Math 2070BC Term 2 Weeks 1 13 Lecture Notes

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1 Math 2070BC Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic generator permutation Symmetric Group symmetric dihedral group k cycle transposition subgroup even permutation alternating group left cosets index Lagrange's Theorem generated finitely generated group homomorphism isomorphism isomorphic Division Theorem Greatest Common Divisor Bézout's Lemma Euclidean Algorithm relatively prime prime Euclid's Lemma The Fundamental Theorem of Arithmetic Unique Factorization congruent modulo m equivalence relation Reflexive Symmetric Transitive Addition Multiplication Chinese Remainder Theorem Q[x] Division Theorem for Bézout's Identity for Polynomials irreducible Unique Factorization for Polynomials ring zero ring integral domain zero divisor cancellation law unit field Reflexive Symmetric Transitive partition quotient set Fraction Field ring homomorphism evaluation map kernel image ideal proper ideal principal ideal Principal Ideal Domain PID congruent modulo I Reflexivity Symmetry Transitivity residue evaluation homomorphism root Root Theorem monic polynomial Greatest Common Divisor GCD

2 Rational Root Theorem primitive Gauss's Lemma content Gauss's Theorem Eisenstein's Criterion subring subfield field extension algebraic minimal polynomial characteristic characteristic zero Frequently Occurring: math finite group domain multiplication identity element inverse commutative number addition bijective matrix nonzero order divide cyclic integer map subset product coefficient multiplicative subgroup relation left coset onetoone prime homomorphism isomorphism isomorphic remainder division additive ring induction divisor positive relatively polynomial degree zero irreducible constant integral unit field ideal root primitive

3 Math 2070 Week 1 Groups Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order Frequently Occurring: group multiplication identity matrix order element inverse abelian number Motivation Finite Designs

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5 Geometric Symmetry in Patterns and Tilings, Clare Horne.... How many ways are there to color a cube, such that each face is either red or green? Answer: 10. Why? How many ways are there to form a triangle with three sticks of equal lengths, colored red, green and blue, respectively? What are the symmetries of an equilateral triangle? Dihedral Group D3

6 By Jim.belk Own work, Public Domain, Link Cayley Table * a b c 2 a a ab ac b ba b bc 2 c ca cb c 2 Cayley Table for D3 r r r s s s r r r r s s s r r r r s s s r r r r s s s s s s s r r r s s s s r r r

7 s s s s r r r Groups Definition 1.1. A group : G G G such that: Associativity for all is a set equipped with a binary operation (typically called group operation or "multiplication"), associative. Existence of an Identity Element. In other words, the group operation is There is an element e G, called an identity element, such that: for all g G. Invertibility a, b, c G G (a b) c = a (b c), g e = e g = g, Each element g G has an inverse g 1 G, such that: g 1 g = g g 1 = e. Remark 1.2. a b = b a Note that we do not require that. We often write ab to denote a b. a, b G Definition 1.3. If ab = ba for all. We say that the group operation is commutative, and that G is an abelian group. Example 1.4. The following sets are groups, with respect to the specified group operations: G = Q {0}, where the group operation is the usual multiplication for rational numbers. The identity is a Q {0} a 1 = 1 a G e = 1 is. The group is abelian., and the inverse of

8 G = Q +, where the group operation is the usual addition for rational numbers. The identity is e = 0. The inverse of a Q with respect to + is a. Note that Q is not a group with respect to multiplication. For in that case, we have, but has no 0 1 Q = 1 inverse such that. e = 1 0 Q Exercise 1.5. Verify that the following sets are groups under the specified binary operation: (Z, +) (R, +) (R, ) (U m, ) m N, where, U m = {1, ξ m, ξ 2 m,, ξ m 1 m }, ξ m = e 2πi/m = cos(2π/m) + i sin(2π/m) C and. The set of bijective functions f : R R, where (i.e. composition of functions). f g := f g Exercise Webwork Problem 2. Webwork Problem 3. Webwork Problem 4. Webwork Problem 5. Webwork Problem 6. Webwork Problem 7. Webwork Problem 8. Webwork Problem 9. Webwork Problem Example 1.7. The set of real matrices with nonzero determinants is a group under matrix multiplication, with identity element: G = GL(2, R) 2 2 e = ( ).

9 In the group G, we have: ( a b Note that there are matrices such that. Hence GL(2, R) 1 c d ) 1 d b = ( ad bc is not abelian. c a ) A, B GL(2, R) AB BA Exercise 1.8. The set (Special Linear Group) of real matrices with determinant SL(2, R) is a group under matrix multiplication. Claim 1.9. The identity element e of a group G is unique. Suppose there is an element e G such that e g = ge = g for all g G. Then, in particular, we have: e e = e e e e = e e = e But since is an identity element, we also have. Hence,. Claim Let G be a group. For all g G, its inverse g 1 is unique. Suppose there exists g G such that g g = gg = e. By the associativity of the group operation, we have: Hence, g 1 is unique. g = g e = g (gg 1 ) = (g g)g 1 = eg 1 = g 1. Let G be a group with identity element e. For g G, n N, let:

10 g n := g g g. n times g n := g 1 g 1 g 1 n times g 0 := e. Claim Let G be a group. 1. For all g G, we have: (g 1 ) 1 = g. a, b G 2. For all, we have: g G n, m Z 3. For all,, we have: (ab) 1 = b 1 a 1. g n g m = g n+m. Exercise. Definition Let G be a group, with identity element e. The order of G G ord g g G n N g n = e n g is the number of elements in. The order of an element is the smallest such that. If no such exists, we say that has infinite order. Theorem Let G be a group with identity element e. Let g be an element of G. If for some, then divides. g n = e n N ord g n Shown in class.

11 Math 2070 Week 2 Groups Keywords: order infinite order cyclic generator permutation Symmetric Group symmetric dihedral group k cycle transposition Frequently Occurring: group identity element order finite cyclic generator permutation bijective map composition symmetric origin cycle product disjoint transposition matrix Definition 2.1. Let G be a group, with identity element e. The order of G G ord g g G n N g n = e n g is the number of elements in. The order of an element is the smallest such that. If no such exists, we say that has infinite order. Theorem 2.2. Let G be a group with identity element e. Let g be an element of G. If for some, then divides. g n = e n N ord g n Shown in class. Exercise 2.3. If G has finite order, then every element of G has finite order.

12 Definition 2.4. A group G is cyclic if there exists g G such that every element of G is equal to g n for some integer n. In which case, we write:, and say that is a generator of. G = g g G Note: The generator of of a cyclic group might not be unique. Example 2.5. (U m, ) is cyclic. Exercise 2.6. A finite cyclic group G has order (i.e. size) n if and only if each of its generators has order n. Exercise 2.7. (Q, +) is not cyclic. Permutations Definition. Let X be a set. A permutation of X is a bijective map σ : X X. Claim 2.8. The set S X of permutations of a set X is a group with respect to, the composition of maps. Let σ, γ be permutations of X. By definition, they are bijective maps from X to itself. It is clear that σ γ is a bijective map from X to itself, hence is a permutation of. So is a well defined binary operation on. σ γ X S X α, β, γ S X α (β γ) = (α β) γ For, it is clear that. Define a map e : X X as follows: e(x) = x, for all x X. e S X e σ = σ e = σ σ S X It is clear that, and that for all. Hence, e is an identity element in S X. Let σ be any element of S X. Since σ : X X is by assumption bijective, there exists a bijective map σ 1 : X X such that

13 σ σ 1 = σ 1 σ = e σ 1 σ the operation.. So is an inverse of with respect to Terminology: We call S X the Symmetric Group on X. Notation: If, where, we denote by. For X = {1, 2,, n} n N S X S n n N, the group S n has n! elements. For n N, by definition an S n σ : X X X = {1, 2,, n} element of is a bijective map, where. We often describe σ using the following notation: σ = ( 1 2 n σ(1) σ(2) σ(n) ) Example 2.9. In S3, is the permutation on which sends to, to itself, and to, i.e.. For we have: (since, for example, σ = ( ) {1, 2, 3} σ(1) = 3, σ(2) = 2, σ(3) = 1 α, β S3 given by: α = ( ), β = ( ), αβ = α β = ( ) ( ) = ( ) β α α β : ). We also have: βα = β α = ( ) ( ) = ( ) Since αβ βα, the group S3 is non abelian. In general, for n > 2, the group S n is non abelian (Exercise: Why?). For the same α S3 defined above, we have: α 2 = α α = ( ) ( ) = ( )

14 and: α 3 = α α 2 = ( ) ( ) = ( ) = e Hence, the order of α is 3. Dihedral Group Consider the subset T of transformations of R 2, consisting of all rotations by fixed angles about the origin, and all reflections over lines through the origin. Consider a regular polygon P with n sides in R 2, centered at the origin. Identify the polygon with its n vertices, which form a subset P = {x1, x2,, x n } R 2 τ(p) = P τ T of. If for some, we say that P is symmetric with respect to τ. Intuitively, it is clear that P is symmetric with respect to n rotations, and reflections in. {r0, r1,, r n 1} n {s1, s2,, s n } T By Jim.belk Own work, Public Domain, Link Theorem The set D n := {r0, r1,, r n 1, s1, s2,, s n } τ γ = τ γ to the group operation defined by transformations). is a group, with respect (composition of

15 Terminology: D n is called a dihedral group. More on Consider the following element in S6: We may describe the action of the notation: where S n (n1n2 n k ) σ = ( ) represents the permutation: Viewing permutations as bijective maps, the "multiplication" by definition the composition. σ : {1, 2,, 6} {1, 2,, 6} σ = (15)(246), n1 n2 n i n i+1 n k n1 using (15)(246) (15) (246) (n1n2 n k ) k 3 (15)(246) We call a cycle. Note that is missing from. This corresponds to the fact that 3 is fixed by σ. is Claim Every non identity permutation in product of disjoint cycles. S n is either a cycle or a Discussed in class. Exercise Disjoint cycles commute with each other. A 2 cycle is often called a transposition, for it switches two elements with each other. Claim Each element of disjoint) transpositions. S n is a product of (not necessarily Sketch of proof:

16 Show that each permutation not equal to the identity is a product of cycles, and that each cycle is a product of transpositions: Example (a1a2 a k ) = (a1a k )(a1a k 1) (a1a3)(a1a2) ( ) = (15)(246) = (15)(26)(24) = (15)(46)(26) next reset Note that a given element σ of S n may be expressed as a product of transpositions in different ways, but: Claim In every factorization of σ as a product of transpositions, the number of factors is either always even or always odd. Exercise. One approach: Show that there is a unique n n matrix, with either 0 or 1 as its coefficients, which sends each standard basis vector e i in R n to e σ(i). Then, use the fact that the determinant of the matrix corresponding to a transposition is, and that the determinant function of matrices is multiplicative. 1

17 Math 2070 Week 3 Groups Keywords: transposition subgroup even permutation alternating group left cosets index Lagrange's Theorem Frequently Occurring: transposition element product disjoint permutation identity number even matrix subgroup group subset nonempty cyclic lagrange equivalence class left coset finite index divide order A 2 cycle is often called a transposition, for it switches two elements with each other. Claim 3.1. Each element of transpositions. S n is a product of (not necessarily disjoint) Sketch of proof: Show that each permutation not equal to the identity is a product of cycles, and that each cycle is a product of transpositions: Example 3.2. (a1a2 a k ) = (a1a k )(a1a k 1) (a1a3)(a1a2) ( ) = (15)(246) = (15)(26)(24) = (15)(46)(26) next reset

18 Note that a given element σ of S n may be expressed as a product of transpositions in different ways, but: Claim 3.3. In every factorization of σ as a product of transpositions, the number of factors is either always even or always odd. Exercise. One approach: Show that there is a unique n n matrix, with either 0 or 1 as its coefficients, which sends each standard basis vector e i in R n to e σ(i). Then, use the fact that the determinant of the matrix corresponding to a transposition is, and that the determinant function of matrices is multiplicative. 1 Exercise Webwork Problem 2. Webwork Problem 3. Webwork Problem 4. Webwork Problem Subgroups Definition 3.5. Let G be a group. A subset H of G is a subgroup of G if it satisfies the following properties: Closure If, then. a, b H ab H Identity The identity element of G lies in H. Inverses If a H, then a 1 H. In particular, a subgroup H is a group with respect to the group operation on G, and the identity element of H is the identity element of G. Example 3.6. For any n Z, nz is a subgroup of (Z, +).

19 Q {0} is a subgroup of (R {0}, ). SL(2, R) GL(2, R) is a subgroup of. The set of all rotations (including the trivial rotation) in a dihedral D n D n group is a subgroup of. Let n N, n 2. We say that σ S n is an even permutation if it is equal to the product of an even number of transpositions. The A n S n subset of consisting of even permutations is a subgroup of S n A n. is called an alternating group. Claim 3.7. A nonempty subset H of a group G is a subgroup of G if and only if, for all, we have. x, y H xy 1 H H G x, y H y 1 H Suppose is a subgroup. For all, we have ; hence, xy 1 H. Conversely, suppose H is a nonempty subset of G, and xy 1 H for all x, y H. Identity Let e be the identity element of G. Since H is nonempty, it contains at least one element h. Since, and by hypothesis h h 1 H, the set H contains e. Inverses Since e H, for all a H we have a 1 = e a 1 H. Closure For all, we know that. Hence, ab = a (b 1 ) 1 H. e = h h 1 a, b H b 1 H Hence, H is a subgroup of G. Claim 3.8. The intersection of two subgroups of a group subgroup of G. G is a Exercise 3.9. Every subgroup of (Z, +) is cyclic. Every subgroup of a cyclic group is cyclic. Lagrange's Theorem

20 Let G be a group, H a subgroup of G. We are interested in knowing how large H is relative to G. We define a relation on G as follows: or equivalently: a b if b = ah for some h H, a b if a 1 b H. Exercise: is an equivalence relation. We may therefore partition G into disjoint equivalence classes with respect to. We call these equivalence classes the left cosets of H. Each left coset of H has the form ah = {ah h H}. We could likewise define right cosets. These sets are of the form Hb, b G. In general, the number of left cosets and right cosets, if finite, are equal to each other Example Let. Let: The set H is a subgroup of G. The left cosets of H in G are as follows: where. In general, for n Z, the left cosets of nz in Z are: G = (Z, +) H = 3Z = {, 9, 6, 3, 0, 3, 6, 9, } i + 3Z := {i + 3k : k Z} 3Z, 1 + 3Z, 2 + 3Z, i + nz, i = 0, 1, 2,, n 1. Definition The number of left cosets of a subgroup H of G is called the index of H in G. It is denoted by: [G : H] Example Let,,. Then, n N G = (Z, +) H = (nz, +) [G : H] = n. Example Let G = GL(n, R). Let:

21 (Exercise: H is a subgroup of G.) Let: Note that. For any, either or Since, we have s 1 g H. So, G = H sh, and [G : H] = 2. Notice that both G and H are infinite groups, but the index of H in G is finite. Example Let,. For each, let: Note that. For each g G, we have: Moreover, for distinct, we have: This implies that s 1 x s y H, hence s y H and s x H are disjoint cosets. We have therefore: The index H = GL + (n, R) := {h G : det h > 0}. s = in this case is infinite. G det s = det s 1 = 1 g G det g > 0 det g < 0. If det g > 0, then g H. If det g < 0, we write: g = (ss 1 )g = s(s 1 g). det s 1 g = (det s 1 )(det g) > 0 det s x = x [G : H] G = GL(n, R) H = SL(n, R) x R s x = x G g = sdet g(s 1 det g g) s det gh x, y R det(s 1 x s y ) = y/x 1. G = x R s x H.

22 Theorem (Lagrange's Theorem) Let G be a finite group. Let H be subgroup of, then divides. More precisely, G = [G : H] H. G H G We already know that the left cosets of H partition G. That is: where if. Hence, i=1 a i H. The theorem follows if we show that the size of each left coset of H is equal to. For each left coset of, pick an element, and define a map G = a1h a2h a [G:H]H, a i H a j H = i j G = [G:H] H S H a S ψ : H S as follows: ψ(h) = ah. We want to show that ψ is bijective. For any s S, by definition of a left coset (as an equivalence class) we have s = ah for some h H. Hence, is surjective. If for some, then ψ ψ(h ) = ah = ah = ψ(h) h, h H h = a 1 ah = a 1 ah = h ψ bijection between two finite sets. Hence, S = H.. Hence, is one to one. So we have a Corollary Let G G divides the order of G. be a finite group. The order of every element of Since is finite, any element of has finite order. Since the order of the subgroup: is equal to divides., it follows from Lagrange's Theorem that Corollary If the order of a group G is prime, then G is a cyclic group. G g G ord g ord g G H = g = {e, g, g 2,, g (ord g) 1 } ord g = H

23 Math 2070 Week 4 Groups Keywords: index Lagrange's Theorem generated finitely generated group homomorphism isomorphism isomorphic Frequently Occurring: left coset subgroup index group finite lagrange element bijective order integer generated rotation homomorphism isomorphism Definition 4.1. The number of left cosets of a subgroup H of G is called the index of H in G. It is denoted by: [G : H] Example 4.2. Let,,. Then, n N G = (Z, +) H = (nz, +) [G : H] = n. Example 4.3. Let G = GL(n, R). Let: H = GL + (n, R) := {h G : det h > 0}. (Exercise: H is a subgroup of G.) Let:

24 Note that. For any, either or Since, we have s 1 g H. So, G = H sh, and [G : H] = 2. Notice that both G and H are infinite groups, but the index of H in G is finite. Example 4.4. Let,. For each, let: Note that. For each g G, we have: Moreover, for distinct, we have: This implies that s 1 x s y H, hence s y H and s x H are disjoint cosets. We have therefore: The index s = in this case is infinite. G det s = det s 1 = 1 g G det g > 0 det g < 0. If det g > 0, then g H. If det g < 0, we write: g = (ss 1 )g = s(s 1 g). det s 1 g = (det s 1 )(det g) > 0 det s x = x [G : H] G = GL(n, R) H = SL(n, R) x R s x = x G g = sdet g(s 1 det g g) s det gh x, y R det(s 1 x s y ) = y/x 1. G = x R s x H. Theorem 4.5. (Lagrange's Theorem) Let G G H G be a finite group. Let be subgroup of, then divides. More precisely, H

25 G = [G : H] H. We already know that the left cosets of H partition G. That is: G = a1h a2h a[g:h]h, a i H a j H = i j G = [G:H] where if. Hence, i=1 a i H. The theorem follows if we show that the size of each left coset of H is equal to H. For each left coset S of H, pick an element a S, and define a map as follows: ψ : H S ψ(h) = ah. We want to show that ψ is bijective. For any s S, by definition of a left coset (as an equivalence class) we have for some. Hence, is surjective. s = ah h H ψ ψ(h ) = ah = ah = ψ(h) h, h H If for some, then h = a 1 ah = a 1 ah = h. Hence, ψ is one to one. So we have a bijection between two finite sets. Hence,. S = H Corollary 4.6. Let G divides the order of G. be a finite group. The order of every element of G Since is finite, any element of has finite order. Since the order of the subgroup: is equal to divides. G g G ord g ord g G Corollary 4.7. H = g = {e, g, g 2,, g (ord g) 1 }, it follows from Lagrange's Theorem that If the order of a group G is prime, then G is a cyclic group. ord g = H Corollary 4.8. If a group G is finite, then for all g G we have: g G = e.

26 Corollary 4.9. Let G be a finite group. Then a nonempty subset H of G is a subgroup of G if and only if it is closed under the group operation of (i.e. for all ). G ab H a, b H It is easy to see that if H is a subgroup, then it is closed under the group operation. The other direction is left as an Exercise. Example Let n be an integer greater than 1. The group A n of even permutations on a set of n elements (see Example 3.6) has order n! 2. A n S n n! View as a subgroup of, which has order. S n = A n (12)A n [S n : A n ] = 2 Exercise: Show that. Hence, we have. It now follows from Lagrange's Theorem ( Theorem 4.5) that: n! A n = =. [S n : A n ] 2 S n Generators Let G be a group, X a nonempty subset of G. The subset of G consisting of elements of the form: g m 1 1 gm 2 2 g m n n, where n N, g i X, m i Z, is a subgroup of G. We say that it is the subgroup of G generated by X. If,. We often write: X = {x1, x2,, x l } l N x1, x2,, x l to denote the subgroup generated by X.

27 Example D n {r0, r1,, r n 1} = r1 In,. x1, x2,, x l G If there exists a finite number of elements such that, we say that G is finitely generated. For example, every cyclic group is finitely generated, for it is generated G = x1, x2,, x l by one element. Every finite group is finitely generated, since we may take the finite generating set X to be G itself. Example Consider G = D3, and its subgroup H = {r0, r1, r2} consisting of its rotations. (We use the convention that anticlockwise rotation by an angle of 2πk/3 H G [G : H] = G / H = 2 G = H gh g G gh = H g H is the ). By Lagrange's Theorem, the index of in is. This implies that for some. Since if, we may conclude that g H. So, g is a reflection. Conversely, for any reflection s D3, the left coset sh is disjoint from H. We have therefore, which implies that G = H s1h = H s2h = H s3h s1h = s2h = s3h s = s i G. In particular, for a fixed, any element in is either a rotation or equal to sr i for some rotation r i. Since H is a cyclic group, generated by the rotation r1, we have D3 = r1, s, where s is any reflection in D3. r k Exercise Webwork Problem 2. Webwork Problem 3. Webwork Problem 4. Webwork Problem 5. Webwork Problem 6. Webwork Problem 7. Webwork Problem 8. Webwork Problem 9. Webwork Problem 10. Webwork Problem 11. Webwork Problem 12. Webwork Problem

28 Group Homomorphisms Definition Let, be groups. A group homomorphism ϕ from G to G is a map ϕ : G G which satisfies: a, b G for all. G = (G, ) G = (G, ) ϕ(a b) = ϕ(a) ϕ(b), Claim If ϕ : G G ϕ(e G ) = e G is a group homomorphism, then: ϕ(g 1 ) = ϕ(g) 1, for all g G. 3., for all,. ϕ(g n ) = ϕ(g) n g G n Z We prove the first claim, and leave the rest as an exercise. Since is the identity element of G, we have e G e G = e G. On the other hand, since ϕ is a group homomorphism, we have: ϕ(e G ) = ϕ(e G e G ) = ϕ(e G ) ϕ(e G ). Since G is a group, ϕ(e G ) 1 exists in G, hence: ϕ(e G ) 1 ϕ(e G ) = ϕ(e G ) 1 (ϕ(e G ) ϕ(e G )) The left hand side is equal to e G, while by the associativity of the right hand side is equal to ϕ(e G ). e G Let ϕ : G G be a homomorphism of groups. The image of ϕ is defined as: The kernel of ϕ is defined as: Claim im ϕ := ϕ(g) := {ϕ(g) : g G} G ker ϕ = {g G : ϕ(g) = e G } G. The image of ϕ is a subgroup of G. The kernel of ϕ is a subgroup of G. Claim 4.17.

29 A group homomorphism ker ϕ = {e G }. ϕ : G G is one to one if and only if Definition Let G, G be groups. A map ϕ : G G is a group isomorphism if it is a bijective group homomorphism. Note that if a homomorphism ϕ is bijective, then ϕ 1 : G G is also a homomorphism, and consequently, ϕ 1 is an isomorphism. If there exists an isomorphism between two groups G and G, we say that the groups G and G are isomorphic. Definition Fix an integer. k Z k k n For any, let denote the remainder of the division of by. Let. We define a binary operation on as follows: n > 0 Z n = {0, 1, 2,, n 1} + Zn Z n k + Zn l = k + l. Exercise Z n = (Z n, + Zn ) is a group, with identity element 0, and j 1 = n j for any j Z n. Example Let. Let be the D n n > 2 subgroup of consisting of all rotations, where r1 denotes the anticlockwise rotation by the angle 2π/n, and r k = r k 1. Then, H is isomorphic to. Z n = (Z n, + Zn ) H = {r0, r1, r2,, r n 1} Define ϕ : H Z n as follows: ϕ(r k ) = k, k {0, 1, 2,, n 1}. For any k Z, let k {0, 1, 2,, n 1} denote the remainder of the division of k by n. By the Division Theorem for Integers, we have:

30 k = nq + k for some integer q Z. It now follows from that, for all, we have: ord r1 = n r i r j = r i 1 rj 1 = r i+j 1 = r nq+ i+j 1 = (r n 1) q r i+j 1 = r i+j. r i, r j H next reset Hence, ϕ(r i r j ) = ) ϕ(r i+j = i + j = i + Zn j = ϕ(r i ) + Zn ϕ(r j ). next reset This shows that ϕ is a homomorphism. It is clear that ϕ is surjective, which then implies that ϕ is one to one, for the two groups have the same size. Hence, ϕ is a bijective homomorphism, i.e. an isomorphism. Example Examples of Group Homomorphisms ϕ : S n ({±1}, ), ϕ(σ) = { 1, σ is an even permutation. 1, σ is an odd permutation. ker ϕ = A n. det : GL(n, R) (R, ) ker det = SL(n, R). Let G be the (additive) group of all real valued continuous functions on. [0, 1] ϕ : G (R, +) ϕ(f) = 1 f(x) dx. 0

31 ϕ : (R, +) (R, ). ϕ(x) = e x.

32 Math 2070 Week 5 Group Homomorphisms Rings Keywords: Division Theorem Greatest Common Divisor Frequently Occurring: group cyclic order isomorphic map element remainder integer Claim 5.1. Any cyclic group of finite order n is isomorphic to. Z n Sketch of Proof: By definition, a cyclic group is equal to for some. Moreover, ord g = ord G. Define a map ϕ : G Z n as follows: G g g G ϕ(g k ) = k, k {0, 1, 2,, n 1}. Show that ϕ is a group isomorphism. (For reference, see the proof of Example 4.21.) Corollary 5.2. If G and G are two finite cyclic groups of the same order, then G is isomorphic to G.

33 Exercise 5.3. (Z, +) An infinite cyclic group is isomorphic to. Exercise 5.4. Let G be a cyclic group, then any group which is isomorphic to G is also cyclic. Product Group Let be groups. The direct product: has a natural group structure where the group operation follows: is defined as The identity element of is, where are the identity elements of A and B, respectively. For any (a, b) A B, we have (a, b) 1 = (a 1, b 1 ), where a 1, b 1 are the inverses of a, b in the groups A, B, respectively. For any collection of groups, we may similarly define a group operation That is: on: The identity element of For any (A, A ), (B, B ) A B := {(a, b) a A, b B} (a, b) (a, b ) = (a A a, b B b ), (a, b), (a, b ) A B. A B e = (e A, e B ) e A, e B A1, A2,, A n A1 A2 A n := {(a1, a2,, a n ) a i A i, i = 1, 2, n}. (a1, a2,, a n ) (a 1, a 2,, a n) = (a1 A 1 a 1, a2 A 2 a 2,, a n An a n A1 A2 A n is: e = (e A 1, e A 2,, e An ). (a1, a2,, a n ) A1 A2 A n, its inverse is: (a1, a2,, a n ) 1 = (a 1 1, a 1 2,, a 1 n ). Exercise 5.5. Z6 is isomorphic to Z2 Z3.

34 Hint: Z2 Z3 (1, 1) Show that is a cyclic group generated by. Example 5.6. The cyclic group Z4 is not isomorphic to Z2 Z2. G = Z2 Z2 2 G = 4 G Each element of is of order at most. Since, cannot be generated by a single element. Hence, G is not cyclic, so it cannot be isomorphic to the cyclic group Z4. Exercise 5.7. Let G isomorphic to G is abelian. Example 5.8. be an abelian group, then any group which is D6 12 D6 = r1, s The group has elements. We have seen that, where r1 is a rotation of order 6, and s is a reflection, which has order 2. So, it is reasonable to ask if D6 is isomorphic to Z6 Z2. The answer is no. For Z6 Z2 is abelian, but D6 is not. Claim 5.9. The dihedral group D3 S3. is isomorphic to the symmetric group We have seen that, where and is any fixed reflection, with: D3 = r, s r = r1 s ord r = 3, ord s = 2, srs = r 1. In particular, any element in D3 may be expressed as r i s j, with,. i {0, 1, 2} j {0, 1} We have also seen that S3 = a, b, where: a = (123), b = (12), ord a = 3, ord b = 2, bab = a 1.

35 Hence, any element in may be expressed as, with, j {0, 1}. Define map ϕ : D3 S3 D3 a i b j i {0, 1, 2} as follows: ϕ(r i s j ) = a i b j, i, j Z We first show that ϕ is well defined: That is, whenever r i s j = r i s j, we want to show that: ϕ(r j s j ) = ϕ(r i s j ). The condition r i s j = r i s j implies that: This holds only if Since and, we have: by Theorem 2.2. Hence, r i i = s j j = e ord r = 3 ord s = 2 3 (i i ), r i i = s j j, since no rotation is a reflection. 2 (j j), ϕ(r i s j )ϕ(r i s j ) 1 = (a i b j )(a i b j ) 1 = a i b j b j a i = a i b j j a i = a i i since ord b = 2. = e since ord a = 3. next reset This implies that. We conclude that ϕ is well defined. We now show that ϕ is a group homomorphism: Given,, we have: ϕ(r i s j ) = ϕ(r i s j ) μ, μ {0, 1, 2} ν, ν {0, 1} ϕ(r μ s ν r μ s ν ) = { ϕ(rμ+μ s ν ), if ν = 0; ϕ(r μ μ s ν+ν ), if ν = 1. = { aμ+μ b ν, if ν = 0; a μ μ b ν+ν = a μ b ν a μ b ν, if ν = 1. = ϕ(r μ s ν )ϕ(r μ s ν ). This shows that ϕ is a group homomorphism. To show that ϕ is a group isomorphism, it remains to show that it is surjective and one to one. It is clear that ϕ is surjective. We leave it as an exercise to show that ϕ is one to one.

36 Example The group: G = {g GL(2, R) g = ( cos θ sin θ ) for some θ R} sin θ cos θ is isomorphic to G = {z C : z = 1}. Here, the group operation on G is matrix multiplication, and the group operation on G is the multiplication of complex numbers. Each element in G is equal to e iθ for some θ R. Define a map ϕ : G G as follows: ϕ (( cos θ sin θ sin θ cos )) = θ eiθ. Exercise: Show that ϕ is a well defined map. Then, show that it is a bijective group homomorphism. Rings Some Results in Elementary Number Theory Theorem (Division Theorem) Let,, then there exist unique q (called the quotient), and r (remainder) in Z, satisfying, such that. 0 r < a b = aq + r a > 0 b 0 We will prove the case,. The other cases are left as exercises. Fix a > 0. First, we prove the existence of the quotient q and remainder for any, using mathematical induction. r b 0 The base step corresponds to the case a, b Z a 0 0 b < a. In this case, if we q = 0 r = b b = qa + r 0 r = b < a let and, then indeed, where. Hence, q and r exist.

37 The inductive step of the proof of the existence of q and r is as follows: Suppose the existence of the quotient and remainder holds for all non negative b < b, we want to show that it must also hold for b. First, we may assume that b a, since the case b < a has already been proved. Let. Then,, so by the inductive hypothesis b = b a 0 b < b b = q a + r q, r Z 0 r < a b = b + a = (q + 1)a + r q = q + 1 r = r b = qa + r 0 r < a we have for some such that. This implies that. So, if we let and, then, where. This establishes the existence of q, r for b. Hence, by mathematical induction, the existence of q, r holds for all b 0. Now we prove the uniqueness of q and r. Suppose, where, with. b = qa + r = q a + r q, q, r, r Z 0 r, r < a qa + r = q a + r r r = (q q)a 0 r, r < a, we have: Then, implies that. Since a > r r = q q a. q q q q = 0 Since is an integer, the above inequality implies that, i.e. q = q, which then also implies that r = r. We have therefore established the uniqueness of q and r. The proof of the theorem, for the case, is now complete. a > 0, b 0 Definition The Greatest Common Divisor gcd(a, b) is the largest positive integer d which divides both a and b (Notation: d a and d b). of a, b Z Note. a 0 gcd(a, 0) = a gcd(0, 0) If, then. is undefined. Lemma b = aq + r (a, b, q, r Z) gcd(b, a) = gcd(a, r) If, then. d a d b d r = b aq d a d r d a d b = qa + r a, b a, r. If two finite sets of integers are If and, then. Conversely, if and, then and. So, the set of common divisors of is the same as the set of the common divisors of the same, then their largest elements are clearly the same. In other words:

38 gcd(b, a) = gcd(a, r). The Euclidean Algorithm b a b0 = b a0 = a b0 = a0q0 + r0 0 r0 < a0. n > 0 b n = a n 1 a n = r n 1 r n Suppose. Let,. Write, where For, let and, where is the remainder of the division of by. That is, If, then that means that, and. Now, suppose. Since is a non negative integer and, eventually, for some n N. Claim b n. a n b n = a n q n + r n, 0 r n < a n. r0 = 0 a b gcd(a, b) = a r0 > 0 r n 0 r n < r n 1 r n = 0 gcd(b, a) = a n By the previous lemma, gcd(b, a) = gcd(b0, a0) = gcd(a0, r0) = gcd(b1, a1) = gcd(a1, r1) = gcd(b2, a2) = = gcd(a n, r n ) = gcd(a n, 0) = a n. next reset Example Find gcd(285, 255). 285 b0 255 b1=a0 30 b2 = 255 a0 = 30 a1=r0 = 15 a2 1 q0 8 q1 2 q r r1 + 0 r2

39 gcd(285, 255) = r1 = 15 So,. next reset

40 Math 2070 Week 6 Elementary Number Theory Keywords: Bézout's Lemma Euclidean Algorithm relatively prime prime Euclid's Lemma The Fundamental Theorem of Arithmetic Unique Factorization congruent modulo m equivalence relation Reflexive Symmetric Transitive Addition Multiplication Chinese Remainder Theorem Frequently Occurring: number remainder integer bézout induction inductive positive divisor relatively prime euclid divide product factorization congruence congruent modulo relation addition multiplication system The Euclidean Algorithm b a b0 = b a0 = a b0 = a0q0 + r0 0 r0 < a0. n > 0 b n = a n 1 a n = r n 1 r n Suppose. Let,. Write, where For, let and, where is the remainder of the division of by. That is, If, then that means that, and. Now, suppose. Since is a non negative integer and, eventually, for some n N. Claim 6.1. b n a n b n = a n q n + r n, 0 r n < a n. r0 = 0 a b gcd(a, b) = a r0 > 0 r n 0 r n < r n 1 r n = 0 gcd(b, a) = a n.

41 By the previous lemma, gcd(b, a) = gcd(b0, a0) = gcd(a0, r0) = gcd(b1, a1) = gcd(a1, r1) = gcd(b2, a2) = = gcd(a n, r n ) = gcd(a n, 0) = a n. next reset gcd(285, 255) Example 6.2. Find. 285 b0 255 b1=a0 30 b2 = 255 a0 = 30 a1=r0 = 15 a2 1 q0 8 q1 2 q r r1 + 0 r2 gcd(285, 255) = r1 = 15 So,. next reset Claim 6.3. (Bézout's Lemma) There exist gcd(a, b) = ax + by. x, y Z such that Sketch of Proof: Approach 1. Recall the notation used in the Euclidean Algorithm in Slide 1. We saw that if, then. We may prove Bézout's Lemma via mathematical induction as follows: First, for integers x l, y l Z r n = 0 induction proof. such that gcd(a, b) = r n 1 0 l < min(n 1, 2) r l = ax l + by l, show that there exist. This is the base step of the

42 n k n 1 0 l < k r l = ax l + by l x l, y l Z We now carry out the inductive step. Suppose. For any integer, suppose for some, for all. Show that: r k = b k a k 1=r k 2 q k a k r k 1 also has the form for some. The desired identity mathemtical induction. r k = ax k + by k x k, y k Z gcd(a, b) = r n 1 = ax n 1 + by n 1 then follows by Approach 2. Consider the set: S = {n Z n = ax + by for some x, y Z}. Show that the the minimum positive element common divisor of a and b. d S is the greatest Exercise 6.4. Find x, y Z such that: gcd(285, 255) = 285x + 255y. Definition 6.5. Two integers are relatively prime if gcd(a, b) = 1. a, b Z Claim 6.6. Two integers a, b Z are relative prime if and only if there exist x, y Z such that. ax + by = 1 a, b gcd(a, b) = 1 x, y Z such that: If are relatively prime, then by definition. So, by Bézout's Lemma there exist ax + by = gcd(a, b) = 1. ax + by = 1 x, y Z a b 1 1 ±1, we concldue that gcd(a, b) = 1. Conversely, suppose for some. Then, any common divisor of and must also be a divisor of. Since is only divisible by

43 Definition 6.7. An integer p 2 is prime if its only proper divisors (i.e. divisors different from ±p ) are ±1. Lemma 6.8. (Euclid's Lemma) a, b p p ab p a p b Let be integers. If is prime and, then and/or. p b p b gcd(p, b) = 1 1 = px + by x, y Z p apx p aby p a = a(px + by) Suppose does not divide (Notation: ), then, which implies that for some. Since and, we have. =1 More generally, Claim 6.9. If a, b are relatively prime and a bc, then a c. Exercise. Claim If a, b are relatively prime and: a c, b c, then: ab c. By assumption, there are s, t Z such that: c = as = bt. a as = bt Claim 6.9 a t gcd(a, b) = 1 t = au u Z c = bt = abu So,, which by implies that, since. Hence, for some, and we have. It follows that ab c.

44 Theorem (The Fundamental Theorem of Arithmetic) Let a 2 positive integer. Then, 1. The integer a is either a prime or a product of primes. 2. Unique Factorization The integer a may be written uniquely as p1, p2,, p l n1, n2,, n l N. a = p n 1 1 pn 2 2 p n l l, be a where are distinct prime numbers, and We prove Part 1 of the theorem by contradiction. 2 Suppose there exist positive integers which are neither primes nor products of primes. Let m be the smallest such integer. Since m is not prime, there are positive integers such that. a, b 1 m = ab a, b < m a b In particular,. So, and must be either primes or products of primes, which implies that m is itself a product of primes, a contradiction. We now prove Part 2 (Unique Factorization) of the theorem by induction. The base step corresponds to the case. Suppose: l = 1 a = p n 1 1 = q m 1 1 qm 2 2 q m k k, p1 q i n1, m i N where is prime, and the 's are distinct primes, and. Then, p1 divides the right hand side, so by Euclid's Lemma p1 divides one of the 's. q i q i Since the 's are prime, we may assume (reindexing if necessary) that p1 = q1. Suppose k > 1. If n1 > m1, then p n 1 m1 1 = q m 2 2 q m k, which implies k that is one of, a contradiction, since the 's are p1 = q1 q2,, q k q i distinct. If n1 m1, then 1 = p m 1 n1 1 q m 2 2 q m k, which is impossible. We k conclude that, and,. k = 1 p1 = q1 n1 = m1 Now we establish the inductive step: Suppose unique factorization is true for all positive integers a which may be written as a = p n 1 1 pn 2 2 p n l, for any. We want to show that it is also true for l l < l any integer a which may be written as a = p n 1 1 pn 2 2 p n l l. In other words, suppose

45 where are prime and. We want to show that, and,, for. If k < l, then by the inductive hypothesis applied to l = k < l, we have k = l, a contradiction. So, we may assume that k l. By Euclid's Lemma, divides, and hence must be equal to, one of the 's. Reindexing if necessary, we may assume that p l = q k. Cancelling p l and q k from both sides of the equation, it is also clear that n l = m k. Hence, we have: Since a = p n 1 1 pn 2 2 p n l l = q m 1 1 q m k k, p i, q i n i, m i N k = l p i = q i n i = m i i = 1, 2,, l l 1 < l p l p n 1 1 pn 2 2 p n l 1 l 1 = qm 1 1 q m k 1 k 1., we may now apply the inductive hypothesis to the integer which is equal to the left hand side of the above equation, and l 1 = k 1 p i = q i n i = m i 1 i l 1 conclude that,,, for. p n l l Since we already know that matches, we have, and 1 i l, n i = m i, for. This establishes the inductive step, and completes the proof. q m k q i k l = k p i = q i Exercise Webwork Problem 2. Webwork Problem 3. Webwork Problem Congruences Definition Let m be a positive integer, then are said to be: a, b Z m (a b) if. congruent modulo m a b mod m, Claim 6.14.

46 The congruence relation is: Reflexive: a a Symmetric: Transitive: mod m ; a b mod m b a mod m is an equivalence relation. In other words, it implies that ; a b mod m b c mod m a c mod m,, imply that. Reflexivity Since, we have. Symmetry m 0 = (a a) a a mod m a b mod m m a b m If, then by definition divides. But if divides a b, it must also divide (a b) = b a, which implies that b a mod m. Transitivity m (a b) m (b c) m ((a b) + (b c)) = (a c) mod m If and, then, which implies that a c. Note. a 0 mod m if and only if m a. Claim a = qm + r a r mod m 0 r < r < m r r mod m 1. If, then. 2. If, then. Exercise. Corollary m 2 a Z m {0, 1, 2,, m 1} Given integer, every is congruent modulo to exactly one of.

47 By Part 1 of the claim, a is congruent mod m to the remainder r of the division of a by m. By definition, the remainder lies in. If r {0, 1, 2,, m 1} a r mod m r {0, 1, 2,, m 1}, for some, then by transitivity, we have r r mod m. By Part 2 of the claim, we have r = r. Theorem Congruence is compatible with addition and multiplication in the following sense: Addition a a mod m b b mod m a + b a + b mod m. If, and, then Multiplication a a mod m b b mod m ab a b mod m If and, then. Addition m (a a ) m (b b ) If and, then: So,. Multiplication m (a a ) + (b b ) = (a + b) (a + b ). a + b a + b If and, then: So, ab a b. mod m m (a a ) m (b b ) m (a a )b + a (b b ) = (ab a b ). mod m Example For,, or. a Z a 2 0, 1 4 mod 8 By, any is congruent modulo to exactly one Corollary 6.16 a Z 8 {0, 1, 2,, 7} Theorem 6.17 a 2 element in. So, by, is congruent modulo 8 to one of:

48 The numbers above a congruent modulo 8 to 0, 1, or 4. The claim follows. {0 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 2, 7 2 } = {0, 1, 4, 9, 16, 25, 36, 49}. Theorem If a and m are relatively prime, then there exists x Z ax 1 mod m such that. Since a and m are relatively prime, by Bézout's Lemma ( Claim 6.3) there exist such that: x, y Z ax + my = 1. This implies that m divides my = 1 ax. So, by definition, we have. ax 1 mod m Theorem (Chinese Remainder Theorem) If m1 and m2 are relatively prime, then the system of congruence relations: x r1 mod m1, x r2 mod m2 has a solution. Moreover, any two solutions are congruent modulo m1m2. Since m1 and m2 are relatively prime, by a previous theorem there exists such that. Let. n Z m1n 1 mod m2 x = m1n(r2 r1) + r1 Since: m1n(r2 r1) 0 mod m1, we have: x r1 mod m1. Moreover, since m1n 1 mod m2, we have: x = m1n(r2 r1) + r1 r2 r1 + r1 r2 mod m2.

49 This shows that the system of congruence relations has at least one solution. If x is another solution to the system, then: x x r1 r1 0 mod m1, x x r2 r2 0 mod m2. m1 (x x ) m2 (x x ) m1, m2 m1m2 (x x ) So, and. Since, are relatively prime, by a previous result we conclude that x x mod m1m2.. In other words, Note. The proof of the Chinese Remainder Theorem as written above is complete. However, it is worthwhile to explain how we come up with the x = m1n(r2 r1) + r1 solution in the first place. Heuristically, the solution may be arrived at as follows: For any q Z, x = m1q + r1 is a solution to the first congruence relation. We want to find q such that m1q + r1 is also a solution to the second congruence relation, that is: m1q + r1 r2 mod m2 or, equivalently, m1q r2 r1 mod m2. ( ) Noting that there exists an n Z such that m1n 1 mod m2, the congruence relation is equivalent to: Hence, ( ) q n(r2 r1) mod m2. x = m1q + r1 is a solution to the system of congruence q m2l + n(r2 r1) l Z l = 0 q = n(r2 r1) x = m1n(r2 r1) + r1 relations if and only if is of the form, where. In particular, gives. Hence, is a solution. Exercise Webwork Problem 2. Webwork Problem 3. Webwork Problem 4. Webwork Problem 5. Webwork Problem

50 6. Webwork Problem 7. Webwork Problem 8. Webwork Problem 9. Webwork Problem 10. Webwork Problem 11. Webwork Problem 12. Webwork Problem 13. Webwork Problem 14. Webwork Problem

51 Math 2070 Week 7 Polynomials Rings Keywords: Division Theorem for Q[x] Greatest Common Divisor Bézout's Identity for Polynomials irreducible Unique Factorization for Polynomials ring zero ring Frequently Occurring: polynomial coefficient degree remainder divide gcd identity zero factorization irreducible product element integer ring additive inverse multiplicative addition multiplication real matrix Polynomials with Rational Coefficients Notation: Q = Set of rational numbers Q[x] = Set of polynomials with rational coefficients = {a0 + a1x + + a n x n n Z 0, a i Q} Q[x] f, g Q[x] Theorem 7.1. Division Theorem for For all, such that f 0 q, r Q[x] deg r < deg f that g = fq + r., there exist unique, satisfying, such

52 We first prove the existence of q and r, via induction on the degree of g. The base step corresponds to the case. In this case, the deg g < deg f q = 0 r = g g = f 0 + g deg r = deg g < deg f choice, works, since, and. Now, we establish the inductive step. Let f be fixed. Given g, suppose for all g with deg g < deg g, there exist q, r Q[x] such that g = fq + r, with deg r < deg f. We want to show that there exist q, r such that, with. Suppose and, where. We may assume that, since the case (i.e. Consider the polynomial: Then,, and by the induction hypothesis we have: for some such that. Hence, g = fq + r deg r < deg f g = a0 + a1x + + a m x m f = b0 + b1x + + b n x n a m, b n 0 m n m < n deg g < deg f) has already been proved. deg g < deg g q, r Q[x] which implies that: This establishes the existence of the quotient and the = r remainder r. Now, we prove the uniqueness of q and r. Suppose, where, with g = fq + r = fq + r deg r, deg r < deg f which implies that: g a = g m x m n f. b n. We have: g = fq + r deg r < deg f a g m x m n f = g = fq + r, b n g = f (q a + m x m n ) + r b n fq + r = fq + r, q = q + x m n The above inequality can hold only if q = q, which in turn implies that r = r. It follows that the quotient q and the remainder r are unique. a m b n q, q, r, r Q[x] deg f(q q ) = deg(r r) < deg f.

53 f, g Q[x] d f Q[x] Definition 7.2. Given, a Greatest Common Divisor of and g is a polynomial in which satisfies the following two properties: 1. d divides both f and g. 2. For any which divides both and, we have e Q[x] f g deg e deg d. Claim 7.3. g = fq + r d g f d f r If, and is a GCD of and, then is a GCD of and. See the proof of Lemma Corollary 7.4. The Euclidean Algorithm applies to. Namely: Suppose. let,, and let be the unique polynomial in for some. For Let g k = f k 1, f k = r k 1. be the remainder such that: for some. Since q0 Q[x] k > 0, let: r k, we have: Q[x] deg g deg f g0 = g f0 = f r0 Q[x] such that: g0 = f0q0 + r0, deg r0 < deg f0, q k Q[x] deg r k < deg f k = deg r k 1 g k = f k q k + r k, deg r0 > deg r1 > deg r2 > deg 0 = (where by convention we let ). Eventually, r n = 0 for some n, and it follows from the previous claim and arguments similar to those used in the case of Z that r n 1 is a GCD of f and g. Example 7.5. x x Q[x] x 3 x 2 x + 1 x 3 + 4x 2 + x 6 Q[x] 1. Find a GCD of and in. 2. Find a GCD of and in.

54 Corollary 7.6. (Bézout's Identity for Polynomials) For any which are not both zero, and a GCD of and, there exist f, g Q[x] d f g u, v Q[x] such that: d = fu + gv. Factorization of Polynomials Definition 7.7. A polynomial p in is irreducible if it satisfies the following conditions: 1., Q[x] deg p > 0 p = ab a, b Q[x] a b 2. if for some, then either or is a constant. Claim 7.8. p Q[x] p f1f2 f1, f2 Q[x] p f1 If is irreducible and, where, then or p f2. Suppose p does not divide f2, then the only common divisors of p and f2 are constant polynomials. In particular, 1 is a GCD of p and f2. Then, by, there exist such that. We have: Corollary 7.6 u, v, Q[x] 1 = pu + f2v f1 = puf1 + f1f2v. Since p divides the right hand side of the above equation, it must divide f1. Theorem 7.9. A polynomial in Q[x] either irreducible or a product of irreducibles. of degree greater than zero is > 0 Suppose there is a nonempty set of polynomials of degree which are neither irreducible nor products of irreducibles. Let p be an element of this set which has the least degree. Since p is not irreducible, there are

55 a, b Q[x] > 0 p = ab a, b deg p of degrees such that. But,, having degrees strictly less than, must be either irreducible or products of irreducibles. This implies that p is a product of irreducibles, a contradiction. Remark: Compare this proof with the proof that every integer is either prime or a product of primes ( Theorem 6.11). Theorem [Unique Factorization for Polynomials] p Q[x] > 0 For any of degree, if: p = f1f2 f n = g1g2 g m, f i, g j Q[x] n = m g j where are irreducible polynomials in, then, and the 's may be reindexed so that f i = λ i g i for some λ i Q, for. i = 1, 2,, n Exercise. See the proof of the unique factorization of integers ( Theorem 6.11). Rings Definition of a Ring Definition A ring (or ) is a set equipped with two operations: which satisfy the following properties: + 1. Properties of : R (R, +, ), + : R R R a + b = b + a a, b R a + (b + c) = (a + b) + c 0 R 1. Commutativity:,. 2. Associativity:. 3. There is an element (called the additive identity element), such that a + 0 = a for all a R. 4. Every element of R has an additive inverse; namely: For all a R, there exists an element of R, usually denoted a, such

56 a + ( a) = 0 that. 2. Properties of : 1. Associativity: a(bc) = (ab)c. 1 R 1 a = a 1 = a a R 2. There is an element (called the multiplicative identity element), such that for all. 3. Distributativity: a (b + c) = a b + a c a, b, c R (a + b) c = a c + b c a, b, c R 1., for all. 2., for all. Note: 1. For convenience's sake, we often write ab for a b. 2. In the definition, commutativity is required of addition, but not of multiplication. 3. Every element has an additive inverse, but not necessarily a multiplicative inverse. That is, there may be an element a R such that for all b R. ab 1 Claim In a ring R unique multiplicative identity element., there is a unique additive identity element and a 0 R 0 + r = r r R = R = 0 1 r = r 0 = 0 r R 1 1 = = = 1 Suppose there is an element such that for all, then in particular. Since is an additive identity, we have. So,. Suppose there is an element such that or all, then in particular. But since is a multiplicative identity element, so. Exercise Prove that: For any r in a ring R, its additive inverse r is unique. That is, if, then. r + r = r + r = 0 r = r Example The following sets, equipped with the usual operations of addition and multiplication, are rings: 1. Z, Q, R 2.

57 Z[x] Q[x] R[x],, (Polynomials with integer, rational, real coefficients, respectively.) 3. Q[ 2] = { n a k ( 2) k a k Q, n N} k=0 = {a + b 2 a, b Q}. 4. For a fixed n, the set of n n matrices with integer coefficients. 5. C[0, 1] = {f : [0, 1] R f is continuous.} 6., the set of real matrices,. M n (R) n n n N The following sets, under the usual operations of addition and multiplication, are not rings: 1. N, no additive identity element, i.e. no 0. 2., nonzero elements have no additive inverses. N {0} GL(n, R) n n n N 3., the set of invertible real matrices,. Claim For all elements in a ring, we have. r R 0r = r0 = 0 By distributativity, 0r = (0 + 0)r = 0r + 0r. 0r Adding (additive inverse of 0r) to both sides, we have: 0 = (0r + 0r) + ( 0r) = 0r + (0r + ( 0r)) = 0r + 0 = 0r. The proof of r0 = 0 is similar and we leave it as an exercise. Claim ( 1)( r) = ( r)( 1) = r For all elements r in a ring, we have. We have: 0 = 0( r) = (1 + ( 1))( r) = r + ( 1)( r). Adding r to both sides, we obtain