ring_theory.pdf

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Ring Theory

Gopikrishnan C. R.

September 28, 2015

Learning Objectives

To identify algebraic spaces endowed with two binary operations - Rings and sub rings

Unity and units.

Multiplicative inverse and division rings.

Skew field and field.

Zero divisors and integral domains.

Characteristic of a ring.

Fermats Little Theorem, Eulers quotient function, Eulers Theorem

Partition of a ring into distinct cells or blocks and induced operation of these cells.

Identify that these cells are coses of multiplication. Ideals.

1 Rings

One of the primary motivations of abstract algebra was to study of roots of polynomials and thespaces in which these roots lie. Every polynomial equation contains two binary operations viz.addition and multiplication. There fore to study the solutions or roots of algebraic polynomial,swe must study the properties of algebraic structures endowed with two binary operations. These

are called rings.Definition 1(Ring). A ringR, +, is a setR together with two binary operations+ and,which we call as addition and multiplication respectively which satisfy the following properties.

1. R, + is an Abelson group.

2. Multiplication is associative.

3. Left and right distributive law hold inR. That is,

a(b + c) = ab + ac

(a + b)c= ac + BC

Example 1. R, +, , Q, +, , C, +,

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Example 2. Zn, +n, n

Notation 1. For a natural numbern, we define,

n.a= a + a + + a (added n times)

For any integern such thatn


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Note 1. There are rings without unity. For instance consider 2Z, +, ; for no a 2Z wehavea.b= b for every b 2Z. Indeed, this is because the unity of multiplication is1 and oneis not even.

There are elements in rings which are not units. For example, take Z, +, . No integerother than1 has an integer multiplicative inverse. Thus except1 all other elements inR are nonunit elements. This in fact substantiate the name unit element (in some sense unit elements

behave like 1).

Definition 5 (Skew Field). A ring R, +, in which the multiplication is abelian, that isa.b = b.a for every a, b R, is called a commutative ring. A ring with unity in which allelements are unit elements is called a division ring or a skew field.

Definition 6 (Field). A commutative division ring or a commutative skew field is called aField.

Example 5. R,C,Q are fields.

Definition 7 (Direct Product of Rings). Let R1 and R2 be rings. Then define R1R2 :=

{(r1, r2) :r1R1, r2 R2}. Define the following operations onR1 R2 as,

(r1, r2) + (r3, r4) = (r1+ r3, r2+ r4)

(r1, r2).(r3, r4) = (r1.r3, r2.r4)

ThenR1 R2, +, . is a ring, called the direct product ofR1 andR2. Note that the first sumin(r1 + r3, r2 + r4)is the addition inR1 and the second sum is the addition inR2 asr1, r3R1andr2, r4 R2. The same also holds for multiplication also. We can extend the same conceptto direct product ofn ringsR1 R2 Rn.

2 Integral Domains

Definition 8 (Zero divisors). Let R be a ring. A non zero element a R is called a zerodivisor if there exists a non zero element b R such thata.b= 0. In this case we shall callaas the left zero divisor andB as the right zero divisor.

Example 6. ConsiderZ14. We have inZ14,

2.7 = 0, 7.4 = 0, 7.8 = 0. (1)

Thus 2, 7, 4 and 8 are zero divisors in Z14. Theory of rings may feel bizarre From our usual

understanding of numbers if xy

= 0 then either x

= 0 and y

= 0. But in rings like Z

14 thereare non zero numbers whose product is still zero. The case ofxy= 0 x = 0 ory= 0 workedin case of zero divisors sinceR contains no zero divisors. This property ofR attributes to thefact thatR is a field, which we shall see in detail further ahead.

Theorem 2. a Zn is a zero divisor iff(a, n)= 1.

Proof. Assume that (a, n) = 1. Then we must prove that ais non zero divisor. Let a.s= 0 forsome s Zn. Since n|0 n|a.s. But (a, n) = 1 n|s. That is s is a multiple ofn. Any multipleofn is zero in Zn. Hence s= 0. Therefore ais a non zero divisor.

Conversely assume that (a, n) =d, d >1. Then,

a.

nd

= a.n

d = a

d.n= 0 (2)

as a

d.n is a multiple of zero and

a

d Zn. Hence a is a zero divisor.

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Corollary 1. Ifp is prime, thenZp has no zero divisors.

Proof. For any a Zp, (a, p) = 1. Hence a is non zero divisor by the above theorem.

Theorem 3. The cancellation laws holds in a ringR iffR contains no zero divisors.

Proof. Suppose that cancellation laws hold in R. That means, a.b= a.c b = c. Let a = 0

and a.b= 0 in R. This implies,

a.b= 0 = a.0 b = 0 (by cancellation law)

Hence R has no zero divisors.Conversely assume that R has no zero divisors. We have to prove that cancellation laws

hold in R. That is, ifa= 0, ab= ac b = c. Let ab= ac. Then

ab= ac a(b c) = 0

b c= 0 (a = 0)

b = c.

Thus cancellation laws hold in R.

Definition 9(Integral Domains). A commutative ringR with unity is called an integral domainif it does not contain any zero divisors.

Theorem 4. Every field is an integral domain.

Proof. Let F is a field. That is F is a commutative ring with unity in which all elements hasa multiplicative inverse. we have to prove that F is an integral domain. Towards this end, itis enough to establish ifab = 0 a = 0 or b = 0. Let ab = 0 and a= 0. Since a = 0, there

exists c such that ac= ca = 1. Thus,

b= 1.b= (ca).b= c(ab) =c.0 = 0

This implies R has no zero divisors.

Theorem 5. Every finite integral domain is a field.

Proof. Let D= 0, 1, a1, , an be a finite integral domain. Let a D and a= 0 and a= 1. Ifa= 1 then the multiplicative ofais a itself. Consider

a.1, a.0, a.a1, , a.an (3)

Let a.ai=a.aj . Sincea = 0 andD contains no zero divisors, cancellation laws implies,ai= aj.Therefore the elements a.1, a.0, a.a1, , a.anare all distinct. Hence D= {a.1, a.0, a.a1, , a.an}.We have 1 D. Thus 1 = a.ai as 1= 0, 1= a. This implies ai is the multiplicative inverse ofa. Hence all elements in D has a multiplicative inverse and D is a field hence.

Note 2. Zp is an integral domain and is also finite. ThusZp is a field.

Definition 10 (Characteristic of a ring). Let for every element a R, where R is a ring,there exists a natural number n such that n.a = 0. Then the smallest such n is called thecharacteristic of the ring denoted by ch(R). If there does not exists any such n, we shall say

the characteristic of the ring is 0.

Example 7.

ch(Zn) = n,ch(R) =ch(Z) = 0. (4)

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Theorem 6. ch(R) =n iffn is the smallest positive integer such thatn.1 = 0.

Proof. Letch(R) =n. That isn is the smallest positive integer such thatn.a = 0. In particularfor a= 1 we have n.1 = 0, and nis the smallest such integer.

Conversely assume that nis the smallest positive integer such that n.1 = 0. Then we musthave for any a R,

n.a= a + a + + a= a(1 + 1 + + 1) =a(n.1) =a.0 = 0

This proves the theorem.

Theorem 7 (Fermats little theorem). Ifa is an integer not dividingp, a prime, thenap a(mod p).

Proof. Let a r (mod p). Then ap rp (mod p). If we can show that rp r (mod p),then we shall obtain the desired result. Note that 0 < r < p. That is r Zp. Zp. We havefor any group G of order n, an = e for every a G. Thus rp = 1 in Zp. That is r

p1 1

(mod p) rp

r (mod p).Definition 11 (Eulers totient function). For any integern >0, define the totient functionas,

(n) =|U(n)| (5)

This(n) is the count of integers relatively prime to n and less thann.

Note 3. 1. If (m,n) = 1, then(mn) =(m)(n).

2. For a primep, (p) = p 1

3. For a primep

,

(pk

) =pk1

(p

1)Theorem 8 (Euler). Ifn is an integer and(a, n) = 1, thena(n) 1 (mod n).

Proof. Let U(n) ={k1, k2, , k(n)}.

where (ki, n) = 1, ki < n. Consider {ak1, ak2, , k(n)}. Ifaki= akj, then by cancellationlaws we have ai=aj. Thus{ak1, ak2, , k(n)} are all distinct. As (a,n) = 1 and (ki, n) =n,we must have (aki, n) = 1. Thus aki mod n U(n). That is U(n) = {ak1, ak2, , ak(n)}mod n. Hence,

k1k2 k(n)=ak1.ak2. a.k(n) mod n.

k1k2 k(n)=a(n)k1k2 k(n) mod n

1 =a(n) mod n

Therefore a(n) 1 (mod n).

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3 Quotient Rings

Let R, +, is a ring. Partition the ring into distinct subsets or cells say, R1, R2, , Rk. Thatis, Ri Rj = for i = j, and

ki=1Ri = R. We impose two operations on this set of subsets

ofR. To add two cells, say R1 and R2, we take an element from R1 called representative fromR1 and an element from R2. We add these two elements (using the addition operation in R),

and find which cell contains the sum. It may happen that the sum varies as the representativechanges. But if irrespective of the choice of the representatives we have the sum lie in a singlecell, then we can uniquely define the sum of the cells R1 and R1 as the cell containing the sumof the representatives from R1 and R2. The same procedure also holds for multiplication. Theoperations thus defined for the cells are called induced operations. If the induced operationsgive the unique outputs then the operations are called well defined operations.

Theorem 9. Partition the ring into distinct cells such that,

Induced operations are well defined

Cells form a ring under these operations.

Then,

Cell containing the additive identity0 ofR is an additive subgroup N ofR, +.

Ifr R, thenrNN andN r N.

Proof. LetNbe cell containing the additive identity 0. To show that Nis an additive subgroup,we must show thatN is closed and the existence of inverse elements. Other two defining axiomsare inherently correct.

Closure : N+ Nis the cell containing the sum of any two representatives from N, sincethe addition is well defined. Taking 0 as the representative we shall obtain N+N= N.This means, for every x, yN, x + y N. Therefore we have the closure property.

Existence of Inverse : Let x N. Let Mbe the cell containing the inverse ofx. If wecan establish that N=M, then we have the existence of inverse hold in N. For that wecompute N+M in two different ways. take representatives 0 N and x1 M. ThusN+M = M. On the other hand N+M = N, by taking representatives x N andx1 M. Since addition is well defined we must have N=M.

This provesNis a subgroup ofR, +. To prove the second part, we take an arbitrary element

r R. Let A be the cell containing r. We compute AN. Take the representative r Aand 0 N. Thus AN = N. This gives rN N, as r A. Similarly we shall obtainN A NN r N.

Theorem 10. LetR can be partitioned into cells with the induced operations well defined. Thenthe cells are precisely the left (also right) cosets ofN(the cell containing 0) with respect to theaddition operation.

Proof. Let r R. Let A be the cell containing r. Then A+ N is the cell containing ther+ 0 (taking representative r R and 0 N). ThusA+ N = A. This impliesr+ N A.To prove the reverse inequality, let B be the cell containing the inverse element of r. Thus

A+ B = N (taking r A and r1 B). Let x A. Then x+ r1 = n for some n N.Therefore x = n r1 x = n+ r = r+ n. Hence x r+ N. Therefore we must haveA r + NA = r + N. A similar argument shows that A= N+ r.

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Theorem 11. If N, + in additive subgroup of R, + for a ring R, and if the induced op-erations of addition and multiplication of cosets r + N for r R are well defined, that is,independent of choice of representatives, then the collection of the these cosets r+ N form aring under these induced operations.

Proof. Let Na denotes the cell containing the element a. SinceNcontains the additive zero,

by the above theorem we must have Na=a + N=N+ a.Step 1: r+ N, + is an additive groupConsider the element r+ N. r+ N+N = Nr + N = Nr+0 = Nr = r+ N. Thus N is theadditive identity. r+ N+ r+ N = Nr+ Nr = Nr+r = N0 = 0 + N = N. Thusr+ Nis the additive inverse of r+ N. Similarly, r+ N+ (s+ N+t + N) = Nr + (Ns+ Nt) =Nr+ (Ns+t) =Nr+(s+t)=N(r+s)+t = Nr+s+ Nt= (Nr+ Ns) + Nt= (r+ N+ s + N) + t + N.Thus induced addition is associative. r+ N+s+N= Nr+ Ns = Nr+s = r + s+N is againa coset. Therefore induced operation is closed. Thus r+ N, + is a group. To prove that it isan abelian group r + N+ s + N=Nr+ Ns=Nr+s = Ns+r = Ns+ Nr =s + N+ r + N. Thusr+ N, + is an additive abelian group.

Step 2 : Multiplication is associative

(r+ N)((s + N)(t + N)) =Nr(NsNt)

=Nr(Nst) =Nr(st)=Nrs(t) = NrsNt

= (NrNs)Nt

= ((r+ N)(s + N))(t + N)

Step 3: Multiplication is distributive over addition

(r+ N)(s + N+ t + N) =Nr(Ns+ Nt)

=Nr(Ns+t) =Nr(s+t)=Nrs+rt

=Nrs+ Nrt = NrNs+ NrNt

= (r+ N)(s + N) + (r+ N)(t + N)

Therefore r+ N, +, is a ring.

Theorem 12. IfN, + is an additive subgroup ofR, + then the induced operations on thecosetsr+ Nare well defined if and only ifrNN andN r N for everyr R.

Proof. To prove that the induced operations are well defined we consider two cosets r + Nand

s+N. Any two arbitrary representatives fromr +Ncan be expressed as r +n1and r +n2wheren1, n2 N. Similarly any two arbitrary representatives from s + Ncan be expressed as s + n3ands + n4wheren3, n4N. On one hand we have,r + n1 + s + n3=r + s + n1+ n3 r + s + N.On the other hand, r+ n2+ s+n4 = r + s+n2+ n4 r + s+N. Thus induced addition isunique irrespective of the representatives. Thus the induced addition is well defined.

To prove the induced multiplication is well defined, consider (r+n1)(s+n3) =rs+rn3+n1s+n1n3. SincerNN, we havern3N. Similarly n1s N. This implies rn3 + n1s + n1n3 N.Thus rs+rn3+ n1s+n1n3rs+N. Similarly we can show that (r+ n2)(s+n4) rs+N.Thus the induced multiplication is also well defined.

Definition 12. An additive subgroup N, + ofR, + whereR is a ring is called a right idealif for anyr R we haveN r N, and is called a left ideal if for anyrR we haveN rN.IfN is both left and right ideal, thenN is called an ideal. {0} is called the trivial ideal. R iscalled the improper ideal. Any idealN such that{0} NR is called a proper ideal.

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Definition 13. If N is an ideal of R. Then the induced operations of cosets of N is welldefined, and it forms a ring under the induced addition and induced multiplication. This is

called the factor ring modNor quotient ring modNdenoted by the symbol R

N.

Theorem 13. If N is an ideal that contain a unit element u of a ring R with unity. ThenN=R.

Proof. Let u N and u be a unit. Since u1 R, we have u1N N, as N is an ideal.This implies u1u = 1 N. If r R, then rN N r.1 N r . This givesR NR = N.

Theorem 14. A field contains no proper ideals.

Proof. All elements in the field are units. Hence by the previous theorem we have the proof.

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