Corollary 1 (Factor Theorem): For a field F, an element is ...



Section 6: Irreducibility and Roots of Polynomials

HW p. 14 # 1-17 at the end of the notes

In this section, we discuss when polynomials are irreducible and discuss the relation between the roots of polynomials and their roots.

Irreducibility of Polynomials

Recall that for a ring R with unity [pic], an element [pic] is a unit of R if it has a multiplicative inverse in R, that is, an element [pic] where [pic]. For certain polynomial rings, only a small class of polynomials are units, as the next theorem indicates.

Theorem 6.1: Let F be a field. Then [pic] is a unit if and only if [pic] is a non-zero constant polynomial.

Proof:

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Facts From Theorem 6.1

1. The theorem shows that rings such as [pic] and [pic] have infinitely many units since in these rings there are an infinite number of constant polynomials.

2. The theorem can be false if F is not a field. For example, the constant polynomial [pic] defined over [pic] is not a unit since [pic]. Also, the polynomial [pic] is a unit in [pic] since [pic].

Definition 6.2: For a field F, a polynomial [pic] is an associate of a polynomial [pic] if [pic] for some non-zero constant [pic].

For example, in [pic], some associates of [pic] are [pic] and [pic] (there are an infinite number of associates for this ring). However, for some rings, there are only a finite number of associates. For example, for the ring [pic], since the only non-zero elements for the field [pic] are 1, 2, 3, and 4, the only associates for the polynomial [pic] are [pic], [pic], [pic], and [pic].

Example 1: List all of the associates for [pic] in [pic].

Solution:

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Note: If [pic], then [pic] is an associate of [pic] Since c is non-zero and an element of a field, then it has a multiplicative inverse [pic]. Hence, we can say that

[pic]

or

[pic]

This allows us to make the following statement:

Fact: [pic] is an associate of [pic] if and only if [pic] is an associate of [pic].

Recall that for the integers [pic], the units are [pic] and 1. Hence, the associates of any integer n has the form [pic], that is, they are the product of n with the units in Z. Hence, Theorem 6.1 above says that the only units in [pic] are the non-zero constant polynomials, Hence, all associates of a polynomial [pic] in [pic] of the form [pic] are the product of [pic] with its units c.

Now if p is prime in Z, its only divisors are its units [pic] and its associates. A similar analogy can be give for polynomials.

Definition 6.3: For a field F, a non-constant polynomial [pic] is said to be irreducible of its only divisors are the constant polynomials (the units of [pic]) and its associates (polynomials of the form [pic], where c is a non-zero constant in F). If [pic] is not irreducible, it is said to be reducible over F.

Another way to define irreducibility is as follows. Since for a polynomial [pic] to be irreducible, we know that its divisors can only be its units or its associates, the degrees of its divisors can only be zero or the same as itself. Thus, we can make this statement:

[pic]

A non-constant polynomial [pic] is irreducible over F if [pic] cannot be expressed as a product [pic] of two non-zero, non-constant polynomials [pic] and [pic] both of lower degree that the degree of [pic].

Fact: Every polynomial of degree 1 in [pic] is irreducible in[pic]. For if we have a first degree polynomial of the form [pic], where [pic], then the only divisors of [pic] can have degree 0 or 1. The degree 0 polynomials can only be constant polynomials, that is, the units of [pic]. Degree 1 polynomials can only be associates of [pic].

Note: A polynomial that is irreducible over a field F may not irreducible over a larger field containing F.

For example, [pic] is irreducible over [pic]. However, [pic],

[pic].

There are other conditions concerning irreducibility of polynomials that are equivalent. The next theorem summarizes these facts.

Theorem 6.4: Let F be a field and [pic] a non-constant polynomial in [pic]. Then the following conditions are equivalent:

(1) [pic] is irreducible.

(2) If [pic] and [pic] are polynomials such that [pic], then [pic] or

[pic].

(3) If [pic] and [pic] are any polynomials such that [pic], then either [pic] or

[pic] is a non-zero constant polynomial.

Proof: (1) [pic] (2). Assume [pic] is irreducible and suppose [pic]. Consider [pic]. If [pic], then Theorem 5.5 says that [pic]and we are done. Suppose [pic]. Then [pic]. Since [pic] is irreducible, then [pic] (an associate of [pic]). Since [pic], [pic]. Hence, [pic] and we are done. A similar argument can be made when considering [pic].

(2) [pic] (3) If [pic], then (2) says that [pic] or [pic]. If, say, [pic], then [pic] and

[pic].

Since [pic] is an integral domain, we can apply the cancellation laws to [pic] on both sides to obtain

[pic].

Hence, [pic] is a unit and since only constant polynomials can be units, [pic]. A similar argument for [pic] holds if we assume [pic].

(3) [pic]1 Suppose [pic] is any divisor of [pic], i.e., suppose [pic]. Then assertion (3) says that either [pic] or [pic] is a constant polynomial. If [pic] is constant, say [pic], then [pic] or [pic]. Hence, [pic] is an associate of [pic]. If [pic] is not constant, then [pic] must be constant. In either case, the divisor [pic] is a non-zero constant or an associate of [pic]. Hence, by definition, [pic] must be irreducible.

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Corollary 6.5: Let F be a field and [pic] an irreducible polynomial in [pic]. If [pic], then [pic] divides at least one of the [pic].

Proof:

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Theorem 6.6: Let F be a field. Then every non-constant polynomial [pic] is a product of irreducible polynomials in [pic]. This factorization is unique in the sense that if

[pic] and [pic],

where each [pic] and [pic] are irreducible, then [pic] (the number of irreducible factors are the same). After the [pic] are reordered and relabeled, if necessary, then each [pic]

is an associate for each [pic] for [pic]

Roots of Polynomials and Reduciblity

To determine if a polynomial is irreducible or reducible, we can examine the roots of the polynomial. To demonstrate, we begin with the following definition.

Definition 6.7: Let R be a commutative ring with [pic]. An element [pic] is said to be a root of the polynomial [pic] if [pic].

Example 2: Find the roots of [pic] over [pic].

Solution:

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Example 3: Find the roots of [pic] over [pic]. Over [pic]

Solution:

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As these examples demonstrate, there is a relation between the zeros of a polynomial and its factors, which is summarized in the following theorem.

Theorem 6.8: (Factor Theorem): For a field F, an element [pic] is a root of [pic] if and only if [pic] is a factor of [pic] in [pic].

Proof:

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Example 4: Determine if [pic] is a factor of [pic] in [pic].

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Example 5: Determine if [pic] is a factor of [pic] in [pic].

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Corollary 6.9: A non-zero polynomial [pic] of degree n has at most n zeros in a field F.

Proof: Let [pic] be a polynomial of degree n is [pic]. If [pic] is a zero of f, the Factor Theorem says that

[pic], where [pic].

If [pic] is a zero of [pic], then

[pic], where [pic].

Continuing the process until we obtain a quotient with no zeros in F gives

[pic], where [pic].

We claim that [pic] are the only possible zeros in F for [pic] where [pic]

for if b is a zero, [pic], then

[pic]

None of the factors of [pic] are 0 by construction. Multiplying non-zero terms in F to get 0 implies F has divisors of 0, which is a contradiction since any field is an integral domain. Hence the [pic] are the only zeros in F for [pic]

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Example 6: Factor [pic] into linear factors in [pic].

Solution:

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Corollary 4.10: Let F be a field and [pic] with [pic].

(1) If [pic] is irreducible in [pic], then [pic] has no roots in F.

(2) If [pic] has degree 2 or 3 and has no roots in F, then [pic] is irreducible in [pic].

Proof: (1) Since [pic] is irreducible if [pic], then [pic] has no linear factors of the form [pic]. Since [pic] has no linear factors, the Factor Theorem (Theorem 4.8) says that [pic]

can have no roots in F.

(2) If [pic] is irreducible, it cannot have a linear factor of degree 1 of the form [pic]. Suppose [pic]. If [pic] is of degree 2 and is reducible, then [pic] must be the product of two linear one degree polynomials, which is impossible. If [pic] is of degree 3 and is reducible, then [pic] must be the product of a second degree irreducible quadratic and a first degree linear factor or three first degree linear factors. This is impossible. Hence, either [pic] or [pic]must be of degree 0, or must be non-zero constants. Thus, [pic] is irreducible.

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Note: Part 2 of Corollary 4.10 is not necessarily true if [pic]. For example, the polynomial

[pic].

is reducible in [pic] even though there are no roots in Q.

Example 7: Determine if [pic] is an irreducible polynomial of [pic]. If not, factor the polynomial as a product of irreducible factors.

Solution:

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Example 8: Determine if [pic] is an irreducible polynomial of [pic]. If not, factor the polynomial as a product of irreducible factors.

Solution:

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Exercises

1. Find a monic associate of

a. [pic] in [pic]. b. [pic] in [pic].

2. List all of the associates for the following of

a. [pic] in [pic].

b. [pic] in [pic].

c. [pic] in [pic].

3. For a prime p, show that a non-zero polynomial in [pic] has exactly [pic] associates.

4. Determine if [pic] is factor of [pic]

a. [pic] and [pic] in [pic].

b. [pic] and [pic] in [pic].

c. [pic] and [pic] in [pic].

d. [pic] and [pic] in [pic].

5. For what value of k is [pic] a factor of [pic] in [pic].

6. For what value of k is [pic] a factor of [pic] in [pic].

7. Show that [pic] divides [pic] if and only if [pic].

8. Determine if the given polynomials are irreducible.

a. [pic] in [pic]. e. [pic] in [pic].

b. [pic] in [pic]. f. [pic] in [pic].

c. [pic] in [pic]. g. [pic] in [pic].

d. [pic] in [pic].

9. The following polynomials can be factored into linear factors over the given field. Find their polynomials.

a. [pic] in [pic].

b. [pic] in [pic].

c. [pic] in [pic].

10. Determine if the following polynomials are irreducible over the given field. If not,

factor the polynomial as a product of irreducible factors.

a. [pic] in [pic].

b. [pic] in [pic].

c. [pic] in [pic].

d. [pic] in [pic].

11. Find all irreducible polynomials of

a. degree 2 in [pic]. c. degree 2 in [pic].

b. degree 2 in [pic]. d. degree 3 in [pic].

12. For Exercise 9, list all of the monic irreducible polynomials for each case.

13. If [pic] is a non-zero root of [pic], then show [pic] is a root of [pic].

14. a. If [pic] and [pic] are associates in [pic], show that they have the same

roots in F.

b. If [pic] have the same roots in F, are they associates in [pic]?

15. Prove the Remainder Theorem: Let [pic], where F is a field, and let [pic]. Show that the remainder [pic] when [pic] is divided by [pic], in accordance with the division algorithm, is [pic].

16. Suppose [pic] are roots of [pic] , where [pic]. Use the Factor

Theorem (Theorem 6.8) to show [pic] and [pic].

17. Prove that [pic] is reducible in [pic], where p is prime, if and only if there exist integers a and b such that [pic] and [pic].

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