polynomial Let R be a ring. A polynomial over R is an expression of the form: f (x) = a0 + a1x + a2x2 +…+ anxn where the ai  R called the coefficients of f (x), and x is called an indeterminate over R.

zero polynomial The polynomial p(x) for which every coefficient is zero is called the zero polynomial and is denoted p(x)=0.

Equality of polynomial Two polynomials f (x) = a0 + a1x + a2x2 +… g (x) = b0 + b1x + b2x2 +… over a ring R are equal if ai = bi for all i = 0,1,2,….

R[x] Let R[x]={ a0 + a1x +…+ anxn | n0,ai∈R for each i} Suppose f (x)=a0 + a1x +… and g(x)=b0 + b1x +… We define the addition and multiplication: Then R[x] is a ring.

Example If and , then Note that In Z [x], In Z3[x],

Theorem Let R be a ring and let x be an indeterminate over R. Then (1) R[x] is a ring. (2) If R is commutative, then R[x] is commutative. (3)If R has a unity 1, then 1 is the unity of R[x].

f (x) = a0 + a1x + a2x2 +… The coefficient a0 is called the constant coefficient of f (x). f (x)= a0 is called a constant polynomial. R is isomorphic to the set of all constant polynomials. We may see R is a subring of R[x].

degree If f(x)= a0 + a1x +…+ anxn and an 0, then n is called the degree of f (x) and is written deg f (x), and the coefficient an is called the leading coefficient of f (x). If the leading coefficient is 1, f (x) is called a monic polynomial. The degree of the zero polynomial is not defined.

Example deg( xx2+2x3) = 3 deg( x+2) =1, deg(5) = 0

Example Determine the degree of (x+1)(x+2)(x+3) in Z3[x], Z6 [x], and Z11[x]. Solution: (x+1)(x+2)(x+3)= x3+6 x2+11 x+6 Since 10 in Z3[x], Z6 [x], and Z11[x]. So the degree of (x+1)(x+2)(x+3) is 3 in Z3[x], Z6 [x], and Z11[x].

Example Determine the degree of (3x+1)(2x+1)(x+1) in Z3[x], Z6 [x], and Z11[x]. Solution: (3x+1)(2x+1)(x+1)= 6x3+11 x2+6 x+1 The degree is 2 in Z3[x], Z6 [x], since 6=0 and 110 in Z3[x], Z6 [x]. The degree is 3 in Z11[x] since 60 in Z11[x].

Example In Z4[x], (1+2x)(2x+1)=1+(2+2)x +22x2=1. If f (x)g(x)  0 in R[x], then deg( f (x)g(x) )  deg( f (x)) + deg(g(x) ). This is equality if R is a domain.

Theorem Let R be a domain. Then (1) R[x] is a domain. (2) If f (x)  0 and g(x)  0, then deg( f (x) g(x) ) = deg( f (x) ) + deg(g(x) ).

Proof Suppose f (x)=a0 + a1x +…+ amxm and g(x)=b0 + b1x +…+ bnxn , where an 0, and bn 0. Then and anbn 0, because R is a domain. So f (x)g(x)  0 and deg( f (x)g(x) ) = m + n .

Theorem R be an integral domain if and only if R[x] is an integral domain.

Exercise 8.1 True or False all Exercise 2,3,4,6,25,27

Division Algorithm Let R be an integral domain and let f (x) and g(x) be nonzero polynomials in R[x]. Assume that f (x)  0. Then uniquely determined polynomials q(x) and r(x) exists such that (1) g(x)=q(x)f (x) + r(x) (2) either r(x) = 0 or deg(r(x))<deg( f (x)).

Theorem Let f (x) and g(x) be nonzero polynomials in F[x], where F is a field. Then a uniquely determined polynomial d(x) exists in F[x] satisfying the followings. (1) d(x) is monic. (2) d(x) divides both f (x) and g(x). (3) If h(x) divides both f (x) and g(x), then h(x) dividees d(x). (4) d(x)=u(x) f (x)+v(x)g(x) for some u(x) and v(x) in F[x]. 15:22

greatest common divisor The monic polynomial d(x) in above Theorem is called the greatest common divisor of f (x) and g(x) in F[x] and is denoted gcd[ f (x),g(x)]. If d(x)=1, f (x) and g(x) are said to be relatively prime in F[x]. 15:22

Remark Let F be a field. If f(x)= h (x)g(x)+ r(x) in F[x], then gcd[f(x),g(x)] = gcd[ r (x),g(x)]. Let F be a field. If f(x)= h (x)g(x) in F[x] and g(x) is monic, then g(x) = gcd[ f (x),g(x)]. 15:22

Example Find the greatest common divisor d(x) of f (x) = x4x2+x1 and g(x)= x3x2+x1 in [x] and express it as a linear combination of these polynomials. Solution: f (x) = (x+1)g(x)+(x2+x) g(x) = (x)(x2+x)+(x1) x2+x = (x)(x1)+0 gcd(f (x) ,g(x))= x1=x f (x) (x2+x 1)g(x) 15:22

Example Find the greatest common divisor d(x) of f (x) = x3+6x2+11x+6 and g(x)= 2x2+6x+4 in [x] and express it as a linear combination of these polynomials. Solution: f (x) = 1/2(x+3)g(x)+0 gcd(f(x),g(x)) = x2+3x+2 = ½ g(x) 15:22

Exercise 8.2 True or False all Exercise 2,7,11

Evaluation Let R be an integral domain, and let f (x)=a0 + a1x +…+ anxn be a polynomial in R[x]. The element f (a)=a0 + a1a +…+ anan of R obtained by substitute a for x is called the evaluation of f (x) at a.

root Let R be a commutative ring and f (x)R[x]. An element a of R is called a root of f (x) if f (a) = 0. The root a is said to have multiplicity m1 if f (x) =(xa)mq(x), where g(a)0.

The polynomial x2+1 has no roots in real set R but has two roots in complex set C. The polynomial x3 x has 6 roots in Z6.

Remainder Theorem If R is a commutative ring and if f (x)R[x] is divided by xa, the remainder is f (a).

Factor Theorem Let R be a commutative ring and let f (x) be a polynomial in R[x]. If aR, then f (a)= 0 if and only if f (x)=(xa)q(x) for some q(x) R[x].

Theorem Let R be an integral domain and let f (x)R[x] be a nonzero polynomial of degree n. Then f (x) has at most n roots in R. Proof. By induction on n = deg( f (x)) n = 0: f (x) has no roots.

proof n = 1: let f (x) = a0+a1x, where a10. Suppose a and b are roots. Then a0+a1a = 0 = a0+a1b. So a1(ba) = 0. This implies a =b since R is a domain.

proof n >1: Suppose aR is a root of f (x). Then f (x) =(xa)g(x) and deg(g(x))= n1. Suppose b is a root of f (x) and b a. Then (ba)g(b)=0. Since R is a domain, we have g(b)=0. That is, b is a root of g(x). Because g(x) has at most n1 roots in R. So f (x) has at most n roots in R.

Fundamental Theorem of Algebra If f (x) is a nonconstant polynomial in C[x], then f (x) has a root in C. 15:22

irreducible If F is a field, a nonzero polynomial p(x) in F [x] is called an irreducible over F if: (1) deg p(x) 1. (2) If p(x) = f (x)g(x) in F [x], then either deg( f (x))=0 or deg(g(x))=0. 15:22

Theorem Let F be a field and deg( p(x)) is 2 or 3, then p(x) is irreducible over F if and only if it has no root in F. 15:22

Example x2+1 is irreducible in R[x] because it has no root in real set R. Determine if p(x)= x3+3x2+x+2 is irreducible over Z5. p(0)=2,p(1)=2,p(2)=4,p(3)=4, and p(4)=3. Hence p(x) bas no root in Z5 and so is irreducible over Z5. 15:22

Example List all polynomials of degree 3 over Z2. What are irreducible? Solution: x3, x3+x2, x3+x2+x, x3+x2+x+1, x3+x2+1, x3+x, x3+x+1, and x3+1 are all polynomials of degree 3 over Z2. x3+x2+1 and x3+x+1 are all irreducible polynomials of degree 3 over Z2. 15:22

Example degree>3 irreducible  no root P(x) = x4+3x2+2 = (x2+1)(x2+2) is not irreducible in [x], but it has no root in . 15:22

Example Suppose D is a subfield of F. If p(x) is irreducible over F, then p(x) is irreducible over D. If p(x) is irreducible over D, then p(x) may not be irreducible over F. p(x) = x22 is irreducible over  but not over . 15:22

Unique Factorization Theorem If F is a field, let f (x) be a nonconstant polynomial in F[x]. Then (1) f (x)=ap1(x)p2(x)…pm(x), where aF and pi(x) is monic and irreducible for all i. (2) The factorization in (1) is unique except for the order of the factors. 15:22

Example Factor f (x) = x31 into irreducible in C[x], R[x], [x], Z5[x], and Z7[x]. Solution: in C[x], in R [x], [x],and Z5[x], f (x)=(x1)(x2+x+1) in Z7[x], f (x)=(x1)(x2)(x4) 15:22

Exercise 8.3 True or False all Exercise 1(a),2(a,e,h),3,4(a,d)

Theorem Let F be a field and let A≠0 be an ideal of F[x]. Then a uniquely determined monic polynomial h(x) exists in F[x] such that A=<h(x)>=h(x)F[x]={h(x)f(x)|f(x)∈F[x]} .

lemma Let F be a field and h(x) be a monic polynomial in F[x] of degree m>0. Then S=F[x]/A  {a0+a1t+…+am-1tm-1|ai∈F} where A=<h(x)>=h(x)F[x] We may write S=F[x]/A ={a0+a1t+…+am-1tm-1|ai∈F} h(t)=0

Example F[x]/<x2>={a+bt|a,b∈F;t2=0} The addition and multiplicationin F[x]/<x2> is (a+bt)+(c+dt)=(a+c)+(b+d)t (a+bt)(c+dt)=ac+(ad+bc)t +bdt2 = ac+(ad+bc)t

lemma h(x) : degree m>0. S=F[x]/<h(x)>  {a0+a1t+…+am-1tm-1|ai∈F} We may write S=F[x]/A ={a0+a1t+…+am-1tm-1|ai∈F, h(t)=0} F[x]/<x2>={a+bt|a,b∈F;t2=0}

If F= Z2={0,1}, then F[x]/<x2>={0,1,t,1+t} is a ring with four elements and + 0 1 t 1+t  0 1 t 1+t 0 0 1 t 1+t 0 0 0 0 0 1 1 0 1+t t 1 0 1 t 1+t t t 1+t 0 1 t 0 t 0 t 1+t 1+t t 1 0 1+t 0 1+t t 1

Z2[x]/<x3+1>={a+bt+ct2|a,b,c∈Z2;t3+1=0} ={0,1,t,t2,1+t,1+t2,t+t2,1+t+t2} Q[x]/<x22>={a+bt|a,b,∈Q;t22=0} {a+b2 |a,b,∈Q}=Q(2)

Theorem Let h(x) be a monic polynomial of degree m>0 in F[x]. Then following conditions are equivalent. (1) F[x]/<h(x)> is a field. (2) h(x) is irreducible over F.

Example 是否存在72 elements的field。 Solution: x2+1 is irreducible over Z7. Consider Z7[x]/< x2+1 >

Problem 一個finite field的order會是多少? Theorem: If F is a finite field, then |F|=pn for some n1, where p is prime.

Problem 那是不是對任一正整數n與任一質數p都有order為pn的field? 是否為unique? exists and unique 如何建構order為pn的field?

Theorem A field of order pn exists and unique up to isomorphic.

Galois field If p is prime and n is a positive integer, the unique field with pn elements is called the Galois field of order pn and is denoted GF(pn). GF(p)= Zp for each prime p. http://episte.math.ntu.edu.tw/people/p_galois/index.html

Problem 如何建構order為pn的field? If f(x) is an irreducible polynomial of degree n in Zp[x], then {a0+a1t+…+an1t n1 | ai∈Zp; f(t)=0} is a field of order pn.

Example Construct GF(8). Solution: p(x)=x3+x+1 is irreducible over Z2. GF(8) = {a+bt+ct2|a,b,c∈Z2; p(t)=0} = {a+bt+ct2|a,b,c∈Z2; t3=t+1 }.

Exercise 8.6 Exercise 9