Chapter Two Ring Theory 2-1 Definitions and examples Definition 2.1.1 A ring R is a set together with two binary operation and (called addition and multiplication defined on R) if satisfying the following axioms:

(1) is an abelian group, (2) For all implies (3) . is associative: (4) the distributive law hold in R: , and Example. , are ring.

Definition 2.1.2. The ring is commutative if multiplication is commutative.

Definition 2.1.3. The ring is said to be ring with identity if

Example: are commutative ring with identity.

Definition 2.1.4. Let be a ring with identity .we say that an element has an inverse a unit element if it has the multiplicative inverse ( i.e  We denoted the set of all unit elements in by

Theorem 2.1.5. Let be a ring with identity. Then is a group.

Proof. Since , then is a non-empty set. Now we prove that the axioms of group are satisfies:

1- let that is each of has inverse multiplication. Hence

and

This implies that is invers of and . Hence the set is closed under multiplication.

2- associative law are holds because is ring.

3- is identity element.

4- If , then is group.

Example.(1) In we see and is an abelian group.

(2) Let be a non-empty set. If is a power set of , then show that

Is a commutative ring with identity.

(3) Let be the square matrix of . Show that

a ring with identity.

Definition 2.1.6. Let be a ring. For all and for all integer define

and define

If R with identity, then

If R with identity and a has a multiplicative inverse, then

Theorem 2.1.7. Let be a ring, for and arbitrary integers n the following hold:

1- 2- 3-

Theorem 2.1.8. Let be a ring and be a zero element. The for all the following hold:

1- 2- 3- 4-

Proof. 1- since

Hence . By cancellation law we get the result.

Similarly we can proof .

2-H.w

3-By (2) we get

4-H.w

Corollary 2.1.9. Let be a ring with identity such that . Then the element and are distinct.

Proof. Since , there exists some nonzero element Now if ,

It follow that , which is contradiction.

Corollary 2.1.10. Let be a ring with identity such that . Then for all the following are hold:

1- and 2-

Definition 2.1.11. Let be a ring and let be a non empty subset of   . If is itself a ring, then is said to a subring of

Remark. Every ring has two trivial subring; for, if 0 denote the zero element of the ring , then both and the ring itself are subrings of

Definition 2.1.12. Let be a ring and . Then is a subring of if and only if

1- –   ( closed under differences) 2- ( closed under multiplication)

Examples.

1- is a subring of and

2- is a subring of . 3- Let R denote the set of all functions The sum and the product of two function are defined by

Suppose is the commutative ring of function of above. Define

Definition 2.1.13. The center of a ring denoted by , is the set

.

Remark. If is comuutaive, then

Theorem 2.1.14. Let be a ring. Then is a subring of .

Proof. Since , then , hence Let  . To prove that For all , then

Therefore and

Therefore hence is a subring of .

2-2 Some type of rings.

Definition 2.2.1. A ring is said to hane divisors of zero if there exist nonzero elements such that the product .

Definition 2.2.2. An integral domain is a commutative ring with identity which does not have divisors of zero.

Examples. are integral domain but is not integral domain.

Theorem 2.2.3. Let be a commutative ring with identity. Then is an integral domain if and only if the cancellation law holds for multiplication.

Proof. We suppose that R is an integral domain . Let such that and

. Hence

Conversely, suppose that the cancellation law holds and .

If the element , then by Theorem 2.1.6 we have , hence

, consequently . That is has no divisors of zero and commutative with identity, we get is an integral domain.

Corollary 2.2.4. Let be an integral domain. Then the only solution of the equation are and .

Proof. Clearly 0 is the solution of the equation

Now , if , since and , hence by cancellation law we get

Definition 2.2.5. A ring (R, +, .) is said to be a division ring if it is a ring with identity in which every nonzero element has a multiplicative inverse.

Definition 2.2.6. A field is a commutative ring with identity in which each nonzero element has an inverse under multiplication.

Examples:

1- are field.

2- (Zn ,+n ,.n) is a field if and only if n is a prime number. 3- is an integral domain but not a field.

Theorem2.2.7. Every field is an integral domain.

Proof. Let ( be a field. Then R is a commutative ring with identity.

Let .

Since R is a field, then the element a has an inverse.. The hypothesis a.b =0 yields

That is contains no divisors of zero. Hence is an integral domain.

Theorem2.2.8. Any finite integral domain is a field.

Proof. Let ( be an integral domain contains n distinct elements say .

Let be any element of , consider the elements These products are all distinct because

If but which is contradiction to are all distinct.

Since  , then for some and has multiplicative inverse and . That is is a field.

Theorem 2.2.9. The ring of integers modulo n is a field if and only if n is a prime number. Proof. Suppose that is a field. To prove that is a prime number. If is not prime, then where It follows

.

Since This means that the system is not an integral domain and hence not a field.

Conversely suppose that n is a prime number. To prove that (Zn ,+n , .n) is a field, enough to show that is an integral domain.

Let and

Hence has no divisors of zero, that is is an integral domain.

Definition 2.2.10. Let be an arbitrary ring. If there exists a positive integer such that for all , then the least positive integer with this property is called the characteristic of the ring. If no such positive integer exists, then we say (R, +, .) has characteristic zero.

Definition 2.2.11. : Let be a ring with identity. Then has characteristic if and only if is the least positive integer for which . Proof: If the ring is of characteristic , it follows that . Were , where then

for every  This mean The characteristic of is less than , which is contradiction. Conversely, Let be the least positive integer in which Let 

Then has characteristic .

Corollary 2.2.12. The characteristic of an integral domain is either zero or a prime.

Proof. Let be a positive characteristic n and assume that is not a prime Then n can be written as with . By Theorem 2.2.11 we have Since by hypothesis is without zero divisors, then either But this contradicts the choice of as the least positive integer such that Hence the characteristic of must be prime.

Example. Show that the characteristic of the ring  is equal two. Since is the zero element of the ring Now for all , then

 –  – . From the definition of characteristic, then the characteristic of is 2.

2-3 Ideals and Quotient rings.

Definition 2.3.1. A subring of the ring is an ideal of if and only if and imply both and

Definition 2.3.2. Let be a ring and a non-empty subset of . Then is an ideal of if and only if (1) imply , (2) imply both

Remark. In a commutative ring , every right ideal is left ideal. Examples.

1) The subring is an ideal of

2) The trivial subrings of the ring ( are both ideals. Any ideal different from is called proper ideal.

3) In the ring for a fixed integer . Then is an ideal of because where

4) is not ideal of but is a subring of

Since and , then . Then is not ideal of

5) Let be the square matrix ring over the field of real number. Then is not an ideal.

Definition 2.3.3. A ring which contains no ideals except trivial ideals is said to be a simple ring.

Definition 2.3.4. Let be a commutative ring with identity. An ideal is called a principal ideal of the ring if generated by a single element and denoted by .

Example. In the ring ( the ideal is a principal ideal generated by 2 and is a principal ideal generated by 3.

Theorem 2.3.5. If is an ideal of the ring then for some nonnegative integer . Proof. If then the theorem is true. Suppose then that that is there exists Since I is an ideal, then – so contains positive integers.

Let be the least positive integer in . We claim Since and is an ideal of then , for all , that is

On the other hand, any integer . By division Algorithm there exists such that , where . Since are members of , it follows that . Our be a least integer implies , and consequently Therefore .

Definition 2.3.6. Let be a commutative ring with identity. A ring is called a principal ideal ring if every ideal is principal.

Theorem 2.3.7. Let be a ring with identity element and be an ideal of containing identity element .Then . Proof . Since is an ideal of , then  Let  , then (because is an ideal of )

Theorem 2.3.8. If is a proper ideal of a ring with identity, then no element of I has a multiplicative inverse; that is . Proof. Suppose to the contrary that there is such that exists. Since is an ideal, then contradiction the hypothesis that is a proper subset of

ITheorem 2.3.9. If and are two ideals of the ring , then

is also an ideal.

Proof. Since and are ideals of the ring , then and

, hence This implies that .

Suppose and . Then and .

As the and are ideals of the ring , it follows from definition

and , and also and .

Hence and which implies that is an ideal of

Quotient rings We now give the analogue of quotient groups for rings. Let be a ring and an ideal of . Let . Let denote the set . The set is called a coset of .For By Theorem 6.1, if and only if Let denote the set Because is a nonempty set. Define the operations + and · on as follows:

and We leave it as an exercise for verify that + and · are binary operations on . Under these binary operations satiesfies the properties of a ring. Let us verify some of these properties. Let . Now

,

This shows that + is associative in . Similarly, + is commutative. Next, note that is the additive identity and for , is the additive inverse of . As in the case of the associativity for +, we can show that · is associative. Next, let us verify one of the distributive law. Now

In a similar manner, we can verify the right distributive property.

Theorem 2.3.10. If is an ideal of , then the ring is ring, known as the quotient ring of by

Definition 2.3.11. An ideal of the ring is a prime ideal if for all implies either .

Example.(1) The ideal of the ring is a prime ideal. (2) A commutative ring with identity is an integral domain if and only if the zero ideal is a prime ide

Theorem 2.3.12. Let be a proper ideal of the ring . Then is a prime ideal if and only if the quotient ring is an integral domain. Proof. First, take to be a prime ideal of . Since is a commutative ring with identity, so is the quotient ring . It remains to show has no divisor of zero. For this, assume that Since is a prime ideal, hence , hencc is without zero divisors. To prove the converse, suppose is an integral domain and . Then we have

By hypothesis , ( contains no divisors of zero, that either . That is is a prime ideal.

Theorem 2.3.13. Let be the ring of integers and . Then the principal ideal is prime if and only if n is a prime number. Prool. First, suppose is a prime ideal of If the integer is not prime, then , where This implies the and such that Is a prime ideal, this implies and this contradication to the hypothesis of and are less than , therefore must be a prime number. Conversely, suppose is a prime number and two integers such that with . Since and sine n is a prime number implies that , therefore is a prime ideal.

Definition 2.3.14. An ideal of the ring is a maximal ideal provided and whenever is an ideal of with , then

Remark. An element is invertible is not belongs to maximal ideal.

Example. 1- is a maxima ideal 2- is not a maximal ideal since 3- is a prime ideal of the ring but is not a maximal ideal since . 4- is a prime ideal of the ring but not a maximal ideal. Theorem 2.3.15. Let be aproper ideal of the commutative ring with identity . Then is a maximal ideal if and only if the quotient ring is a field. Proof. Let be a maximal ideal of Since is a commutative ring with identity, then the quotient ring is also a commutative ring with identity. It remains to show that every non-zero elemnt in has inverse. . Since is an ideal of , the is an ideal of and By suppose is a maximal ideal, then .

Since . That is . Therefore has an inverse, consequently is a field. Conversely, suppose is It field and is any ideal of such that

Since , then there exist an element . Since is a field, then has an inverse say , therefore

Since Hence is a maximal ideal.

Definition 2.3.16. A ring is called a local ring if has only one maximal ideal.

Definition 2.3.17. The radical of a ring , denoted by , is the set

If , then we say is a ring without radical or is a semi• simple ring.

Example. In , find

Remark. is an ideal of

Definition 2.3.18. An ideal of a ring is said to be a primary ideal if with implies for some positive integer n.

Example. An ideal of is a primary.

Definition 2.3.19. An element of a ring is said to be a nilpotent if there exists a positive integer n such that

Theorem 2.3.19. Let be an ideal of a ring . Then is a primary if and only if every zero divisor of the quotient ring is nilpotent.

Proof. Suppose is a primary ideal and is a zero divisor in .

That is there exists a npnzero element such that

Since is a primary, then there exists a positive integer n such that Hence is nilpotent element in .

Conversely, suppose every zero divisor is nilpotent.

Let such that with . We must to sow that for some .

If , it is trivial.

If . Since , hence is divisor of zero.

By hypothesis is a nilpotent element, that is there exist a positive integer n such that consequently is primary.

2.4. Homomorhpisms

Definition 2.4.1. Let and be two rings and a function from into ; in symbols, . Then is said to be a (ring) homomorphism from into if and only if 1- 2- for every

Example. Let be the function defined by

Hence is a ring homomorphism.

Example. Let be the function defined by

Hence is not a ring homomorphism.

Theorem 2.4.2. Let be a homomorphism from the ring into the ring . Then the following hold: . 1) , where is the zero element of 2) 3) The triple is a subring of If, in addition, and are rings with identity elements 1 and 1', respectively, and , then 4) 5) for each invertible element

Proof. Similar of Theorem 8.4

Theorem 2.4.3. Let be a homomorphism from the ring into the ring . Then

1- If is a subring of , then is a subring of .

2- If is a subring of the ring , then is a subring of . 3- If is an ideal of the ring , then is an ideal of . 4- If and is an ideal of , then is an ideal of . Proof. 1-

Since  then .

Let .

Now , Since , and

, Since

Therefore by Definition 2.1.12, we get is a subring of .

3- By part (2) is a subring of . To show that is an ideal of , such that

Now suppose Since is a homomorphism and is a subring of , then we have , Since is an ideal of . Therefore , and Let . Since is an ideal of , then . Hence since is a homomrphism, we get and

Therefore is an ideal of ( .

Example. defined by is a homomorphism and but is not an ideal of

Definition 2.4.4. Let be a homomorphism from the ring into the ring . Then kerenel of , denoted by , is the set

Theorem 2.4.5. If is a homomorphism from the ring into the ring , then is an ideal of

Proof. Since is an ideal and , then by Theorem 2.4.3 is an ideal of the ring .

Theorem 2.4.5. Let be a homomorphism from the field on to the field Then either is the trivial homomorphism or else and are isomorphic. Proof. By The Theorem 2.4.4 is an ideal of the field . Since is a field has no ideal other than itself and . Hence either the set or else . If , then and this contradication for , hence and this implies that is one-to-one. Therefore is an isomorphism, consequently

Definition 2.4.6. We said that is a subfield of the field is meant any subring of which is itself a field.

Example. The ring of rational numbers is a subfield of the field

Is equivalent to

The triple will be a subfield of the field provided (1) is a subgroup of the additive group and (2) is a subgroup of the multiplicative group .

Definition 2.4.7. A ring is imbedded in a ring if there exists some subring of such that .

The field of quotient of an integral domain. Let D be an integral domain. D D = {(a, b) : a, b D}. Let S be the subset of D D given by S = {(a, b) : a, b D, b ≠ 0}. Define a relation on S as follows: Two elements (a, b) and (c, d) in S are equivalent (denoted by (a, b) (c, d)) if ad =bc.

Lemma 2.4.8. The relation is an equivalence relation. Proof.

(1) Reflexive: (a, b) (c, d), since multiplication in D is commutative. (2) Symmetric: Suppose that (a, b) (c, d), then ad =bc. Hence cb =da, consequently (c, d) (a, b). (3) Transitive: suppose that (a, b) (c, d) and (c, d) (e, f). Then ad =bc and cf = de, so afd = fad = fbc =bfc = bde = bed (D is commutative)

since d ≠ 0 and D is an integral domain, hence afd = bed af = be

(a, b) (e, f). From (1), (2) and (3) we get that is an equivalence

relation.

Hence it gives a partition of S in to equivalence class. We write the equivalence class of .

Let Define addition and multiplication on F as follows:

[ and

Now we show that the operations defined above is well-defined.

First note that if , then Since D is an integral domain, then , so both .

To show that the multiplication (.`) is well-defined, suppose that

and (c,d) ~(c1,d1). We have two show that [(a,b)] [(c,d)] =[(a1,b1) ]

[(c1,d1)] . ab1 = a1b and cd1 = c1d  ab1 cd1 = b1a d1c  ab1 cd1 = b1a d1c . This means that

(a1 c1 ,b1 d1) ~ (ac,bd) which means that the multiplication is well-defined .

Theorem 2.4.9. Let D be an integral domain and  .

Define a relation on S as follows: . Then

 is a field with addition and multiplication defined as follows: [(

Proof.

(1) + is a commutative :

(2) It is easy to show that + is a associative (3) [(0,1)] is identity for addition in F: (4) is an additive inverse of in F: (since because ) Thus .

(5) It is easy to show that multiplication is a associative. (6) [(1,1)] is identity for multiplication in F :

(7) Multiplication is commutative (8) The distributive law hold in F (9) Let  and hence because if , then , consequently which is a contradiction . Thus a is a non zero element in F. Now

is a multiplicative inverse of

Hence F is a field. This field called the field of quotients of R. the quotient field of an integral domain (D,+, .`) is the smallest field containing D as a subring.

Example. The field of quotients of Z, is the ring of integers is Q

Theorem 2.4.10. The integral domain (R, +, .) can be imbedded in its of quotients

(F, +’, .’).

Proof. Consider the subset F’ of F consisting of all elements of the form where 1 is the multiplicative identity of (R, +, .):

Now it is must be show that is a subring.

Let be onto mapping defined by for each . Since the condition implies , we see is one to one. Now we show that f is homomorphism: and

Accordingly

Theorem 2.4.11. (First isomorphism theorem) If is a homomorphism from the ring onto the ring Then

.

Proof. Put We define a function by

We must show that is well defined, suppose . Therefore But f is homomorphism, then . Hence is well defined. Now to show that is a homomorphism, suppose that

Hence is a homomorphism. Let Since is a homomorphism, therefore . Hence is one-to-one. Finally, for all there exists such that

Hence is onto. Therefore is an isomorphism and .

Remark. If is not onto, then

Theorem 2.4.12. (second isomorphism theorem) If is a subring of the ring and is an ideal of , then

Proof. Similarly to prove Theorem 9.3

Theorem 2.4.13. (Third isomorphism theorem) If and are two ideals of the ring and , then is an ideal of the ring and

Proof. Similarly to prove Theorem 9.4.

Chapter Three Polynomial rings

The polynomial ring R[x] in indeterminate x with coefficients from R is the set of all n n-1 formal sums an x + an-1 x +…. + a1 x +a0 with n ≥ 0 and ai R, That is n n-1 R[x] = { an x + an-1 x +…. + a1 x +a0 : n ≥ 0 and ai R }. n If an≠0 , then the polynomial is of degree n , an x is the leading term, and an the leading coefficient. Addition of polynomial is component wise

+ = ( where an and bn may be zero in order for addition of polynomials of different degree to be defined ). Multiplication performed by first defined a xi b xi = ab xi+j and then extended to all polynomials by distributive law , in general

( ) × ( ) = ) . n n Two polynomials p(x) = a0+a1x+…+anx and q(x) = b0+b1x+…+bnx , are equal if ai = bi for each i. The ring R appears in R[x] as the constant polynomials. If g(x) is a polynomial over a ring R, then degree g(x) denoted deg g(x). If R[x] has unity 1 and you must have x( x = (0, 1, 0, 0, …) ) 2+ x2 in Z[x] (i.e (2, 0, 1, 0, 0, …)). If R is a ring with two determinates, then we can form (R[x])[y] = R[x, y]

The ring R[x1, x2, . …, xn] of polynomials in the n indeterminate x with coefficients in R.

Theorem 3.1. The triple (R[x], +, .) forms a ring, known as the ring of polynomials over R.

Examples. (1) Let f(x) =1+3x+2x5 a polynomial, then the leading coefficient of f(x) = 2, deg f(x) = 3.

(2) In Z2[x]. If f(x) = x + 1, then we have (x + 1) + (x + 1) = 2x + 2 = 0, and (x+1)2 = x2 + 2x + 1 = x2 + 1

Remark . Let R be a ring.

(1) If R is a commutative , then so is R[x].

(2) If R is a ring with identity 1R, then R[x] is a ring with identity and the identity

element is 1R[x] = 1R + 0R + …

(3) don’t write 1R when appear as coefficient for a polynomial, as follows: x3 + x2 + 2x + 2 (1.x3 + 1.x2 + 2x + 2).

(4) In general if f(x) and g(x) are two polynomials over a ring R, then (1) deg( f(x) + g(x))  max {deg f(x) , deg g(x) }. (2) deg (f(x) . g(x))  deg f(x) + deg g(x).

4 2 Example. Let f(x) =1+5x+2x and g(x) = 1+4x be two polynomials in Z8[x]. The leading coefficient of f(x) = 2, deg f(x) = 4 and deg g(x) = 2. f(x). g(x) = 1+4x2+5x +4x3+ 2x4  deg((f(x)+ deg g(x))=7  deg((f(x). deg g(x))=4, and f(x) + g(x) = 2+5x+4x2+2x4, deg (f(x) + g(x) ) = 4.

Theorem 2.3 . Let (R , + , .) be an integral domain and f(x) , g(x) be two nonzero elements of (R[x] , + , .) then : deg (f(x).g(x)) = deg f(x) + deg g(x) .

Proof : suppose f(x) , g(x) ϵ R[x] with deg f(x) =n and deg g(x) =m , so that

n f(x) = ao + a1x + … + anx , an ≠0

m g(x) = bo + b1x + … + bmx , bm ≠ 0 from the definition of multiplication

n+m f(x) . g(x) = ao . bo + (ao .b1+ a1 . bo ) x + … + (an . bm) x since an ≠0 and bm ≠ 0 and R is an integral domain , then an . bm ≠ 0 accordingly, f(x) . g(x) ≠ 0 and deg(f(x).g(x)) = n+m = deg f(x) +deg g(x)

Corollary 2.4 . Let (R, + , .) be an integral domain. Then (R[x], + , .) is an integral domain.

proof . We have if (R , + .) is a commutative ring with identity , then so is (R[x] , + , .) . To see that (R, +, .) has no divisors , let f(x) ≠ 0 , g(x) ≠ 0 in R[x] . Then deg (f(x).g(x)) = deg f(x) + deg g(x)>0 , hence the product cannot be the zero polynomial

Theorem 2.5. (Division algorithm) n n-1 Let (R, +, .) be a commutative ring with identify and let f(x) = an x + an-1 x +… +a0 m m-1 and g(x) = bm x + bn-1 x +… +b0 , be two elements in R[x], with both an , bn non zero elements of R and m > 0 and the leading coefficient of g(x) is invertible. Then there are unique polynomials q(x) and r(x) in R[x] such that f(x)=q(x) g(x) + r(x), with r(x) = 0 or degree r(x) < degree g(x).

Examples.

4 3 2 - 2 (1) Consider f(x) = x - 3x + x 3x + 1 in Z5[x] and let g(x)= x +2x -6. To find q(x) and r(x), divide f(x) by g(x),

x2  x  3 x2  2x  3 x4  3x3  2x2  x4  2x3  3x2  x3  x2  4x  x3  2x2  3x  3x2  2x 1  3x2  x  4 x  3

So x4 - 3 x3 + x2 - 3x + 1 = (x2 - 2x + 3) (x2 – 3x + 3) + (3x + 5), remembering that the 2 coefficients are in Z5 . Then q(x) = x -3x +3, and r(x) = 3x+5.

4 3 (2)Consider f(x) = x + 3x + 2x + 4 in Z5[x] and let g(x)= x - 1. To find q(x) and r(x), divide f(x) by g(x),

n n-1 Definition 2.6. Let (R, +, .) be a ring with identity and f(x) = an x + an-1 x +… +a0  n n-1 R[x] . Then if r R we define f(r) by f(r) = an r + an-1 r +… +a0  R.

3 2 Example. Let f(x) = x +4x +3Q(x) . Then f(2) = 8+4(4)+3=27

Definition 2.7. Let R be a commutative ring and f(x) a polynomial over R. Any element r  R such that f(r) = 0 is a zero of f(x) in R( or r is a root of f(x) ).

Definition 2.8. Let R be a commutative ring with identity and f(x), g(x) be non zero polynomials in R[x]. Then g(x) is said to be a factor of f(x), if there exists a non zero polynomial h(x)  R[x] such that f(x) = h(x) g(x).

Example. Let f(x) = (x-1)(x+5) be a polynomial of Z[x]. Then (x-1) is a factor of f(x).

Proposition 2.9. Let f(x) be a polynomial over a commutative ring with identity and a be an element in R. Then a is a root of f(x) if and only if (x-a) is a factor of f(x). Proof. Suppose (x-a) is a factor of f(x). Then there exists a polynomial q(x) such that f(x) =q(x) (x-a). Then f(a)= q(a) 0 which implies f(a) = 0, and a is a root of f(x). Conversely, suppose f(a) =0. By division algorithm, there exist q(x) and r(x) in R[x] such that f(x) =q(x) (x-a)+ r(x) , deg r(x) < deg (x-a) or r(x) =0. Since deg (x-a) =1, then r(x) is a constant polynomial. But 0= f(a) = q(a)(a - a)+r(a) = r(a).Hence r(x) =0, consequently f(x) = q(x) (x-a).

Definition 2.10. The element r is a root of multiplicity m of f(x) if (x-a)m| f(x) but (x- a)m+1 f(x). A zero of multiplicity 1 is called simple zero.

Theorem 2.11. (Fundamental theorem of algebra): If f(x) is a non constant polynomial over the field of complex numbers, then f(x) has at least one root in C.

Theorem 2.12. Let R be an integral domain, f(x) be a non zero polynomial over R. If deg f(x) = n, then f(x) has at most n distinct roots R. Proof. We proved by induction on the degree of f(x). When deg f(x) = 0, then there exists 0 ≠ a0 R such that f(x) = a0. This means f(x) has no root in R.

If deg f(x) = 1, then there exists 0 ≠ a1 R such that f(x) = a0 +a1x. This means f(x) has at most one root in R; indeed, if a1 is invertible, - .a0 is the only root of f(x). Now, suppose the theorem is true for all polynomials of degree n-1≥1, and let deg f(x) = n. If r is a root of f(x), then there exists q(x) R[x] such that f(x) = (x-r)q(x), where q(x) of degree n-1. Any root t of f(x) distinct from r must be a root of q(x), by substitution, we have f(t) = (r – t)q(t) = 0. Since R has no zero divisors, then q(t) = 0. From hypotheses , q(x) has at most n-1 distinct roots. As the only roots of f(x) are r and those of q(x). That is f(x) cannot have more than n distict roots in R.

The following example shows that the condition that R is an integral domain is the last theorem is necessary.

Example. Consider the ring R =Z2Z2 .Clearly R is not an integral domain. Now consider the polynomial f(x) = x2 + x .

It is not difficult to show that every element of Z2 Z2 is a root of f(x). where

Z2 Z2 = {(0, 0), (1, 0), (0, 1), (1, 1)}. So f(x) has four roots.