IV.Rings [Compatibility Mode].pdf

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Rings,Fields

TS. Nguyễn Viết Đông

Rings,Fields

• 1. Rings, Integral Domains and Fields,

• 2. Polynomial and Euclidean Rings

• 3. Quotient Rings

1. Rings, Integral Domains and Fields

• 1.1.Rings

• 1.2. Integral Domains and Fields

• 1.3.Subrings and Morphisms of Rings

1. Rings, Integral Domains and Fields

• 1.1.Rings

• A ring (,, R + ・) is a set R, together with two binary

operations + and ・ on R satisfying the following axioms.

For any elements a, b, c א R,

(i) (a + b) + c = a + (b + c). (associativity of addition)

(ii) a + b = b + a. (commutativity of addition)

(iii) there exists 0 א R, called the zero, such that

a + 0 = a. (existence of an additive identity)

(i ) v there exists (−a) א R such that a + (−a) = 0.( i ex stence of

an additive inverse)

(v) (a ・ b) ・ c = a ・ (b ・ c). (associativity of

multiplication)

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1. Rings, Integral Domains and Fields

(vi) there exists 1 א R such that

1・ a = a ・ 1 = a. (existence (existence of multiplicative multiplicative identity) identity)

(vii) a ・ (b + c) = a ・ b+a ・ c

and (b + c)・ a=b ・ a+c ・ a.(distributivity)

• Axioms (i)–(iv) are equivalent to saying that (R,+) is an

abelian group.

• The ring (,, R + ・) is ca ed ll a co utat ve mmutative ring if, in

addition,

(viii) a ・ b=b ・ a for all a, b א R. (commutativity of

multiplication)

1. Rings, Integral Domains and Fields

• The integers under addition and multiplication satisfy all of

the axioms above,so that (Z,+, ・) is a commutative ring.

Also, (Q, +,・), (R,+, ・), and (C,+, ・) are all commutative commutative

rings. If there is no confusion about the operations, we write

only R for the ring (R,+, ・). Therefore, the rings above

would be referred to as Z,Q,R, or C. Moreover, if we refer to

a ring R without explicitly defining its operations, it can be

assumed that they are addition and multiplication.

• Many authors authors do not require require a ring to have a multiplicative multiplicative

identity, and most of the results we prove can be verified to

hold for these objects as well. We must show that such an

object can always be embedded in a ring that does have a

multiplicative identity.

1. Rings, Integral Domains and Fields

• Example 1.1.1. Show that (Zn,+, ・) is a commutative ring,

where addition and multiplication on congruence classes,

mod l u o n, are d fi d e ne by the equations

[x] + [y] = [x + y] and [x] ・ [y] = [xy].

• Solution. It iz well know that (Zn,+) is an abelian group.

Since multiplication on congruence classes is defined in

terms of representatives, it must be verified that it is well

defined. Suppose that [x] = [x’] and [y] = [y’], so that x ≡ x’

and y ≡ y’ mod n. This implies that x = x’ + kn

and y = y '+ ln for some k, l א Z. Now x ・ y = (x’ + kn) ・

(y’ + ln) = x ・ y + (ky’ + lx’ + kln)n, so x ・ y ≡ x’ ・ y’

mod n and hence [x ・ y] = [x’ ・ y’]. This shows that

multiplication is well defined.

1. Rings, Integral Domains and Fields

The remaining axioms now follow from the definitions of

addition addition and multiplication multiplication and from the properties properties of the

integers. The zero is [0], and the unit is [1]. The left

distributive law is true, for example, because

[x] ・([y] + [z]) = [x] ・ [y + z] = [x ・ (y + z)]

= [x ・ y+x ・ z] by distributivity in Z

= [x ・ y] + [x ・ z] = [x] ・ [y] + [x] ・ [z].

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Example. The “ linear equation” on Zm

[x]m + [a]m = [b]m

where [a]m and [b]m are given, has a unique solution:

[x] = [b ] [a] = [b a] m = [b ]m – [a]m = [b – a]m

Let m = 26 so that the equation [x]26 + [3]26 = [b]26 has a

unique solution for any [b]26 in Z26 .

It follows that the function [x]26 →[x]26 + [3]26 is a

bijection of Z26 j to itself.

We can use this to define the Caesar’s encryption: the

English letters are represented in a natural way by the

elements of Z26: A →[0]26 , B →[1]26 , ..., Z →[25]26

For simplicity, we write: A →0, B → 1, ..., Z →25

These letters are encrypted so that A is encrypted by

the letters represented by [0]26 + [3]26 = [3]26, i.e. D.

Similarly B is encrypted by the letters represented by

[1]26 + [3]26 = [4]26, i.e. E, ... and finally Z is encrypted

by [25]26 + [3]26 = [2]26, i.e. C.

In this way the message “MEET YOU IN THE PARK”

is encrypted as

MEET YOU I N THE PARK

12 4 4 19 24 14 20 8 13 19 7 4 15 0 17 10

1 17 23 11 16 22 10 7 18 3 20 13

P H H W B R X L Q W K H S D U N

15 7 7 22

To decrypt a message, we use the inverse function:

[x]26 → [x]26 – [3]26 = [x – 3]26

P H H W is represented by 15 7 7 22

However this simple encryption method is easily detected

And hence decrypted by 12 4 4 19

M E E T The corresponding decrypted message is

However this simple encryption method is easily detected.

We can improve the encryption using the function

f : [x]26 →[ax + b]26

where a and b are constants chosen so that this function is a

bijection

First we choose an invertible element a in Z26 i.e. there

exists a’ in Z26 such that

We write [a’ ]26 = [a]26–1 if it exists.

[a]26 [a’ ]26 = [a a’ ]26 = [1]26

[ ]26 [ ]26

The solution of the equation

[a]26 [x]26 = [c]26

is [x]26 = [a]26–1 [c]26 = [a’c]26

We also say that the solution of the linear congruence We also say that the solution of the linear congruence

a x ≡ c (mod 26)

is x ≡ a’c (mod 26)

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Example. Let a = 7 and b = 3, then the inverse of [7]26 is

Now the inverse function of f is given by

[x]26 → [a’(x – b)]26

Example. Let a 7 and b 3, then the inverse of [7]26 is

[15]26 since [7]26 [15]26 = [105]26 = [1]26

Now the letter M is encrypted as

[12]26 →[7 ⋅12 + 3]26 = [87]26 = [9]26

which corresponds to I. Conversely I is decrypted as

[9]26 →[15 ⋅ (9 – 3) ]26 = [90]26 = [12]26

which corresponds to M.

To obtain more secure encryption method, more

sophisticated modular functions can be used

1. Rings, Integral Domains and Fields

• Example 1.1.2. Show that (Q(√2),+, ・) is a

commutative commutative ring where Q(√2) ={a + b√2 א R|a, b א Q}.

Solution. The set Q(√2) is a subset of R, and the addition

and multiplication is the same as that of real numbers.

First, we check that + and ・ are binary operations on

Q(√2). If a, b, c, d א Q, we have

(a + b√2) + (c + d√2) = (a + c) + (b + d)√2 א Q(√2)

since (a + c) and (b + d) א Q. Also,

(a + b√2) ・ (c + d√2) = (ac + 2bd) + (ad + bc)√2 א Q(√2)

since (ac + 2bd) and (ad + bc) א Q.

1. Rings, Integral Domains and Fields

• We now check that axioms (i)–(viii) of a commutative ring are valid in Q(√2).

(i) Addition of real numbers is associative.

(ii) Addition of real numbers is commutative.

(iii) The zero is 0 = 0 + 0√2 א Q(√2).

(iv) The additive inverse of a+b√2 is (−a) + (−b)√2 א Q(√2),

since (−a) and (−b) א Q.

(v) Multiplication of real numbers is associative.

(vi) The multiplicative multiplicative identity identity is 1 = 1 + 0√2 א Q(√2).

(vii) The distributive axioms hold for real numbers and hence hold for elements of Q(√2).

(viii) Multiplication of real numbers is commutative.

1. Rings, Integral Domains and Fields

• 1.2. Integral Domains and Fields

• One very useful pp y ro erty of the familiar number systems is

the fact that if ab = 0, then either a=0or b = 0. This

property allows us to cancel nonzero elements because if

ab = ac and a ≠ 0, then a(b − c) = 0, so b = c. However, this property does not hold for all rings. For example, in Z4, we

have [2] ・ [2] = [0], and we cannot always cancel since

[2] ・ [1] = [2] ・ [3], but [1]≠[3].

• If (R +, , ・) is a commutative commutative ring, a nonzero nonzero element element a א R

is called a zero divisor if there exists a nonzero element b א

R such that a ・ b = 0. A nontrivial commutative ring is

called an integral domain if it has no zero divisors.

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1. Rings, Integral Domains and Fields

• A field is a ring in which the nonzero elements form

an abelian g p rou under multiplication. In other words,

a field is a nontrivial commutative ring R satisfying

the following extra axiom.

(ix) For each nonzero element a א R there exists a−1 א

R such that a ・ a−1 = 1.

• The rings Q,R, and C are all fields, but the integers

do not form a field.

• Proposition 1.2.1. Every field is an integral domain;

that is, it has no zero divisors.

1. Rings, Integral Domains and Fields

• Theorem 1.2.2. A finite integral domain is a field.

• Proof. Let D = {x0, x1, x2,..., xn} be a finite integral integral domain

with x0 as 0 and x1 as 1. We have to show that every nonzero

element of D has a multiplicative inverse.

If xi is nonzero, we show that the set xiD = {xix0, xix1, xix2,...

, xixn} is the same as the set D. If xixj = xixk, then, by the

cancellation property, xj = xk.Hence all the elements xix0, xix1,

x x x x are distinct and x D is a subset of D with the ix2,...,xixn are distinct, and xiD is a subset of D with the

same number of elements. Therefore, xiD = D. But then there

is some element, xj , such that xixj = x1 = 1.

Hence xj = xi -1, and D is a fiel

1. Rings, Integral Domains and Fields

• Theorem 1.2.3. Zn is a field if and only if n is prime.

• Proof. Suppose Suppose that n is prime and that [a] ・ [b] = [0]

in Zn. Then n|ab. So n|a or n|b by Euclid’s Lemma .

Hence [a] = [0] or [b] = [0], and Zn is an integral

domain. Since Zn is also finite, it follows from Theorem

1.2.2 that Zn is a field.

Suppose that n is not prime. Then we can write n = rs,

where r and s are i t n egers such th t a 1 < r < n and

1 < s < n. Now [r] = [0] and [s] = [0] but [r] ・ [s] = [rs]

= [0]. Therefore, Zn has zero divisors and hence is not a

field.

1. Rings, Integral Domains and Fields

Example 2.1.2. Is (Q(√2),+, ・) an integral domain or a field?

Solution. From Example 1.1.2 we know that Q(√2) is a

commutative ring. Let a + b√2 be a nonzero element, so that at

least one of a and b is not zero. Hence a − b√2 ≠ 0 (because √2

is not in Q), so we have

This is an element of Q(√2), and so is the inverse of a + b√2.

Hence Q(√2) is a field (and an integral domain).

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1. Rings, Integral Domains and Fields

• 1.3.SUBRINGS AND MORPHISMS OF RINGS

• If (R +, , ・) is a ring, a nonempty nonempty subset S of R is called a

subring of R if for all a, b א S:

(i) a + b א S.

(ii) −a א S.

(iii) a ・ b א S.

(iv) 1 א S.

• Conditions (i) and (ii) imply that (S,+) is a subgroup of (R,+)

and can be replaced by the condition a − b א S.

1. Rings, Integral Domains and Fields

• For example, Z,Q, and R are all subrings of C. Let D be the

set of n × n real diagonal matrices. Then D is a subring of the

ring of all n × n realmatrices, Mn g (R), because the sum, n( ), ,

difference, and product of two diagonal matrices is another

diagonal matrix. Note that D is commutative even though

Mn(R) is not.

• Example1.3.1. Show that Q(√2) = {a + b√2|a, b א Q} is a

subring of R .Solution. Let a + b√2, c + d√2 א Q(√2). Then

(i) (a + b√2) + (c + d√2) = (a + c) + (b + d)√2 א Q(√2).

(ii) −(a + b√2) = (−a) + (−b)√2 א Q(√2).

(iii) (a + b√2) ・ (c + d√2) = (ac + 2bd) + (ad + bc)√2 א

Q(√2).

(iv) 1 = 1 + 0√2 א Q(√2).

1. Rings, Integral Domains and Fields

• A homomorphism between two rings is a function between

their underlying sets that preserves the two operations of

addi it on and mul i li i ltiplication and also the element 1. Many

authors use the term morphism instead of homomorphism.

• More precisely, let (R,+, ・) and (S,+, ・) be two rings. The

function

f :R → S is called a ring morphism if for all a, b א R:

(i) f (a + b) = f (a) + f (b).

(ii) f (a ・ b) = f (a) ・ f (b).

(iii) f (1) = 1.

• A ring isomorphism is a bijective ring morphism. If there is

an isomorphism between the rings R and S, we say R and S

are isomorphic rings and write R ≅ S.

1. Rings, Integral Domains and Fields

• Example 1.3.2. Show that f :Z24 → Z4, defined by f ([x]24)

= [x]4 is a ring morphism.

• Proof. Since the function is defined in terms of

representatives of equivalence classes, we first check that

it is well defined. If [x]24 = [y]24, then x ≡ y mod 24 and

24|(x − y). Hence 4|(x − y) and [x]4 = [y]4, which shows that

f is well defined.

We now check the conditions for f to be a ring morphism.

(i) f ([x]24 + [y]24) = f ([x + y]24) = [x + y]4 = [x]4 + [y]4 ( ) ([ ] . 24 [y]24) ([ y]24) [ y]4 [ ] [y]4

(ii) f ([x]24 ・ [y]24) = f ([xy]24) = [xy]4 = [x]4 ・ [y]4.

(iii) f ([1]24) = [1]4

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2. Polynomial and Euclidean Rings

• 2.1.Polynomial Rings

• 2.2. Euclidean Rings

2. Polynomial and Euclidean Rings

• 2.1.Polynomial Rings

• If R is a commutative ring, a polynomial p(x) in the indeterminate x

over the ring R is an expression of the form

p(x) = a0 + a1x+a2x2 + ・・・+anxn, where a0, a1, a2,...,an א R and n א N. The element ai is called the coefficient of xi in p(x). If the

coefficient of xi is zero, the term 0xi may be omitted, and

if the coefficient of xi is one, 1xi may be written simply as xi .

Two polynomials f (x) and g(x) are called equal when they are

identical, that is, when the coefficient of xn is the same in each

p y ol nomial for every n .

In particular,

a0 + a1x+a2x2 + ・・・+anxn = 0

is the zero polynomial if and only if a0 = a1 = a2 = ・・ = an = 0

2. Polynomial and Euclidean Rings

• If n is the largest integer for which an ≠ 0, we say that

p(x) has degree n and write degp(x) = n. If all the

coefficients of p(x) are zero, then p(x) is called the zero

polynomial, and its degree is not defined. The set of all

polynomials in x with coefficients from the

commutative ring R is denoted by R[x]. That is,

R[x] = {a0 + a1x+a2x2 + ・・・+anxn|ai א R, n א N}.

• This forms a ring (R[x] + (R[x],+, ・) called the polynomial polynomial

ring with coefficients from R when addition and

multiplication of the polynomials

2. Polynomial and Euclidean Rings

• For example, in Z5[x], the polynomial ring with

coefficients coefficients in the integers integers modulo 5, we have

(2x3 + 2x2 + 1) + (3x2 + 4x + 1) = 2x3 + 4x + 2

and

(2x3 + 2x2 + 1) ・ (3x2 + 4x + 1) = x5 + 4x4 + 4x + 1.

When working in Zn[x], the coefficients, but not the

exponents, are reduced

• Proposition 2.2.2 If R is an integral domain and p(x)

and q(x) are nonzeropolynomials in R[x], then

deg(p(x) ・ q(x)) = deg p(x) + deg q(x)

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2. Polynomial and Euclidean Rings

• 2.2. Euclidean Rings

• An integral domain R is called a Euclidean ring if for each

nonzero element a א R, there exists a nonnegative integer δ(a)

such that:

(i) If a and b are nonzero elements of R, then δ(a) ≤ δ(ab).

(ii) For every pair of elements a, b א R with b ≠ 0, there exist

elements q, r א R such that

a = qb + r where r = 0 or δ(r) < δ(b). (division algorithm)

Ring Z of integers is a euclidean ring if we take δ( ) b = | |, b the

absolute value of b, for all b א Z. A field is trivially a

euclidean ring when δ(a) = 1 for all nonzero elements a of the

field.

Ring of polynomials, with coefficients in a field, is a euclidean

ring when we take δ(g(x)) to be the degree of the polynomial

g(x).

2. Polynomial and Euclidean Rings

• EUCLIDEAN ALGORITHM

• The division algorithm allows us to generalize the concepts of

di i v sors and greatest common di i v sors to any euclidean ring.

Furthermore, we can produce a euclidean algorithm that will enable

us to calculate greatest common divisors.

• If a, b, q are three elements in an integral domain such that a = qb,

we say that b divides a or that b is a factor of a and write b|a. For example, (2 + i)|(7 + i) in the gaussian integers, Z[i], because

7 + i = (3 − i)(2 + i).

Proposition 2.2.1. Let a, b, c be elements in an integral domain R.

(i) If a|b and a|c, then a|(b + c).

(ii) If a|b, then a|br for any r א R.

(iii) If a|b and b|c, then a|c.

2. Polynomial and Euclidean Rings

• By analogy with Z, if a and b are elements in an integral

domain R, then the element element g א R is called a greatest greatest common

divisor of a and b, and is written g = gcd(a, b), if the following

hold:

(i) g|a and g|b.

(ii) If c|a and c|b, then c|g.

The element l א R is called a least common multiple of a and

b, and is written l = l ( cm a, b), if the f ll i o ow ng h ld o :

(i) a|l and b|l.

(ii) If a|k and b|k, then l|k.

2. Polynomial and Euclidean Rings

• Euclidean Algorithm.

Let a, b be elements of a euclidean ring R and let b be

nonzero. By repeated use of the division algorithm, we can

write

a = bq1 + r1 where δ(r1) < δ(b)

b = r1q2 + r2 where δ(r2) < δ(r1)

r1 = r2q3 + r3 where δ(r3) < δ(r2)

...

rk−2 = rk−1qk + rk where δ(rk) < δ(rk−1)

rk−1 = rkqk+1 + 0.

If r1 = 0, then b = gcd(a, b); otherwise, rk = gcd(a, b).

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2. Polynomial and Euclidean Rings

Furthermore, elements s, t א R such that gcd(a, b) = sa + tb

can be found by starting with the equation rk = rk−2 − rk−1qk

and successi l ve y working up the sequence of equations

above, each time replacing ri in terms of ri−1 and ri−2.

• Example 2.1.1. Find the greatest common divisor of 713 and 253 in Z and find two integers s and t such that

713s + 253t = gcd(713, 253).

Solution. By the division algorithm,

we have(i) 713 = 2 · 253 + 207 a = 713 b = 253 r we have(i) 713 2 253 + 207 a 713, b 253, r1 = 207

(ii) 253 = 1 · 207 + 46 r2 = 46

(iii) 207 = 4 · 46 + 23 r3 = 23

46 = 2 · 23 + 0. r4 = 0

2. Polynomial and Euclidean Rings

• The last nonzero remainder is the greatest common

divisor. Hence

gcd(713, 253) = 23.

We can find the integers s and t by using equations (i)–

(iii). We have

23 = 207 − 4 · 46 from equation (iii)

= 207 − 4(253 − 207) from equation (ii)

= 5 · 207 − 4 · 253

= 5 · (713 − 2 · 253) − 4 · 253 from equation (i)

= 5 · 713 − 14 · 253.

• Therefore, s = 5 and t = −14.

2. Polynomial and Euclidean Rings

• Example 2.2.2. Find the inverse of [49] in the field Z53

• Solution Let [x] = [49]−1 Solution. Let [x] [49] in Z53 Then [49] · [x] = [1]; 53. Then [49] [x] [1];

that is, 49x ≡ 1 mod 53. We can solve this congruence

by solving the equation 49x − 1 = 53y, where y א Z. By

using the euclidean algorithm we have

53 = 1 · 49 + 4 and 49 = 12 · 4 + 1.

Hence

gcd(49, 53) = 1 = 49 − 12 · 4 = 49 − 12(53 − 49)

= 13 · 49 − 12 · 53.

Therefore, 13 · 49 ≡ 1 mod 53 and [49]−1 = [13] in Z53.

3.Ideals and quotient rings

• 3.1.Ideals

• 3.2.Quotient rings

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3.Ideals and quotient rings

• 3.1. Ideals.

A nonempty nonempty subset I of a ring R is called an ideal of R if

the following conditions are satisfied for all x, y א I

and r א R:

• (i) x − y א I .

• (ii) x ・ r and r ・ x א I .

Condition (i) implies that (I,+) is a subgroup of (R,+).

In any ring R, R itself is an ideal, and {0} is an ideal.

• Proposition 3.1.1. Let a be an element of commutative

ring R. The set {ar|r א R} of all multiples of a is an

ideal of R called the principal ideal generated by a.

This ideal is denoted by (a).

3.Ideals and quotient rings

• For example, (n) = nZ, consisting of all integer multiples of

n, is the principal ideal generated by n in Z.

• The set of all polynomials in Q[x] that contain x2 − 2 as a

factor is the principal ideal (x2 − 2) = {(x2 − 2) ・ p(x)|p(x)

א Q[x]} generated by x2 − 2 in Q[x].

• The set of all real polynomials that have zero constant term is the principal ideal (x) = {x ・ p(x)|p(x) א R[x]}

generated by x in R[x]. It is also the set of real polynomials

with 0 as a root.

• The set of all real polynomials, in two variables x and y, that have a zero constant term is an ideal of R[x, y]. However,

this ideal is not principal

3.Ideals and quotient rings

• However, every ideal is principal in many commutative

rings; these are called rings; these are called principal ideal rings principal ideal rings.

• Theorem 3.1.1. A euclidean ring is a principal ideal ring.

• Corollary 3.1.2. Z is a principal ideal ring, so is F[x], if

F is a field.

• Proposition 3.1.3. Let I be ideal of the ring R. If I

contains the identity 1 then I is the entire ring R contains the identity 1, then I is the entire ring R.

3.Ideals and quotient rings

• 3.2. Quotient rings.

• Theorem Theorem 3.2.1. Let I be an ideal in the ring R. Then the set of

cosets forms a ring (R/I,+, ・) under the operations defined by

(I + r1) + (I + r2) = I + (r1 + r2)

and

(I + r1)(I + r2) = I + (r1r2).

This ring (R/I,+, ・) is called the quotient ring (or factor ring)

of R by I

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3.Ideals and quotient rings

Example 3.2.1. If I = {0, 2, 4} is the ideal generated by 2 in

Z6, find the tables for the quotient ring Z6/I .

Solution. There are two cosets of Z6 by I: namely,

I = {0, 2, 4} and I + 1 = {1, 3, 5}. Hence

Z6/I = {I, I + 1}.

The addition and multiplication tables given in Table 10.1

show that the quotient ring Z6/I is isomorphic to Z2.

3.Ideals and quotient rings

• Theorem 3.2.2. Morphism Theorem for Rings. If f :R → S is

a ring morphism, then R/Kerf is isomorphic to Imf .

• This result is also known as the first isomorphism theorem

for rings.

• Proof. Let K = Kerf . It follows from the morphism theorem

for groups, that ψ: R/K → Imf, defined by

ψ(K + r) = f (r),

is a group isomorphism. Hence we need only prove that ψ

is a ring morphism morphism. We have

ψ{(K + r)(K + s)} = ψ{K + rs} = f (rs) = f (r)f(s)

= ψ(K + r)ψ(K + s

3.Ideals and quotient rings

• Example 3.2.1. Prove that Q[x]/(x2 − 2) ≅ Q(√2).

• Solution. Consider the ring morphism ψ:Q[x] → R defined √ g p ψ Q[ ] by ψ(f (x)) = f (√2) . The kernel is the set of polynomials

containing x2 − 2 as a factor, that is, the principal ideal

(x2 − 2). The image of ψ is Q(√2) so by the morphism

theorem for rings, Q[x]/(x2 − 2) ≅ Q(√2).

• In this isomorphism, the element

a0 + a1x א Q[x]/(x2 − 2)

is mapped to a + a √2 א Q(√2) Addition and multiplication

is mapped to a0 + a1√2 א Q(√2). Addition and multiplication

of the elements a0 + a1x and b0 + b1x in Q[x]/(x2 − 2)

correspond to the addition and multiplication of the real

numbers a0 + a1√2 and b0 + b1√2.