1. 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.

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

3. 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,….

4. 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.

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

6. 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].

7. 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].

8. 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.

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

10. 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].

11. 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].

12. 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.

13. 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) ).

14. 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 .

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

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

17. 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)).

18. 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].
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19. 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].
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20. 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)].
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21. 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)
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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)
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23. Exercise 8.2
True or False
all
Exercise
2,7,11

24. 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.

25. 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.

26. 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.

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

28. 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].

29. 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.

30. 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.

31. 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.

32. Fundamental Theorem of Algebra
If f (x) is a nonconstant polynomial in C[x], then f (x) has a root in C.
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33. 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.
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34. 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.
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35. 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.
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36. 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.
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37. 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 .
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38. 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 .
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39. 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.
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40. 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)
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41. Exercise 8.3
True or False
all
Exercise
1(a),2(a,e,h),3,4(a,d)

42. 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]} .

43. 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

44. 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

45. 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}

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

47. 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)

48. 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.

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

50. 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.

51. 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?

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

53. 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.

54. 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.

55. 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 }.

56. Exercise 8.6
Exercise 9