< x , , xn > G/N Nghĩa ký hiệu Set of natural numbers Set of integers Set of rational numbers Set of real numbers Set of complex numbers integers modulo m n Group of invertible elements of Zn Center of a group G The number of elements of S Order of a group G Index of G over H normal subgroup Direct sum Submodule generated by P Submodule generated by x1 , , xn Quotient group of G over N ii PREFACE iii iv CONTENTS NOTATIONS ii PREFACE iii Chapter Binary Operations Groups Subgroups cyclic groups Normal subgroups and quotient groups Group homomorphisms Direct product of groups 1 11 15 19 25 33 37 37 42 47 52 52 Rings Rings and Fields Subrings, Ideals and quotient rings Ring homomorphisms Characteristic of rings Field of fractions of a integral domain Application 53 References 55 v Chương Groups §1 BINARY OPERATIONS Definition 1.1 Let X be a non-empty set A binary operation ∗ on X is a map ∗: X ×X →X (x, y) → x ∗ y, For binary operations ∗ : X × X → X, two notations are in widely used: the additive notation and the multiplicative notation In the additive notation, x ∗ y is denoted by x + y; then ∗ is an addition In the multiplicative notation, x ∗ y is denoted by x.y or by xy; then ∗ is a multiplication In this chapter we mostly use the multiplicative notation Addition (+) and multiplication (.) are two familiar binary operations on appropriate sets For addition, x + y is called the sum of x and y; for multiplication, x.y or xy is called the product of x and y Example 1.2 (1) The ordinary addition on N (or Z, Q, R, C ) is a binary operation on N (or Z, Q, R, C ) (2) The ordinary multiplication on N (or Z, Q, R, C ) is a binary operation on N (or Z, Q, R, C ) (3) Denote X X by the set of the maps from X to X For f, g ∈ X X , the composition map f ◦ g of g and f is a binary operation on X X Chương Chapter (4) Addition and multiplication of matrices also provide binary operations on the set Mn (R) of all n × n matrices with coefficients in a field K, for any given integer n > Definition 1.3 Let ∗ be a binary operation on X (1) The binary operation ∗ is said to be associative if for all elements x, y, z ∈ X we have (x ∗ y) ∗ z = x ∗ (y ∗ z) (2) The binary operation ∗ is said to be commutative if for all elements x, y ∈ X we have x ∗ y = x ∗ y (3) An element x ∈ X is said to be cancelable on the left (right) if for all y, z ∈ X, x ∗ y = x ∗ z =⇒ y = z (resp y ∗ x = z ∗ x =⇒ y = z) (4) An element x ∈ X is said to be cancelable if it is cancelable on the left and right (5) The binary operation ∗ is said to be cancelable on X if for all x ∈ X, x is cancelable on X The ordinary addition and multiplication on N (or Z, Q, R, C ) in Example 1.2 are associative and commutative The operation ◦ on X X is associative but it is not commutative By definition, associativity states that products with three terms, or more, not depend on the placement of parentheses Definition 1.4 Let ∗ be a binary operation on X (1) An element e of X is called an identity of ∗ on X if e ∗ x = x = x ∗ e for all x ∈ X (2) Let e be an identity of ∗ on X An element x of X is said to be left invertible (or right invertible) if there exists x ∈ X such that x ∗ x = e (resp x ∗ x = e) (3) Let e be an identity of ∗ on X An element x of X is said to be invertible if x is left invertible and right invertible; that is, there exists x ∈ X such that x ∗ x = x ∗ x = e Then the such element x is called the inverse of x and denote by x−1 ✎ Cao Huy Linh -College of Education-Hue University § Binary Operations Readers will easily show that an identity element, if it exists, is unique The identity of additional operation is usually called the neutral element In the multiplicative (additional) notation, we usually denote the identity (resp.neutral) element, if it exists, by (resp 0) Definition 1.5 (1) A non-empty set X along with a binary operation ∗ on its is said to be a semi-group if the operation ∗ is associative, written by (X, ∗) (2) A semi-group (X, ∗) is said to be a monoid if the operation ∗ has an identity e (3) A monoid (X, ∗) is said to be a group if every element of X is invertible A group X is said to be an Abel group if the operation ∗ is commutative Example 1.6 (1) (N, +) is a semi-group (2) (N0 , +) and (Z, ·) are monoids but they are not groups (3) (Z, +) is a group; more exactly, (Z, +) is a Abel group Definition 1.7 A non-empty set X along with two binary operations addition (+) and multiplication (·) (1) X is said to be a ring if it satisfies three properties (1) (X, +) is a Albel group (2) (X, ·) is a semi-group (3) The multiplication · is distributive to the addition + on the left and right; that is, x(y + z) = xy + xz and (x + y)z = xz + yz If the operation · on the ring X is commuatative (resp has identity), we say X is a commutative ring (resp with identity) (2) A ring X with identity (with the neutral element denotes by 0) is said to be a skew-field if every non-zero element x in X is invertible (3) A commutative skew-field X is said to be a field Another way to say that a commutative ring X with identity is said to be a field if every non-zero element x in X is invertible ✎ Cao Huy Linh -College of Education-Hue University Chương Chapter Example 1.8 Let + and · be the ordinary addition and multiplication on the number sets N, N0 , Z, Q, R, C Then (1) (Z, +, ·) is a commutative ring (2) (Q, +, ·), (R, +, ·) and (C, +, ·) are fields Let X be a monoid, and xl , , xn be elements of G (where n is an integer > 1) We define their product inductively: n xi = x1 · · · xn = (x1 · · · xn−1 ) · xn i=1 xn = x1 · · · xn which xi = x for i = n and x0 = e If (X, +) be a monoid, then we define nx = x1 + · · · + xn where xi = · · · xn for i = n EXERCISES 1.1 Let X X denote the set of the maps from X to X and operation of the maps Show that the composite a) f ∈ X X satisfies the cancellation law on the left if and only if f is injective b) f ∈ X X satisfies the cancellation law on the right if and only if f is surjective c) f ∈ X X satisfies the cancellation law if and only if f is bijective 1.2 Let ∗ be a binary operation on X such that x ∗ y = y for all x, y ∈ X An element e of X is called the left (resp right) identity if e ∗ x = x (resp x ∗ e = x for all x ∈ X Show that a) (X, ∗) is a simi-group which having the left identity b) If X has at least two elements, then it doesn’t have the right identity ✎ Cao Huy Linh -College of Education-Hue University 41 § Rings and Fields EXERCISES 1.1 In the definition of a ring with identity, show that one may omit the requirement that the addition be commutative 1.2 Let A be an additional Abel group Denote by End(A) the set of group homomorphisms from A to itself We define two binary operation on End(A) that, for all f, g ∈ End(A), (f + g)(x) = f (x) + g(x) for all x ∈ A, f g(x) = f (g(x)) for all x ∈ A Show that End(A) along with two above operations is a ring 1.3 A unit of a ring R with identity is an element u of R such that uv = vu = for some v ∈ R Show that v is unique (given u) Show that the set of all units of R is a group under multiplication 1.4 Let R be a ring with identity Show that u is a unit of R if and only if xu = uy = 1R for some x, y ∈ R 1.5 Show that an element x ∈ Zn is a unit of Zn if and only if x and n are relatively prime 1.6 Find all units of the ring of Gauss integer numbers √ 1.7 Show that complex numbers of the form a + ib 2, where a and b are integers, constitute a ring Find the units 1.8 Prove that Zm is a field if and only if m is prime 1.9 Let R be a ring Show that R1 = R × Z, with operations (x, m) + (y, n) = (x + y, m + n), and (x, m)(y, n) = (xy + nx + my, mn), is a ring with identity 1.10 A ring R is said to be von Neumann regular if for all a ∈ R, there exists an element x ∈ R such that axa = a Prove that the ring R1 in Exercise 1.9 is not von Neumann regular ✎ Cao Huy Linh -College of Education-Hue University 42 Chương Rings 1.11 Let R be a ring such that every element x in R satisfies x2 = x Prove that R is commutative 1.12 Let R be a ring such that every element x in R satisfies x2 = −x Prove that R is commutative 1.13 Let R be a ring such that every element x in R satisfies x3 = x Prove that R is commutative 1.14 Let R be a non zero ring such that the equation ax = b has a solution in R for all a ∈ R \ {0R } and b ∈ R Prove that R is a skew-field 1.15 Let R be a non-zero ring such that for all a ∈ R \ {0R } there exists a unique b ∈ R such that aba = a Prove that R is a skew-field 1.16 Prove that a finite ring R with identity such that every non-zero element of R is a non-zero divisor is a skew-field §2 SUBRINGS, IDEALS AND QUOTIENT RINGS Subrings of a ring R are subsets of R that inherit a ring structure from R Definition 2.1 A subring of a ring R is a non-empty subset S of R such that S is a subgroup of (R, +), is closed under multiplication (x, y ∈ S implies xy ∈ S) Example 2.2 (1) If R is a ring, {0R } and R are subrings of R We say {0R } is called the trivial subring of R and R is called the improper subring of R (2) The ring of integers Z is a subring of the ring of rational numbers Q (3) If m ∈ Z, mZ is a subring of Z Proposition 2.3 Let R be a ring and S be a non-empty subset of R The following conditions are equivalent (i) S is a subring of R (ii) For all x, y ∈ S: x + y ∈ S, −x ∈ S, and xy ∈ S ✎ Cao Huy Linh -College of Education-Hue University § Subrings, Ideals and quotient rings 43 (ii) For all x, y ∈ S: x − y ∈ S and xy ∈ S Proof The proof is left as an exercise for the reader Example 2.4 Let R be a ring and the center of R is defined by Z(R) = {x ∈ R | xa = ax for all a ∈ R} Then, Z(R) is a subring of R Indeed, as 0R x = x0R = 0, 0R ∈ Z(R) It follows Z(R) = ∅ Next, we take any x, y ∈ Z(R) Then, for all a ∈ R, (x − y)a = xa − ya = ax − ay = a(x − y) Thus x − y ∈ Z(R) In addition, (xy)a = x(ya) = x(ay) = (xa)y = (ax)y = a(xy) This implies xy ∈ Z(R) So, Z(R) is a subring of R Definition 2.5 Let R be a ring A subring I of (R) is said to be a left (right) ideal of R if for all r ∈ R and for all x ∈ I, we have rx ∈ I (resp xr ∈ I) A subset I of R is said to be an ideal of R if I is both left and right ideal of R From now, for short we say an ideal instead of a left ideal Example 2.6 (1) If R is a ring, {0R } and R are ideals of R We say {0R } is called the trivial ideal of R and R is called the improper ideal of R (2) If m ∈ Z, mZ is a subring of Z (3) In the ring of rational numbers Q, Z is not an ideal of Q Proposition 2.7 Let R be a ring and ∅ = I ⊆ R Then the following conditions are equivalent (1) I is an ideal of R (2) For all x, y ∈ I and for all a ∈ R: x + y ∈ I, −x ∈ I, ax ∈ R (3) For all x, y ∈ I and for all a ∈ R: x − y ∈ I, ax ∈ R Proof The proof is left as an exercise for the reader Example 2.8 Let R be a ring and z ∈ R Then Rz = {xz | x ∈ R} is a left ideal of R Indeed, since 0R = 0.z ∈ Rz, Rz = ∅ For all x, y ∈ Rz and for all a ∈ R, there exists x1 , y1 ∈ R such that x = x1 z and y = y1 z We ✎ Cao Huy Linh -College of Education-Hue University 44 Chương Rings have x − y = x1 z − y1 z = (x1 − y1 )z ∈ Rz On the other hand, ax = a(x1 z) = (ax1 )z ∈ Rz This implies that Rz is a left ideal of R Similarly, zR is a right ideal of R Proposition 2.9 The intersection of a non-empty family of left (right) ideals is a left (right) ideal of R Proof We need to show for one side ideal and another side is similar Assume that Λ = ∅ and Ij are left ideals of R for all j ∈ Λ Set I = j∈Λ Ij We need to prove that I is a left ideal of R Indeed in Chapter 1, I is a subgroup of the additional group (R, +) Now, for all a ∈ R and x ∈ I, then x ∈ Ij for all j ∈ Λ It follows ax ∈ Ij for all j ∈ Λ Therefore ax ∈ ∩j∈J Ij = I This implies that I is a left ideal of R Now, let X be a subset of a ring R The family of left ideals of R that contain X is non-empty, because R is an element of this family So, the intersection of the family of left ideals of R that contain X is a left ideal of R Definition 2.10 Let X be a subset of a ring R The intersection of the family of left (right) ideals of R that contain X is called the left (reght) ideal of R generated by X; this ideal is denoted by (X (res X)) Remark 2.11 Let X be a subset of a ring R (i) (X is the smallest left ideal of R that contains X (ii) (∅ = ∅) = {0R } (iii) If X = {x1 , , xr }, we can write (x1 , , xr instead by ({x1 , , xr } and x1 , , xr ) instead by {x1 , , xr }) Example 2.12 Let R be a ring with identity, z ∈ R and X = {z} Then (z = Rz and z) = zR Definition 2.13 Let R be a ring (1) If I1 , , Ir are left ideals of R, then we define r I1 + · · · + Ir = ( Ii i=1 ✎ Cao Huy Linh -College of Education-Hue University 45 § Subrings, Ideals and quotient rings (2) If I1 , , Ir are right ideals of R, then we define r I1 + · · · + Ir = Ii ) i=1 Example 2.14 Let R be a ring with identity, z1 , , zr ∈ R Then (z1 , , zr = Rz1 + · · · + Rzr and z1 , , zr ) = z1 R + · · · + zr R Now, let R be a ring and I both sides ideal of R Then, I is a normal subgroup of additional group R We have a quotient group (R/I, +) with elements x¯ = x + I for x ∈ R and x¯ + y¯ = x + y Notice that x¯ = y¯ iff x − y ∈ I Define a multiplication on R/I as follows x¯.¯ y = x.y (that means (x + I)(y + I) = xy + I) Proposition 2.15 The additional group (R/I, +) along with above multiplication becomes a ring Proof The proof is left as an exercise for the reader Definition 2.16 The ring R/I as in Proposition 2.15 is called a quotient ring of R for I Example 2.17 Take R = Z and I = mZ is both sides ideal of Z Then Z/mZ = Zm EXERCISES 2.1 Show that every intersection of subrings of a ring R is a subring of R 2.2 Show that intersection of the family of subrings of R that contains a subset X of R is the smallest subring of R This subring is called the subring of R generated by X, it is denoted by [X] ✎ Cao Huy Linh -College of Education-Hue University 46 Chương Rings 2.3 Show that the subring [X] of R generated by X is the set of all sums of products of elements of X and opposites of such products 2.4 Let A be a ring and P ⊆ A Put Z(P ){x ∈ A | xp = px for all p ∈ P } Prove that a) Z(P ) is a subring of R b) If P1 ⊂ P2 , then Z(P2 ) ⊂ Z(P1 ) c) Z(Z(Z(P ))) = Z(P ) for all P ⊂ A d) Z(P ) = Z([P ]) for all P ⊂ A 2.5 Let I and J be ideals of a ring R Show that union of I and J is an ideal of R if and only if I ⊆ J or J ⊆ I 2.6 Let R be a ring (without identity) and z ∈ R Show that (z = Rz + Zz and z) = zR + zZ 2.7 Find an example a ring R and a left ideal I of R but I is not a right ideal of R 2.8 An element x of a ring is nilpotent when xn = for some n > Show that the nilpotent elements of a commutative ring R constitute an ideal of R 2.9 Let R be a ring and m ∈ Z Show that the subset I = {x ∈ A | mx = 0R } is an ideal of R 2.10 Let R be a ring and X = {xy − yx | x, y ∈ R} Show that R/(X) is a commutative ring ✎ Cao Huy Linh -College of Education-Hue University 47 § Ring homomorphisms §3 RING HOMOMORPHISMS In this section, we will study maps from a ring R into a ring S that presrves two binary operations between R and S Definition 3.1 Let R and S be rings A map f : R −→ S is said to be a ring homomorphism if (i) f (x + y) = f (x) + f (y) for all x, y ∈ R; (ii) f (x.y) = f (x).f (y) for all x, y ∈ R Proposition 3.2 Let f : R −→ S be a ring homomorphism (i) f (0R ) = 0S ; (ii) f (−x) = −f (x) for all x ∈ R; (iii) f (x − y) = f (x) − f (y) Proof The proof is left as an exercise for the reader Example 3.3 (1) Let R, S be rings The map f : R −→ S x −→ f (x) = 0S is a ring homomorphism and it is called a trivial homomorphism (2) Let S be a subring of a ring R The map j : S −→ R x −→ j(x) = x is a ring homomorphism and it is called a inclusive homomorphism In particular, if S = R then j = IdR is identification homomorphism (3) Let I be a two-sides ideal of a ring R The map p : R −→ R/I x −→ p(x) = x¯ = x + I is a ring homomorphism and it is called a canonical projection (or canonical epimorphism) ✎ Cao Huy Linh -College of Education-Hue University 48 Chương Rings Proposition 3.4 Let f : R −→ S be a ring homomorphism (i) If R1 is a subring of R, then f (R1 ) is a subring of S; (ii) If S1 is a subring of S, then f −1 (S1 ) is a subring of R Proof (i) Since 0R ∈ R1 , 0S = f (0R ) ∈ f (R1 ) It follows f (R1 ) = ∅ For all y1 , y2 ∈ f (R1 ), there exist x1 , x2 ∈ R1 such that y1 = f (x1 ) and y2 = f (x2 ) Then, y1 − y2 = f (x1 ) − f (x2 ) = f (x1 − x2 ) and y1 y2 = f (x1 ).f (x2 ) = f (x1 x2 ) Since R1 is a subring of R, x1 − x2 ∈ R1 and x1 x2 ∈ R1 Hence y1 − y2 ∈ f (R1 ) and y1 y2 ∈ f (R1 ) This implies that f (R1 ) is a subring of S (ii) It is clear that f −1 (S1 ) = ∅ Suppose that x1 , x2 ∈ f −1 (S1 ) Then f (x1 ) ∈ S1 and f (x2 ) ∈ S1 Since S1 is a subring of S, f (x1 − x2 ) = f (x1 ) − f (x2 ) ∈ S1 and f (x1 x2 ) = f (x1 ).f (x2 ) ∈ S1 Hence x1 −x2 ∈ f −1 (S1 ) and x1 x2 ∈ f −1 (S1 ) This implies f −1 (S1 ) is a subring of R The above proposition tell us that ring homomorphisms preserve structure of subrings Remark 3.5 Let f : R −→ S be a ring homomorphism (i) If I is a left ideal of R, then f (I) is not necessary a left ideal of S For example, taking the inclusive homomorphism j : Z −→ Q Then Z is an ideal of Z, but j(Z) = Z is not an ideal of Q (ii) If J is a left of S, then f −1 (J) is a left ideal of R Readers can prove this assertion ✎ Cao Huy Linh -College of Education-Hue University § Ring homomorphisms 49 Proof By Proposition 3.4, f −1 (J) is a subring of R For all a ∈ R and x ∈ f −1 (J), we have f (x) ∈ J Since J is a left ideal of S, f (ax) = f (a).f (x) ∈ J Hence ax ∈ f −1 (J) So, f −1 (J) is a left ideal of R Definition 3.6 Let f : R −→ S be a ring homomorphism Then, (i) Im(f ) = f (R) is called the image of f ; (ii) Ker(f ) = f −1 (0S ) = {x ∈ R | f (x) = 0S } is called the kernel of f Notice that Im(f ) is a subring of S and Ker(f ) is a two-sides ideal of R Definition 3.7 (i) A ring homomorphism f : R −→ S is said to be monomorphic (res epimorphic, isomorphic) if f is injective (res surjective, bijective) (ii) Two rings R and S are called isomorphic to each other if there exists a ring isomorphism f : R −→ S; then denote by R ∼ = S The following theorem give a criteria for a ring homomorphism to be monomorphic (res epimorphic, isomorphic) Theorem 3.8 Let f : R −→ S be a ring homomorphism (i) f is monomorphic if and only if Ker(f ) = {0R } (ii) f is epimorphic if and only if Im(f ) = S Proof (i) Suppose that the ring homomorphism f is monomorphic Then For all x ∈ Ker(f ), we have f (x) = 0S = f (0R ) Since f is injective, x = 0R It follows Ker(f ) = {0R } On the other hand, suppose that Ker(f ) = {0R } Taking any x1 , x2 ∈ R such that f (x1 ) = f (x2 ) Then f (x1 −x2 ) = f (x1 )−f (x2 ) = 0S Thus x1 −x2 ∈ Ker(f ) By hypothesis Ker(f ) = {0R }, hence x1 − x2 = 0R This implies that x1 = x2 So, f is injetive; that means f is monomorphism (ii) Easy and the proof is left for the reader Corollary 3.9 Let f : R −→ S be a ring homomorphism Then, f is isomorphic if and only if Ker(f ) = {0R } and Im(f ) = S Theorem 3.10 Let f : R → S be a ring homomorphism and g : R −→ T be a ring epimorphism such that Ker(g) ⊆ Ker(f ) There exists a unique ring homomorphism h : T −→ S such that f = hg Moreover, ✎ Cao Huy Linh -College of Education-Hue University 50 Chương Rings (i) If Ker(g) = Ker(f ), then h is monomorphic; (ii) If f is epimorphic, then h is epimorphic Proof For any y ∈ T , there exists x ∈ R such that y = g(x) (by hypothesis g is epimorphic) Then, set h(y) = f (x) This defines a map h : T → S Indeed, for any y1 , y2 ∈ T , there exist x1 , x2 ∈ R such that y1 = g(x1 ), y2 = g(x2 ) Suppose that y1 = y2 , then g(x1 ) = g(x2 ) It follows g(x1 − x2 ) = 0T Hence x1 − x2 ∈ Ker(g) By hypothesis Ker(g) ⊆ Ker(f ), x1 − x2 ∈ Ker(f ) That means f (x1 − x2 ) = f (x1 ) − f (x2 ) = 0S This implies that h(y1 ) = f (x1 ) = f (x2 ) = h(y2 ) So, h is a map Next, we need to prove h is a ring homomorphism and f = hg Indeed, for all y1 , y2 ∈ T Suppose y1 = g(x1 ) and y2 = g(x2 ) One has y1 + y2 = g(x1 ) + g(x2 ) = g(x1 + x2 ) and y1 y2 = g(x1 ).g(x2 ) = g(x1 x2 ) Then, h(y1 + y2 ) = f (x1 + x2 ) = f (x1 ) + f (x2 ) = h(y1 ) + h(y2 ) and h(y1 y2 ) = f (x1 x2 ) = f (x1 ).f (x2 ) = h(y1 ).h(y2 ) Hence h is a ring homomorphism Moreover, for any x ∈ R, set y = g(x) We have hg(x) = h(g(x)) = h(y) = f (x) It follows that hg = f Finally, soppose that there is a ring homomorphism h : T → S such that f = h g Then, for all y ∈ T , there exists x ∈ R such that y = g(x) we have h (y) = h (g(x)) = h g(x) = f (x) = h(y) This implies that h = h (i) Now, suppose that Ker(g) = Ker(f ) For all y ∈ Ker(h) and y = g(x), h(y) = f (x) = 0S Hence x ∈ Ker(f ) = Ker(g) This implies y = g(x) = 0T It follows that h is monomorphic (ii) Suppose that f is epimorphic Then, Im(h) = h(T ) = h(g(R)) = hg(R) = f (R) = Im(f ) = S So, h is epimorphic ✎ Cao Huy Linh -College of Education-Hue University 51 § Ring homomorphisms From the above theorem, we obtain the following corollary Corollary 3.11 d Let f : R −→ S be a ring homomorphism Then, (i) If f is epimorphic, then R/ Ker(f ) ∼ = S (ii) R/ Ker(f ) ∼ = S Proof EXERCISES 3.1 Let f : R −→ R be a ring homomorphism (it is called a ring endomorphism) and B = {x ∈ R | f (x) = x} Show that B is a subring of R 3.2 Let Mn (R) be the ring of n by n matrices with coefficients in the ring R If Ck is the subset of Mn (R) consisting of matrices that are except perhaps in column k,show that Ck is a left ideal of Mn (R) Similarly,if Rk consists of matrices that are except perhaps in row k,then Rk is a right ideal of Mn (R) 3.3 Let R be a commutative ring whose only proper ideals are {0} and R Show that R is a field 3.4 a) Let f : R −→ S be an epimorphism of rings and let I be a left ideal of R Show that f (I) is a left ideal of S b) Find a ring homomorphism f : R −→ S of commutative rings and a left ideal I of R such that f (I) is not a left ideal of S 3.5 Let f : R −→ S be a homomorphism of rings and let J be an ideal of S Show that f −1 (J) is an ideal of R 3.6 Let R be a ring and let I be an ideal of R Show that every ideal of R/I is the quotient J/I of a unique ideal J of R that contains I ✎ Cao Huy Linh -College of Education-Hue University 52 Chương Rings 3.7 Let I ⊆ J be ideals of a ring R Show that (R/I)/(J/I) ∼ = R/J 3.8 Let R be a ring and I, J be ideals of R Denote by I + J is the smallest ideal of R that contains I ∪ J Show that I + J = {x + y | x ∈ I, y ∈ J.} 3.9 Let R be a ring and I, J be ideals of R Show that I + J/J ∼ = I/(I ∩ J 3.10 Let S be a subring of a ring R and let I be an ideal of R Show that S + I = {x + y | x ∈ S, y ∈ I} is a subring of R , I is an ideal of S + I , S ∩ I is an ideal of S , and (S + I)/I ∼ = S/(S ∩ I) §4 CHARACTERISTIC OF RINGS §5 FIELD OF FRACTIONS OF A INTEGRAL DOMAIN ✎ Cao Huy Linh -College of Education-Hue University Chương Application 53 54 Chương Application ✎ Cao Huy Linh -College of Education-Hue University BIBLIOGRAPHY [1] Bùi Huy Hiền - Phan Doãn Thoại - Nguyễn Hữu Hoan, Bài tập đại số số học, tập II Nhà xuất Giáo dục, 1985 [2] Lê Thanh Hà, Các cấu trúc đại số Nhà xuất Giáo dục, 1999 [3] Lê Thanh Hà, Đa thức Nhân tử hóa Nhà xuất Giáo dục, 2000 [4] Lê Thanh Hà, Môđun Đại số Nhà xuất Giáo dục, 2002 [5] Nguyễn Hữu việt Hưng, Đại số đại cương Nhà xuất Giáo dục, 1998 [6] Nguyễn Xuân Tuyến - Lê văn Thuyết, Đại số trừu tượng Nhà xuất Giáo dục, 2005 [7] M.F.Atiyah-I.G.Macdonald Introduction to Commutative Algebra Addison-wesley Publishing Company, 1969 [8] M.Hall, The Theory of Groups New York: Macmillan, 1959 [9] S.Mac Lane, G.Birkhoff, A survey of Modern Algebra The Macmillan Company, New York, 1967 [10] S.Lang, Algebra Addison-Wesley, 1971 [11] A.I.Kostrikin Introduction d’algèbre Mir-Moscou, 1976 55