ch 16 to 17: Polynomial Ring and Factorization of Polynomials

0.0(0)
studied byStudied by 0 people
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/25

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

26 Terms

1
New cards

Let R be a commutative ring. The ring of polynomials over R in the indeterminate x is

the set of formal symbols R[x] = {anxn + an-1xn-1 + … + a1x + a0 | ai belonging to R, n is a nonnegative integer}

2
New cards

Two elements anxn +an-1 xn-1 + … + a1x + a0 and bmxm + bm-1xm-1 + … + b1x + b0 of R[x] are considered equal if and only if

ai = bi for all nonnegative integers i (Define ai = 0 when i > n and bi = 0 when i > m

3
New cards

Let R be a commutative ring and let f(x) = anxn + an-1xn-1 + … + a1x + a0 and g(x) = bmxm + bm-1xm-1 + … + b1x + b0 belong to R[x]. Then

f(x) + g(x) =

(as + bs) xs + (as-1 + bs-1) xs-1 + … + (a1 + b1) x + a0 + b0, where s is the maximum of m and n, ai = 0 for i > n, and bi = 0 for i > m

4
New cards

Let R be a commutative ring and let f(x) = anxn + an-1xn-1 + … + a1x + a0 and g(x) = bmxm + bm-1xm-1 + … + b1x + b0 belong to R[x]. Then

f(x) g(x) =

c m+n x m+n + c m+n-1 x m+n-1 + … + c1x + c0 , where ck = akb0 + ak-1 b1 + … + a1 b k-1 + a0 bk for k = 0, … , m + n

5
New cards

Theorem 16.1

If D is an integral domain, then D[x] is

an integral domain

6
New cards

Theorem 16.2 - Division Algorithm for F[x]

Let F be a field and let f(x), g(x) be elements of F[x] with g(x) ≠ 0. Then there exist

unique polynomials q(x) and r(x) in F[x] such that f(x) = g(x) q(x) + r(x) and either r(x) = 0 or deg r(x) < deg g(x)

7
New cards

Corollary 1 - Remainder Theorem

Let F be a field, a be an element of F and f(x) be an element of F[x]. The f(a) is

the remainder in the division of f(x) by x - a

8
New cards

Corollary 2 - Factor Theorem

Let F be a field, a be an element of F, and f(x) be an element of F[x]. Then a is a zero of f(x) if and only if

x - a is a factor of f(x)

9
New cards

Theorem 16.3

A polynomial of degree n over a field has

at most n zeros, counting multiplicity

10
New cards

A principal ideal domain is

an integral domain R in which every ideal has the form < a > = {ra | r belonging to R} for some a in R

11
New cards

Theorem 16.4

Let F be a field. The F[x] is

a principal ideal domain

12
New cards

Theorem 16.5

Let F be a field, I a nonzero ideal in F[x], and g(x) an element of F[x]. Then, I = < g (x) > if and only if

g(x) is a polynomial of minimum degree in I

13
New cards

Let D be an integral domain. A polynomial f(x) from D[x] that is neither the zero polynomial nor a unit in D[x] is said to be irreducible over D if,

whenever f(x) is expressed as a product f(x) = g(x) h(x), with g(x) and h(x) from D[x], then g(x) or h(x) isa unit in D[x]

14
New cards

A nonzero, nonunit element of D[x] that is not irreducible over D is

reducible over D

15
New cards

Theorem 17.1

Let F be a field. If f(x) is an element of F[x] and deg f(x) is 2 or 3, then f(x) is reducible over F if and only if

f(x) has a zero in F

16
New cards

The content of a nonzero polynomial an xn + an-1 x n-1 + … + a0, where the a’s are integers, is

the greatest common divisor of the integers an , an-1 , …, a0

17
New cards

A primitive polynomial is

an element of Z[x] with content 1

18
New cards

Gauss’s Lemma

The product of two primitive polynomials is

primitive

19
New cards

Theorem 17.2

Let f(x) be an element of Z[x]. If f(x) is reducible over Q, then it is

reducible over Z

20
New cards

Theorem 17.3

Let p be a prime and suppose that f(x) be an element of Z[x] with deg f(x) ≥ 1. Let f’(x) be the polynomial in Zp [x] obtained from f(x) by reducing all the coefficients of f(x) modulo p.

If f’(x) is irreducible over Zp and deg f’(x) = deg f(x), then f(x) is

irreducible over Q

21
New cards

Theorem 17.4 - Eisenstein’s Criterion

Let f(x) = an xn + an-1 xn-1 + … + a0 be an element of Z[x].

If there is a prime p such that p cannot divide an , p can divide an-1 , … , p can divide a0 and p2 cannot divide a0 , then f(x) is

irreducible over Q

22
New cards

Theorem 17.5

Let F be a field and let p(x) be an element of F[x]/ Then < p(x) > is a maximal ideal in F[x] if and only if

p(x) s irreducible over F

23
New cards

Corollary 1

Let F be a field and p(x) be an irreducible polynomial F. The F[x] / < p(x) > is

a field

24
New cards

Corollary 2

Let F be a field and let p(x), a(x), b(x) be an element of F[x]. If p(x) is irreducible over F and p(x) | a(x) b(x), then

p(x) | a(x) or p(x) | b(x)

25
New cards

Theorem 17.6 pt 1

Every polynomial in Z[x] that is not the zero polynomial or a unit in Z[x] can be written in the form

b1 b2 … bs p1 (x) p2 (x) … pm (x), where bi’s are irreducible polynomials of degree 0 and pi (x)’s are irreducible polynomials of positive degrees.

26
New cards

Theorem 17.6 pt 2

If b1 b2 … bs p1 (x) p2 (x) … pm (x) = c1 c2 … ct q1(x) q2 (x) … qn (x), where the bi’s and ci’s are irreducible polynomials of degree 0 and pi (x)’s and qi’s are irreducible polynomials of positive degree, then

s = t, m = n, and , after renumbering the c’s and q(x)’s , we have bi = ± ci for i = 1, … , s and pi (x) = ± qi (x) for i = 1, …, m