s∗(sa − sa) = 0. Show that ∼ is an equivalence relation on R × S. (b) Let a/s denote the equivalence class of (a, s) ∈ R × S and let S−1R be the set of all equivalence classes with respect to ∼. Deﬁne the operations of addition and multiplication on S−1R by a s b t b t + a s = = at + bs st ab st, EXERCISES 305 respectively. Prove that these operations are well-deﬁned on S−1R and that S−1R is a ring with identity under these operations. The ring S−1R is called the ring of quotients of R with respect to S. (c) Show that the map ψ : R → S−1R deﬁned by ψ(a) = a/1 is a ring homomorphism. (d) If R has no zero divisors and 0 /∈ S, show that ψ is one-to-one. (e) Prove that P is a prime ideal of R if and only if S = R \ P is a multiplicative subset of R. (f) If P is a prime ideal of R and S = R \ P, show that the ring of quotients S−1R has a unique maximal ideal. Any ring that has a unique maximal ideal is called a local ring. References and Suggested Readings [1] Atiyah, M. F. and MacDonald, I. G. Introduction to Commutative Algebra. Westview Press, Boulder, CO, 1994. [2] Zariski, O. and Samuel, P. Commutative Algebra, vols. I and II. Springer, New York, 1975, 1960. Sage Sage supports distinctions between �
�plain” rings, domains, principal ideal domains and ﬁelds. Support is often very good for constructions and computations with PID’s, but sometimes problems get signiﬁcantly harder (computationally) when a ring has less structure that that of a PID. So be aware when using Sage that some questions may go unanswered for rings with less structure. 19 Lattices and Boolean Algebras The axioms of a ring give structure to the operations of addition and multiplication on a set. However, we can construct algebraic structures, known as lattices and Boolean algebras, that generalize other types of operations. For example, the important operations on sets are inclusion, union, and intersection. Lattices are generalizations of order relations on algebraic spaces, such as set inclusion in set theory and inequality in the familiar number systems N, Z, Q, and R. Boolean algebras generalize the operations of intersection and union. Lattices and Boolean algebras have found applications in logic, circuit theory, and probability. 19.1 Lattices Partially Ordered Sets We begin by the study of lattices and Boolean algebras by generalizing the idea of inequality. Recall that a relation on a set X is a subset of X × X. A relation P on X is called a partial order of X if it satisﬁes the following axioms. 1. The relation is reﬂexive: (a, a) ∈ P for all a ∈ X. 2. The relation is antisymmetric: if (a, b) ∈ P and (b, a) ∈ P, then a = b. 3. The relation is transitive: if (a, b) ∈ P and (b, c) ∈ P, then (a, c) ∈ P. 306 19.1 LATTICES 307 We will usually write a b to mean (a, b) ∈ P unless some symbol is naturally associated with a particular partial order, such as a ≤ b with integers a and b, or X ⊆ Y with sets X and Y. A set X together with a partial order is called a partially ordered set, or poset. Example 1. The set of integers (or rationals or reals) is a poset where a ≤ b has the usual meaning for two integers
a and b in Z. Example 2. Let X be any set. We will deﬁne the power set of X to be the set of all subsets of X. We denote the power set of X by P(X). For example, let X = {a, b, c}. Then P(X) is the set of all subsets of the set {a, b, c}: ∅ {a} {b} {c} {a, b} {a, c} {b, c} {a, b, c}. On any power set of a set X, set inclusion, ⊆, is a partial order. We can represent the order on {a, b, c} schematically by a diagram such as the one in Figure 19.1. {a, b, c} {a, b} {a, c} {b, c} {a} {b} {c} ∅ Figure 19.1. Partial order on P({a, b, c}) Example 3. Let G be a group. The set of subgroups of G is a poset, where the partial order is set inclusion. Example 4. There can be more than one partial order on a particular set. We can form a partial order on N by a b if a \| b. The relation is certainly reﬂexive since a \| a for all a ∈ N. If m \| n and n \| m, then m = n; hence, the 308 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS relation is also antisymmetric. The relation is transitive, because if m \| n and n \| p, then m \| p. Example 5. Let X = {1, 2, 3, 4, 6, 8, 12, 24} be the set of divisors of 24 with the partial order deﬁned in Example 4. Figure 19.2 shows the partial order on X. 8 4 2 24 1 12 6 3 Figure 19.2. A partial order on the divisors of 24 Let Y be a subset of a poset X. An element u in X is an upper bound of Y if a u for every element a ∈ Y. If u is an upper bound of Y such that u v for every other upper bound v of Y, then u is called a least upper bound or supremum of Y. An element l in
X is said to be a lower bound of Y if l a for all a ∈ Y. If l is a lower bound of Y such that k l for every other lower bound k of Y, then l is called a greatest lower bound or inﬁmum of Y. Example 6. Let Y = {2, 3, 4, 6} be contained in the set X of Example 5. Then Y has upper bounds 12 and 24, with 12 as a least upper bound. The only lower bound is 1; hence, it must be a greatest lower bound. As it turns out, least upper bounds and greatest lower bounds are unique if they exist. Theorem 19.1 Let Y be a nonempty subset of a poset X. If Y has a least upper bound, then Y has a unique least upper bound. If Y has a greatest lower bound, then Y has a unique greatest lower bound. 19.1 LATTICES 309 Proof. Let u1 and u2 be least upper bounds for Y. By the deﬁnition of the least upper bound, u1 u for all upper bounds u of Y. In particular, u1 u2. Similarly, u2 u1. Therefore, u1 = u2 by antisymmetry. A similar argument show that the greatest lower bound is unique. On many posets it is possible to deﬁne binary operations by using the greatest lower bound and the least upper bound of two elements. A lattice is a poset L such that every pair of elements in L has a least upper bound and a greatest lower bound. The least upper bound of a, b ∈ L is called the join of a and b and is denoted by a ∨ b. The greatest lower bound of a, b ∈ L is called the meet of a and b and is denoted by a ∧ b. Example 7. Let X be a set. Then the power set of X, P(X), is a lattice. For two sets A and B in P(X), the least upper bound of A and B is A ∪ B. Certainly A ∪ B is an upper bound of A and B, since A ⊆ A ∪ B and B ⊆ A ∪ B. If C is some other set containing both A and B, then C must contain A ∪ B; hence, A ∪ B is the least upper bound of A and B. Similarly
, the greatest lower bound of A and B is A ∩ B. Example 8. Let G be a group and suppose that X is the set of subgroups of G. Then X is a poset ordered by set-theoretic inclusion, ⊆. The set of subgroups of G is also a lattice. If H and K are subgroups of G, the greatest lower bound of H and K is H ∩ K. The set H ∪ K may not be a subgroup of G. We leave it as an exercise to show that the least upper bound of H and K is the subgroup generated by H ∪ K. In set theory we have certain duality conditions. For example, by De Morgan’s laws, any statement about sets that is true about (A ∪ B) must also be true about A ∩ B. We also have a duality principle for lattices. Principle of Duality. Any statement that is true for all lattices remains true when is replaced by and ∨ and ∧ are interchanged throughout the statement. The following theorem tells us that a lattice is an algebraic structure with two binary operations that satisfy certain axioms. Theorem 19.2 If L is a lattice, then the binary operations ∨ and ∧ satisfy the following properties for a, b, c ∈ L. 1. Commutative laws: a ∨ b = b ∨ a and a ∧ b = b ∧ a. 2. Associative laws: a ∨ (b ∨ c) = (a ∨ b) ∨ c and a ∧ (b ∧ c) = (a ∧ b) ∧ c. 310 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS 3. Idempotent laws: a ∨ a = a and a ∧ a = a. 4. Absorption laws: a ∨ (a ∧ b) = a and a ∧ (a ∨ b) = a. Proof. By the Principle of Duality, we need only prove the ﬁrst statement in each part. (1) By deﬁnition a ∨ b is the least upper bound of {a, b}, and b ∨ a is the least upper bound of {b, a}; however, {a, b} = {b, a}. (2) We will show that
a ∨ (b ∨ c) and (a ∨ b) ∨ c are both least upper bounds of {a, b, c}. Let d = a ∨ b. Then c d ∨ c = (a ∨ b) ∨ c. We also know that a ∨ b) ∨ c. A similar argument demonstrates that b (a ∨ b) ∨ c. Therefore, (a ∨ b) ∨ c is an upper bound of {a, b, c}. We now need to show that (a ∨ b) ∨ c is the least upper bound of {a, b, c}. Let u be some other upper bound of {a, b, c}. Then a u and b u; hence, d = a ∨ b u. Since c u, it follows that (a ∨ b) ∨ c = d ∨ c u. Therefore, (a ∨ b) ∨ c must be the least upper bound of {a, b, c}. The argument that shows a ∨ (b ∨ c) is the least upper bound of {a, b, c} is the same. Consequently, a ∨ (b ∨ c) = (a ∨ b) ∨ c. (3) The join of a and a is the least upper bound of {a}; hence, a ∨ a = a. (4) Let d = a ∧ b. Then a a ∨ d. On the other hand, d = a ∧ b a, and so a ∨ d a. Therefore, a ∨ (a ∧ b) = a. Given any arbitrary set L with operations ∨ and ∧, satisfying the conditions of the previous theorem, it is natural to ask whether or not this set comes from some lattice. The following theorem says that this is always the case. Theorem 19.3 Let L be a nonempty set with two binary operations ∨ and ∧ satisfying the commutative, associative, idempotent, and absorption laws. We can deﬁne a partial order on L by a b if a ∨ b = b. Furthermore, L is a lattice with respect to if for all a, b ∈ L, we deﬁne the least upper bound and greatest lower bound of a and b by a ∨ b and a ∧ b, respectively. Proof
. We ﬁrst show that L is a poset under. Since a ∨ a = a, a a and is reﬂexive. To show that is antisymmetric, let a b and b a. Then a ∨ b = b and b ∨ a = a. By the commutative law. Finally, we must show that is transitive. Let a b and b c. Then a ∨ b = b and b ∨ c = c. Thus, a ∨ c = a ∨ (b ∨ c) = (a ∨ b) ∨ c = b ∨ c = c, 19.2 BOOLEAN ALGEBRAS 311 or a c. To show that L is a lattice, we must prove that a ∨ b and a ∧ b are, respectively, the least upper and greatest lower bounds of a and b. Since a = (a ∨ b) ∧ a = a ∧ (a ∨ b), it follows that a a ∨ b. Similarly, b a ∨ b. Therefore, a ∨ b is an upper bound for a and b. Let u be any other upper bound of both a and b. Then a u and b u. But a ∨ b u since (a ∨ b) ∨ u = a ∨ (b ∨ u) = a ∨ u = u. The proof that a ∧ b is the greatest lower bound of a and b is left as an exercise. 19.2 Boolean Algebras Let us investigate the example of the power set, P(X), of a set X more closely. The power set is a lattice that is ordered by inclusion. By the deﬁnition of the power set, the largest element in P(X) is X itself and the smallest element is ∅, the empty set. For any set A in P(X), we know that A ∩ X = A and A ∪ ∅ = A. This suggests the following deﬁnition for lattices. An element I in a poset X is a largest element if a I for all a ∈ X. An element O is a smallest element of X if O a for all a ∈ X. Let A be in P(X). Recall that the complement of A is A = X \ A = {x : x ∈ X
and x /∈ A}. We know that A ∪ A = X and A ∩ A = ∅. We can generalize this example for lattices. A lattice L with a largest element I and a smallest element O is complemented if for each a ∈ X, there exists an a such that a ∨ a = I and a ∧ a = O. In a lattice L, the binary operations ∨ and ∧ satisfy commutative and associative laws; however, they need not satisfy the distributive law a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c); however, in P(X) the distributive law is satisﬁed since A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) for A, B, C ∈ P(X). We will say that a lattice L is distributive if the following distributive law holds: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) for all a, b, c ∈ L. 312 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS Theorem 19.4 A lattice L is distributive if and only if a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) for all a, b, c ∈ L. Proof. Let us assume that L is a distributive lattice. a ∨ (b ∧ c) = [a ∨ (a ∧ c)] ∨ (b ∧ c) = a ∨ [(a ∧ c) ∨ (b ∧ c)] = a ∨ [(c ∧ a) ∨ (c ∧ b)] = a ∨ [c ∧ (a ∨ b)] = a ∨ [(a ∨ b) ∧ c] = [(a ∨ b) ∧ a] ∨ [(a ∨ b) ∧ c] = (a ∨ b) ∧ (a ∨ c). The converse follows directly from the Duality Principle. A Boolean algebra is a lattice B with a greatest element I and a smallest element O such that B is both distributive and complemented. The power set of X, P(X), is our prototype
for a Boolean algebra. As it turns out, it is also one of the most important Boolean algebras. The following theorem allows us to characterize Boolean algebras in terms of the binary relations ∨ and ∧ without mention of the fact that a Boolean algebra is a poset. Theorem 19.5 A set B is a Boolean algebra if and only if there exist binary operations ∨ and ∧ on B satisfying the following axioms. 1. a ∨ b = b ∨ a and a ∧ b = b ∧ a for a, b ∈ B. 2. a ∨ (b ∨ c) = (a ∨ b) ∨ c and a ∧ (b ∧ c) = (a ∧ b) ∧ c for a, b, c ∈ B. 3. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) and a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) for a, b, c ∈ B. 4. There exist elements I and O such that a ∨ O = a and a ∧ I = a for all a ∈ B. 5. For every a ∈ B there exists an a ∈ B such that a ∨ a = I and a ∧ a = O. 19.2 BOOLEAN ALGEBRAS 313 Proof. Let B be a set satisfying (1)–(5) in the theorem. One of the idempotent laws is satisﬁed since a = a ∨ O = a ∨ (a ∧ a) = (a ∨ a) ∧ (a ∨ a) = (a ∨ a) ∧ I = a ∨ a. Observe that I ∨ b = (I ∨ b) ∧ I = (I ∧ I) ∨ (b ∧ I) = I ∨ I = I. Consequently, the ﬁrst of the two absorption laws holds, since a ∨ (a ∧ b) = (a ∧ I) ∨ (a ∧ b) = a ∧ (I ∨ b) = a ∧ I = a. The other idempotent and absorption laws are proven similarly. Since B also satisﬁes
(1)–(3), the conditions of Theorem 19.3 are met; therefore, B must be a lattice. Condition (4) tells us that B is a distributive lattice. For a ∈ B, O ∨ a = a; hence, O a and O is the smallest element in B. To show that I is the largest element in B, we will ﬁrst show that a ∨ b = b is equivalent to a ∧ b = a. Since a ∨ I = a for all a ∈ B, using the absorption laws we can determine that a ∨ I = (a ∧ I) ∨ I = I ∨ (I ∧ a) = I or a I for all a in B. Finally, since we know that B is complemented by (5), B must be a Boolean algebra. Conversely, suppose that B is a Boolean algebra. Let I and O be the greatest and least elements in B, respectively. If we deﬁne a ∨ b and a ∧ b as least upper and greatest lower bounds of {a, b}, then B is a Boolean algebra by Theorem 19.3, Theorem 19.4, and our hypothesis. Many other identities hold in Boolean algebras. Some of these identities are listed in the following theorem. Theorem 19.6 Let B be a Boolean algebra. Then 314 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS 1. a ∨ I = I and a ∧ O = O for all a ∈ B. 2. If a ∨ b = a ∨ c and a ∧ b = a ∧ c for a, b, c ∈ B, then b = c. 3. If a ∨ b = I and a ∧ b = O, then b = a. 4. (a) = a for all a ∈ B. 5. I = O and O = I. 6. (a ∨ b) = a ∧ b and (a ∧ b) = a ∨ b (De Morgan’s Laws). Proof. We will prove only (2). The rest of the identities are left as exercises. For a ∨ b = a ∨ c and a ∧ b = a ∧ c, we have b = b ∨ (b ∧ a) = b ∨ (a ∧ b)
= b ∨ (a ∧ c) = (b ∨ a) ∧ (b ∨ c) = (a ∨ b) ∧ (b ∨ c) = (a ∨ c) ∧ (b ∨ c) = (c ∨ a) ∧ (c ∨ b) = c ∨ (a ∧ b) = c ∨ (a ∧ c) = c ∨ (c ∧ a) = c. Finite Boolean Algebras A Boolean algebra is a ﬁnite Boolean algebra if it contains a ﬁnite number of elements as a set. Finite Boolean algebras are particularly nice since we can classify them up to isomorphism. Let B and C be Boolean algebras. A bijective map φ : B → C is an isomorphism of Boolean algebras if φ(a ∨ b) = φ(a) ∨ φ(b) φ(a ∧ b) = φ(a) ∧ φ(b) for all a and b in B. 19.2 BOOLEAN ALGEBRAS 315 We will show that any ﬁnite Boolean algebra is isomorphic to the Boolean algebra obtained by taking the power set of some ﬁnite set X. We will need a few lemmas and deﬁnitions before we prove this result. Let B be a ﬁnite Boolean algebra. An element a ∈ B is an atom of B if a = O and a ∧ b = a for all nonzero b ∈ B. Equivalently, a is an atom of B if there is no nonzero b ∈ B distinct from a such that O b a. Lemma 19.7 Let B be a ﬁnite Boolean algebra. If b is a nonzero element of B, then there is an atom a in B such that a b. Proof. If b is an atom, let a = b. Otherwise, choose an element b1, not equal to O or b, such that b1 b. We are guaranteed that this is possible since b is not an atom. If b1 is an atom, then we are done. If not, choose b2, not equal to O or b1, such that b2 b1. Again, if b2 is
an atom, let a = b2. Continuing this process, we can obtain a chain O · · · b3 b2 b1 b. Since B is a ﬁnite Boolean algebra, this chain must be ﬁnite. That is, for some k, bk is an atom. Let a = bk. Lemma 19.8 Let a and b be atoms in a ﬁnite Boolean algebra B such that a = b. Then a ∧ b = O. Proof. Since a ∧ b is the greatest lower bound of a and b, we know that a ∧ b a. Hence, either a ∧ b = a or a ∧ b = O. However, if a ∧ b = a, then either a b or a = O. In either case we have a contradiction because a and b are both atoms; therefore, a ∧ b = O. Lemma 19.9 Let B be a Boolean algebra and a, b ∈ B. The following statements are equivalent. 1. a b. 2. a ∧ b = O. 3. a ∨ b = I. Proof. (1) ⇒ (2). If a b, then a ∨ b = b. Therefore, a ∧ b = a ∧ (a ∨ b) = a ∧ (a ∧ b) = (a ∧ a) ∧ b = O ∧ b = O. 316 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS (2) ⇒ (3). If a ∧ b = O, then a ∨ b = (a ∧ b) = O = I. (3) ⇒ (1). If a ∨ b = I, then a = a ∧ (a ∨ b) = (a ∧ a) ∨ (a ∧ b) = O ∨ (a ∧ b) = a ∧ b. Thus, a b. Lemma 19.10 Let B be a Boolean algebra and b and c be elements in B such that b c. Then there exists an atom a ∈ B such that a b and a c. Proof. By Lemma 19.9, b ∧ c = O. Hence, there exists an atom a such that a b ∧ c. Consequently, a b and a c. Lemma 19.11 Let b ∈ B and a
1,..., an be the atoms of B such that ai b. Then b = a1 ∨ · · · ∨ an. Furthermore, if a, a1,..., an are atoms of B such that a b, ai b, and b = a ∨ a1 ∨ · · · ∨ an, then a = ai for some i = 1,..., n. Proof. Let b1 = a1 ∨ · · · ∨ an. Since ai b for each i, we know that b1 b. If we can show that b b1, then the lemma is true by antisymmetry. Assume b b1. Then there exists an atom a such that a b and a b1. Since a is an atom and a b, we can deduce that a = ai for some ai. However, this is impossible since a b1. Therefore, b b1. Now suppose that b = a1 ∨ · · · ∨ an. If a is an atom less than ba1 ∨ · · · ∨ an) = (a ∧ a1) ∨ · · · ∨ (a ∧ an). But each term is O or a with a ∧ ai occurring for only one ai. Hence, by Lemma 19.8, a = ai for some i. Theorem 19.12 Let B be a ﬁnite Boolean algebra. Then there exists a set X such that B is isomorphic to P(X). Proof. We will show that B is isomorphic to P(X), where X is the set of atoms of B. Let a ∈ B. By Lemma 19.11, we can write a uniquely as a = a1 ∨ · · · ∨ an for a1,..., an ∈ X. Consequently, we can deﬁne a map φ : B → P(X) by φ(a) = φ(a1 ∨ · · · ∨ an) = {a1,..., an}. 19.3 THE ALGEBRA OF ELECTRICAL CIRCUITS 317 Clearly, φ is onto. Now let a = a1 ∨· · ·∨an and b = b1 ∨· · ·∨bm be elements in B, where each a
i and each bi is an atom. If φ(a) = φ(b), then {a1,..., an} = {b1,..., bm} and a = b. Consequently, φ is injective. The join of a and b is preserved by φ since φ(a ∨ b) = φ(a1 ∨ · · · ∨ an ∨ b1 ∨ · · · ∨ bm) = {a1,..., an, b1,..., bm} = {a1,..., an} ∪ {b1,..., bm} = φ(a1 ∨ · · · ∨ an) ∪ φ(b1 ∧ · · · ∨ bm) = φ(a) ∪ φ(b). Similarly, φ(a ∧ b) = φ(a) ∩ φ(b). We leave the proof of the following corollary as an exercise. Corollary 19.13 The order of any ﬁnite Boolean algebra must be 2n for some positive integer n. 19.3 The Algebra of Electrical Circuits The usefulness of Boolean algebras has become increasingly apparent over the past several decades with the development of the modern computer. The circuit design of computer chips can be expressed in terms of Boolean algebras. In this section we will develop the Boolean algebra of electrical circuits and switches; however, these results can easily be generalized to the design of integrated computer circuitry. A switch is a device, located at some point in an electrical circuit, that controls the ﬂow of current through the circuit. Each switch has two possible states: it can be open, and not allow the passage of current through the circuit, or a it can be closed, and allow the passage of current. These states are mutually exclusive. We require that every switch be in one state or the other: a switch cannot be open and closed at the same time. Also, if one switch is always in the same state as another, we will denote both by the same letter; that is, two switches that are both labeled with the same letter a will always be open at the same time and closed at the same time. Given two switches, we can construct two fundamental types of circuits. Two switches a and b are in series
if they make up a circuit of the type that is illustrated in Figure 19.3. Current can pass between the terminals A and B in a series circuit only if both of the switches a and b are closed. We 318 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS will denote this combination of switches by a ∧ b. Two switches a and b are in parallel if they form a circuit of the type that appears in Figure 19.4. In the case of a parallel circuit, current can pass between A and B if either one of the switches is closed. We denote a parallel combination of circuits a and b by a ∨ b. A a b B Figure 19.3. a ∧ b A a b B Figure 19.4. a ∨ b We can build more complicated electrical circuits out of series and parallel circuits by replacing any switch in the circuit with one of these two fundamental types of circuits. Circuits constructed in this manner are called series-parallel circuits. We will consider two circuits equivalent if they act the same. That is, if we set the switches in equivalent circuits exactly the same we will obtain the same result. For example, in a series circuit a ∧ b is exactly the same as b ∧ a. Notice that this is exactly the commutative law for Boolean algebras. In fact, the set of all series-parallel circuits forms a Boolean algebra under the operations of ∨ and ∧. We can use diagrams to verify the diﬀerent axioms of a Boolean algebra. The distributive law, a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c), is illustrated in Figure 19.5. If a is a switch, then a is the switch that is always open when a is closed and always closed when a is open. A circuit that is always closed is I in our algebra; a circuit that is always open is O. The laws for a ∧ a = O and a ∨ a = I are shown in Figure 19.6. Example 9. Every Boolean expression represents a switching circuit. For example, given the expression (a ∨ b) ∧ (a ∨ b) ∧ (a ∨ b), we can construct the circuit in Figure 19.7. Theorem 19.14 The set of all circuits is a Boolean algebra. 19.3 THE ALGEBRA OF ELECTRICAL C
IRCUITS 319 a b c a a b c Figure 19.5. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) a a a a Figure 19.6. a ∧ a = O and a ∨ a = I We leave as an exercise the proof of this theorem for the Boolean algebra axioms not yet veriﬁed. We can now apply the techniques of Boolean algebras to switching theory. Example 10. Given a complex circuit, we can now apply the techniques of Boolean algebra to reduce it to a simpler one. Consider the circuit in Figure 19.7. Since (a ∨ b) ∧ (a ∨ b) ∧ (a ∨ b) = (a ∨ b) ∧ (a ∨ b) ∧ (a ∨ b) = (a ∨ b) ∧ (a ∨ b) = a ∨ (b ∧ b) = a ∨ O = a, a b a b a b Figure 19.7. (a ∨ b) ∧ (a ∨ b) ∧ (a ∨ b) 320 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS we can replace the more complicated circuit with a circuit containing the single switch a and achieve the same function. Historical Note George Boole (1815–1864) was the ﬁrst person to study lattices. In 1847, he published The Investigation of the Laws of Thought, a book in which he used lattices to formalize logic and the calculus of propositions. Boole believed that mathematics was the study of form rather than of content; that is, he was not so much concerned with what he was calculating as with how he was calculating it. Boole’s work was carried on by his friend Augustus De Morgan (1806–1871). De Morgan observed that the principle of duality often held in set theory, as is illustrated by De Morgan’s laws for set theory. He believed, as did Boole, that mathematics was the study of symbols and abstract operations. Set theory and logic were further advanced by such mathematicians as Alfred North Whitehead (1861–1947), Bertrand Russell (1872–1970), and David Hilbert (1862–1943). In Principia Mathematica, Whitehead and Russell attempted to show the connection between
mathematics and logic by the deduction of the natural number system from the rules of formal logic. If the natural numbers could be determined from logic itself, then so could much of the rest of existing mathematics. Hilbert attempted to build up mathematics by using symbolic logic in a way that would prove the consistency of mathematics. His approach was dealt a mortal blow by Kurt G¨odel (1906–1978), who proved that there will always be “undecidable” problems in any suﬃciently rich axiomatic system; that is, that in any mathematical system of any consequence, there will always be statements that can never be proven either true or false. As often occurs, this basic research in pure mathematics later became indispensable in a wide variety of applications. Boolean algebras and logic have become essential in the design of the large-scale integrated circuitry found on today’s computer chips. Sociologists have used lattices and Boolean algebras to model social hierarchies; biologists have used them to describe biosystems. Exercises 1. Draw the lattice diagram for the power set of X = {a, b, c, d} with the set inclusion relation, ⊆. 2. Draw the diagram for the set of positive integers that are divisors of 30. Is this poset a Boolean algebra? 3. Draw a diagram of the lattice of subgroups of Z12. 4. Let B be the set of positive integers that are divisors of 36. Deﬁne an order on B by a b if a \| b. Prove that B is a Boolean algebra. Find a set X such that B is isomorphic to P(X). EXERCISES 321 5. Prove or disprove: Z is a poset under the relation a b if a \| b. 6. Draw the switching circuit for each of the following Boolean expressions. (a) (a ∨ b ∨ a) ∧ a (b) (a ∨ b) ∧ (a ∨ b) (c) a ∨ (a ∧ b) (d) (c ∨ a ∨ b) ∧ c ∧ (a ∨ b) 7. Draw a circuit that will be closed exactly when only one of three switches a, b, and c are closed. 8. Prove or disprove that the two circuits shown are equivalent. Let X be a ﬁnite set containing
n elements. Prove that P(X) = 2n. Conclude that the order of any ﬁnite Boolean algebra must be 2n for some n ∈ N. 10. For each of the following circuits, write a Boolean expression. If the circuit can be replaced by one with fewer switches, give the Boolean expression and draw a diagram for the new circuit 11. Prove or disprove: The set of all nonzero integers is a lattice, where a b is deﬁned by a \| b. 12. Prove that a ∧ b is the greatest lower bound of a and b in Theorem 19.3. 322 CHAPTER 19 LATTICES AND BOOLEAN ALGEBRAS 13. Let L be a nonempty set with two binary operations ∨ and ∧ satisfying the commutative, associative, idempotent, and absorption laws. We can deﬁne a partial order on L, as in Theorem 19.3, by a b if a ∨ b = b. Prove that the greatest lower bound of a and b is a ∧ b. 14. Let G be a group and X be the set of subgroups of G ordered by set-theoretic inclusion. If H and K are subgroups of G, show that the least upper bound of H and K is the subgroup generated by H ∪ K. 15. Let R be a ring and suppose that X is the set of ideals of R. Show that X is a poset ordered by set-theoretic inclusion, ⊆. Deﬁne the meet of two ideals I and J in X by I ∩ J and the join of I and J by I + J. Prove that the set of ideals of R is a lattice under these operations. 16. Let B be a Boolean algebra. Prove each of the following identities. (a) a ∨ I = I and a ∧ O = O for all a ∈ B. (b) If a ∨ b = I and a ∧ b = O, then b = a. (c) (a) = a for all a ∈ B. (d) I = O and O = I. (e) (a ∨ b) = a ∧ b and (a ∧ b) = a ∨ b (De Morgan’s laws). 17. By drawing the appropriate
diagrams, complete the proof of Theorem 19.14 to show that the switching functions form a Boolean algebra. 18. Let B be a Boolean algebra. Deﬁne binary operations + and · on B by a + b = (a ∧ b) ∨ (a ∧ b) a · b = a ∧ b. Prove that B is a commutative ring under these operations satisfying a2 = a for all a ∈ B. 19. Let X be a poset such that for every a and b in X, either a b or b a. Then X is said to be a totally ordered set. (a) Is a \| b a total order on N? (b) Prove that N, Z, Q, and R are totally ordered sets under the usual ordering ≤. 20. Let X and Y be posets. A map φ : X → Y is order-preserving if a b implies that φ(a) φ(b). Let L and M be lattices. A map ψ : L → M is a lattice homomorphism if ψ(a ∨ b) = ψ(a) ∨ ψ(b) and ψ(a ∧ b) = ψ(a) ∧ ψ(b). Show that every lattice homomorphism is order-preserving, but that it is not the case that every order-preserving homomorphism is a lattice homomorphism. 21. Let B be a Boolean algebra. Prove that a = b if and only if (a∧b)∨(a∧b) = O for a, b ∈ B. EXERCISES 323 Table 19.1. Boolean polynomials 22. Let B be a Boolean algebra. Prove that a = 0 if and only if (a∧b)∨(a ∧b) = b for all b ∈ B. 23. Let L and M be lattices. Deﬁne an order relation on L × M by (a, b) (c, d) if a c and b d. Show that L × M is a lattice under this partial order. Programming Exercises A Boolean or switching function on n variables is a map f : {O, I}n → {0, I}. A Boolean polynomial is a special type of Boolean function: it is any
type of Boolean expression formed from a ﬁnite combination of variables x1,..., xn together with O and I, using the operations ∨, ∧, and. The values of the functions are deﬁned in Table 19.1. Write a program to evaluate Boolean polynomials. References and Suggested Readings [1] Donnellan, T. Lattice Theory. Pergamon Press, Oxford, 1968. [2] Halmos, P. R. “The Basic Concepts of Algebraic Logic,” American Mathe- matical Monthly 53 (1956), 363–87. [3] Hohn, F. “Some Mathematical Aspects of Switching,” American Mathematical Monthly 62 (1955), 75–90. [4] Hohn, F. Applied Boolean Algebra. 2nd ed. Macmillan, New York, 1966. [5] Lidl, R. and Pilz, G. Applied Abstract Algebra. 2nd ed. Springer, New York, 1998. [6] Whitesitt, J. Boolean Algebra and Its Applications. Dover, Mineola, NY, 2010. Sage Sage has a full suite of functionality for both posets and lattices, all as part of its excellent support for combinatorics. There is little in this chapter that cannot be investigated with Sage. 20 Vector Spaces In a physical system a quantity can often be described with a single number. For example, we need to know only a single number to describe temperature, mass, or volume. However, for some quantities, such as location, we need several numbers. To give the location of a point in space, we need x, y, and z coordinates. Temperature distribution over a solid object requires four numbers: three to identify each point within the object and a fourth to describe the temperature at that point. Often n-tuples of numbers, or vectors, also have certain algebraic properties, such as addition or scalar multiplication. In this chapter we will examine mathematical structures called vector spaces. As with groups and rings, it is desirable to give a simple list of axioms that must be satisﬁed to make a set of vectors a structure worth studying. 20.1 Deﬁnitions and Examples A vector space V over a ﬁeld F is an abelian group with a scalar product α · v or αv
deﬁned for all α ∈ F and all v ∈ V satisfying the following axioms. • α(βv) = (αβ)v; • (α + β)v = αv + βv; • α(u + v) = αu + αv; • 1v = v; where α, β ∈ F and u, v ∈ V. The elements of V are called vectors; the elements of F are called scalars. It is important to notice that in most cases two vectors cannot be 324 20.1 DEFINITIONS AND EXAMPLES 325 multiplied. In general, it is only possible to multiply a vector with a scalar. To diﬀerentiate between the scalar zero and the vector zero, we will write them as 0 and 0, respectively. Let us examine several examples of vector spaces. Some of them will be quite familiar; others will seem less so. Example 1. The n-tuples of real numbers, denoted by Rn, form a vector space over R. Given vectors u = (u1,..., un) and v = (v1,..., vn) in Rn and α in R, we can deﬁne vector addition by u + v = (u1,..., un) + (v1,..., vn) = (u1 + v1,..., un + vn) and scalar multiplication by αu = α(u1,..., un) = (αu1,..., αun). Example 2. If F is a ﬁeld, then F [x] is a vector space over F. The vectors in F [x] are simply polynomials. Vector addition is just polynomial addition. If α ∈ F and p(x) ∈ F [x], then scalar multiplication is deﬁned by αp(x). Example 3. The set of all continuous real-valued functions on a closed interval [a, b] is a vector space over R. If f (x) and g(x) are continuous on [a, b], then (f + g)(x) is deﬁned to be f (x) + g(x). Scalar multiplication is deﬁned by (αf )(
x) = αf (x) for α ∈ R. For example, if f (x) = sin x and g(x) = x2, then (2f + 5g)(x) = 2 sin x + 5x2. √ √ 2 : a, b ∈ Q}. Then V is a vector Example 4. Let V = Q( √ √ √ space over Q. If u = a+b 2 2, then u+v = (a+c)+(b+d) is again in V. Also, for α ∈ Q, αv is in V. We will leave it as an exercise to verify that all of the vector space axioms hold for V. 2 ) = {a + b 2 and v = c+d Proposition 20.1 Let V be a vector space over F. Then each of the following statements is true. 1. 0v = 0 for all v ∈ V. 2. α0 = 0 for all α ∈ F. 3. If αv = 0, then either α = 0 or v = 0. 326 CHAPTER 20 VECTOR SPACES 4. (−1)v = −v for all v ∈ V. 5. −(αv) = (−α)v = α(−v) for all α ∈ F and all v ∈ V. Proof. To prove (1), observe that 0v = (0 + 0)v = 0v + 0v; consequently, 0 + 0v = 0v + 0v. Since V is an abelian group, 0 = 0v. The proof of (2) is almost identical to the proof of (1). For (3), we are done if α = 0. Suppose that α = 0. Multiplying both sides of αv = 0 by 1/α, we have v = 0. To show (4), observe that v + (−1)v = 1v + (−1)v = (1 − 1)v = 0v = 0, and so −v = (−1)v. We will leave the proof of (5) as an exercise. 20.2 Subspaces Just as groups have subgroups and rings have subrings, vector spaces also have substructures. Let V be a vector space over a ﬁeld F, and W a subset of V. Then W is a subspace of V
if it is closed under vector addition and scalar multiplication; that is, if u, v ∈ W and α ∈ F, it will always be the case that u + v and αv are also in W. Example 5. Let W be the subspace of R3 deﬁned by W = {(x1, 2x1 + x2, x1 − x2) : x1, x2 ∈ R}. We claim that W is a subspace of R3. Since α(x1, 2x1 + x2, x1 − x2) = (αx1, α(2x1 + x2), α(x1 − x2)) = (αx1, 2(αx1) + αx2, αx1 − αx2), W is closed under scalar multiplication. To show that W is closed under vector addition, let u = (x1, 2x1 + x2, x1 − x2) and v = (y1, 2y1 + y2, y1 − y2) be vectors in W. Then u + v = (x1 + y1, 2(x1 + y1) + (x2 + y2), (x1 + y1) − (x2 + y2)). Example 6. Let W be the subset of polynomials of F [x] with no odd-power terms. If p(x) and q(x) have no odd-power terms, then neither will p(x)+q(x). Also, αp(x) ∈ W for α ∈ F and p(x) ∈ W. 20.3 LINEAR INDEPENDENCE 327 Let V be any vector space over a ﬁeld F and suppose that v1, v2,..., vn are vectors in V and α1, α2,..., αn are scalars in F. Any vector w in V of the form n w = αivi = α1v1 + α2v2 + · · · + αnvn i=1 is called a linear combination of the vectors v1, v2,..., vn. The spanning set of vectors v1, v2,..., vn is the set of vectors obtained from all possible linear combinations of v1, v2,..
., vn. If W is the spanning set of v1, v2,..., vn, then we often say that W is spanned by v1, v2,..., vn. Proposition 20.2 Let S = {v1, v2,..., vn} be vectors in a vector space V. Then the span of S is a subspace of V. Proof. Let u and v be in S. We can write both of these vectors as linear combinations of the vi’s: u = α1v1 + α2v2 + · · · + αnvn v = β1v1 + β2v2 + · · · + βnvn. Then u + v = (α1 + β1)v1 + (α2 + β2)v2 + · · · + (αn + βn)vn is a linear combination of the vi’s. For α ∈ F, αu = (αα1)v1 + (αα2)v2 + · · · + (ααn)vn is in the span of S. 20.3 Linear Independence Let S = {v1, v2,..., vn} be a set of vectors in a vector space V. If there exist scalars α1, α2... αn ∈ F such that not all of the αi’s are zero and α1v1 + α2v2 + · · · + αnvn = 0, then S is said to be linearly dependent. If the set S is not linearly dependent, then it is said to be linearly independent. More speciﬁcally, S is a linearly independent set if implies that α1v1 + α2v2 + · · · + αnvn = 0 α1 = α2 = · · · = αn = 0 for any set of scalars {α1, α2... αn}. 328 CHAPTER 20 VECTOR SPACES Proposition 20.3 Let {v1, v2,..., vn} be a set of linearly independent vectors in a vector space. Suppose that v = α1v1 + α2v2 + · · · + αnvn = β1v1 + β2v2 + · · · + β
nvn. Then α1 = β1, α2 = β2,..., αn = βn. Proof. If v = α1v1 + α2v2 + · · · + αnvn = β1v1 + β2v2 + · · · + βnvn, then (α1 − β1)v1 + (α2 − β2)v2 + · · · + (αn − βn)vn = 0. Since v1,..., vn are linearly independent, αi − βi = 0 for i = 1,..., n. The deﬁnition of linear dependence makes more sense if we consider the following proposition. Proposition 20.4 A set {v1, v2,..., vn} of vectors in a vector space V is linearly dependent if and only if one of the vi’s is a linear combination of the rest. Proof. Suppose that {v1, v2,..., vn} is a set of linearly dependent vectors. Then there exist scalars α1,..., αn such that α1v1 + α2v2 + · · · + αnvn = 0, with at least one of the αi’s not equal to zero. Suppose that αk = 0. Then vk = − α1 αk v1 − · · · − αk−1 αk vk−1 − αk+1 αk vk+1 − · · · − αn αk vn. Conversely, suppose that vk = β1v1 + · · · + βk−1vk−1 + βk+1vk+1 + · · · + βnvn. Then β1v1 + · · · + βk−1vk−1 − vk + βk+1vk+1 + · · · + βnvn = 0. The following proposition is a consequence of the fact that any system of homogeneous linear equations with more unknowns than equations will have a nontrivial solution. We leave the details of the proof for the end-of-chapter exercises. 20.3 LINEAR INDEPENDENCE 329 Proposition 20.5 Suppose that a vector space V is spanned by n vectors. If m > n, then any set of m vectors in
V must be linearly dependent. A set {e1, e2,..., en} of vectors in a vector space V is called a basis for V if {e1, e2,..., en} is a linearly independent set that spans V. Example 7. The vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1) form a basis for R3. The set certainly spans R3, since any arbitrary vector (x1, x2, x3) in R3 can be written as x1e1 + x2e2 + x3e3. Also, none of the vectors e1, e2, e3 can be written as a linear combination of the other two; hence, they are linearly independent. The vectors e1, e2, e3 are not the only basis of R3: the set {(3, 2, 1), (3, 2, 0), (1, 1, 1)} is also a basis for R3. Example 8. Let Q( 2, 1 − {1 + √ √ 2 } are both bases of Q( 2 ). √ 2 ) = {a + b √ 2 : a, b ∈ Q}. The sets {1, √ √ 2 } and From the last two examples it should be clear that a given vector space has several bases. In fact, there are an inﬁnite number of bases for both of these examples. In general, there is no unique basis for a vector space. However, every basis of R3 consists of exactly three vectors, and every basis of Q( 2 ) consists of exactly two vectors. This is a consequence of the next proposition. √ Proposition 20.6 Let {e1, e2,..., em} and {f1, f2,..., fn} be two bases for a vector space V. Then m = n. Proof. Since {e1, e2,..., em} is a basis, it is a linearly independent set. By Proposition 20.5, n ≤ m. Similarly, {f1, f2,..., fn} is a linearly independent set, and the last proposition implies that m ≤ n. Consequently, m = n. If {e1, e2,..
., en} is a basis for a vector space V, then we say that the dimension of V is n and we write dim V = n. We will leave the proof of the following theorem as an exercise. Theorem 20.7 Let V be a vector space of dimension n. 1. If S = {v1,..., vn} is a set of linearly independent vectors for V, then S is a basis for V. 2. If S = {v1,..., vn} spans V, then S is a basis for V. 330 CHAPTER 20 VECTOR SPACES 3. If S = {v1,..., vk} is a set of linearly independent vectors for V with k < n, then there exist vectors vk+1,..., vn such that {v1,..., vk, vk+1,..., vn} is a basis for V. Exercises 1. If F is a ﬁeld, show that F [x] is a vector space over F, where the vectors in F [x] are polynomials. Vector addition is polynomial addition, and scalar multiplication is deﬁned by αp(x) for α ∈ F. 2. Prove that Q( 2 ) is a vector space. √ √ √ 3. Let Q( 2, 3 ) be the ﬁeld generated by elements of the form a + b 2, 3, 3 ) is a vector space of dimension 4 2 + c √ √ where a, b, c are in Q. Prove that Q( over Q. Find a basis for Q( 3 ). √ √ 2, √ √ 4. Prove that the complex numbers are a vector space of dimension 2 over R. 5. Prove that the set Pn of all polynomials of degree less than n form a subspace of the vector space F [x]. Find a basis for Pn and compute the dimension of Pn. 6. Let F be a ﬁeld and denote the set of n-tuples of F by F n. Given vectors u = (u1,..., un) and v = (v1,..., vn) in F n and α in F, deﬁ
ne vector addition by u + v = (u1,..., un) + (v1,..., vn) = (u1 + v1,..., un + vn) and scalar multiplication by αu = α(u1,..., un) = (αu1,..., αun). Prove that F n is a vector space of dimension n under these operations. 7. Which of the following sets are subspaces of R3? If the set is indeed a subspace, ﬁnd a basis for the subspace and compute its dimension. (a) {(x1, x2, x3) : 3x1 − 2x2 + x3 = 0} (b) {(x1, x2, x3) : 3x1 + 4x3 = 0, 2x1 − x2 + x3 = 0} (c) {(x1, x2, x3) : x1 − 2x2 + 2x3 = 2} (d) {(x1, x2, x3) : 3x1 − 2x2 2 = 0} EXERCISES 331 8. Show that the set of all possible solutions (x, y, z) ∈ R3 of the equations Ax + By + Cz = 0 Dx + Ey + Cz = 0 forms a subspace of R3. 9. Let W be the subset of continuous functions on [0, 1] such that f (0) = 0. Prove that W is a subspace of C[0, 1]. 10. Let V be a vector space over F. Prove that −(αv) = (−α)v = α(−v) for all α ∈ F and all v ∈ V. 11. Let V be a vector space of dimension n. Prove each of the following statements. (a) If S = {v1,..., vn} is a set of linearly independent vectors for V, then S is a basis for V. (b) If S = {v1,..., vn} spans V, then S is a basis for V. (c) If S = {v1,..., vk} is a set of linearly independent vectors for V with k < n, then there exist
vectors vk+1,..., vn such that {v1,..., vk, vk+1,..., vn} is a basis for V. 12. Prove that any set of vectors containing 0 is linearly dependent. 13. Let V be a vector space. Show that {0} is a subspace of V of dimension zero. 14. If a vector space V is spanned by n vectors, show that any set of m vectors in V must be linearly dependent for m > n. 15. Linear Transformations. Let V and W be vector spaces over a ﬁeld F, of dimensions m and n, respectively. If T : V → W is a map satisfying T (u + v) = T (u) + T (v) T (αv) = αT (v) for all α ∈ F and all u, v ∈ V, then T is called a linear transformation from V into W. (a) Prove that the kernel of T, ker(T ) = {v ∈ V : T (v) = 0}, is a subspace of V. The kernel of T is sometimes called the null space of T. (b) Prove that the range or range space of T, R(V ) = {w ∈ W : T (v) = w for some v ∈ V }, is a subspace of W. (c) Show that T : V → W is injective if and only if ker(T ) = {0}. (d) Let {v1,..., vk} be a basis for the null space of T. We can extend this basis to be a basis {v1,..., vk, vk+1,..., vm} of V. Why? Prove that {T (vk+1),..., T (vm)} is a basis for the range of T. Conclude that the range of T has dimension m − k. 332 CHAPTER 20 VECTOR SPACES (e) Let dim V = dim W. Show that a linear transformation T : V → W is injective if and only if it is surjective. 16. Let V and W be ﬁnite dimensional vector spaces of dimension n over a ﬁeld F. Suppose that T : V → W is a vector

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