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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 1  0 , an element u  R is a unit of R if it has a
multiplicative inverse in R, that is, an element u 1  R where uu 1  1 . 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 f ( x)  F [ x] is a unit if and only if f (x) is a nonzero constant polynomial.
Proof:
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Facts From Theorem 6.1
1. The theorem shows that rings such as Q[x ] and R[ x ] 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
f ( x)  3 defined over Z [x] is not a unit since 1 / 2  Z . Also, the polynomial 3x  1
is a unit in Z 9 [ x] since (3x  1)(6 x  1)  18 x 2  9 x  1  0  0  1  1 .
mod 9
Definition 6.2: For a field F, a polynomial f ( x)  F [ x] is an associate of a polynomial
g ( x)  F [ x] if f ( x)  cg ( x) for some non-zero constant c  F .
1 2
( x  2)
2
(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 Z 5 [ x] , since the only
For example, in Q[x ] , some associates of g ( x)  x 2  2 are 3( x 2  2) and
non-zero elements for the field Z 5  {0, 1, 2, 3, 4} are 1, 2, 3, and 4, the only associates
for the polynomial g ( x)  x 2  2 are (1)( x 2  2)  x 2  2 , (2)( x 2  2)  2 x 2  4 ,
(3)( x 2  2)  3x 2  6  3x 2  1 , and (4)( x 2  2)  4 x 2  8  3x 2  3 .
mod 5
mod 5
Example 1: List all of the associates for x 2  x  2 in Z 7 [ x] .
Solution:
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Note: If f ( x)  cg ( x) , then f (x) is an associate of g (x). Since c is non-zero and an
element of a field, then it has a multiplicative inverse c 1  F . Hence, we can say that
c 1cg ( x)  c 1 f ( x)
or
g ( x)  c 1 f ( x)
This allows us to make the following statement:
Fact: f (x) is an associate of g (x ) if and only if g (x ) is an associate of f (x) .
Recall that for the integers Z , the units are  1 and 1. Hence, the associates of any
integer n has the form  n , that is, they are the product of n with the units in Z. Hence,
Theorem 6.1 above says that the only units in F [x ] are the non-zero constant
polynomials, Hence, all associates of a polynomial g (x ) in F [x ] of the form cg (x) are
the product of g (x ) with its units c.
Now if p is prime in Z, its only divisors are its units  1 and its associates. A similar
analogy can be give for polynomials.
Definition 6.3: For a field F, a non-constant polynomial p( x)  F [ x] is said to be
irreducible of its only divisors are the constant polynomials (the units of F [x ] ) and its
associates (polynomials of the form cp (x) , where c is a non-zero constant in F). If p (x )
is not irreducible, it is said to be reducible over F.
Another way to define irreducibility is as follows. Since for a polynomial p (x ) 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:
A non-constant polynomial p( x)  F [ x] is irreducible over F if p ( x ) cannot be
expressed as a product g ( x )h( x ) of two non-zero, non-constant polynomials g (x ) and
h(x) both of lower degree that the degree of p (x ) .
Fact: Every polynomial of degree 1 in F [x ] is irreducible in F [x ] . For if we have a first
degree polynomial of the form ax  b , where a, b  F , then the only divisors of ax  b
can have degree 0 or 1. The degree 0 polynomials can only be constant polynomials, that
is, the units of F [x ] . Degree 1 polynomials can only be associates of ax  b .
Note: A polynomial that is irreducible over a field F may not irreducible over a larger
field containing F.
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For example, f ( x)  x 2  2 is irreducible over Q[x ] . However, R[ x ] ,
f ( x)  x 2  2  ( x  2 )( x  2 ) .
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 p (x ) a non-constant polynomial in F [x ] . Then the
following conditions are equivalent:
(1) p (x ) is irreducible.
(2) If e(x ) and f (x) are polynomials such that p( x) | e( x)  f ( x) , then p ( x) | e( x) or
p ( x) | f ( x) .
(3) If s (x ) and t (x) are any polynomials such that p ( x)  s ( x)t ( x) , then either s (x ) or
t (x) is a non-zero constant polynomial.
Proof: (1)  (2). Assume p (x ) is irreducible and suppose p( x) | e( x)  f ( x) . Consider
gcd( p( x), e( x)) . If gcd( p ( x), e( x))  1 , then Theorem 5.5 says that p ( x) | f ( x) and we
are done. Suppose gcd( p( x), e( x))  d ( x)  1 . Then d ( x) | p( x) . Since p (x ) is
irreducible, then d ( x)  cp( x) (an associate of p (x ) ). Since d ( x) | e( x ) ,
e( x)  k ( x)d ( x)  ck ( x) p( x) . Hence, p ( x) | e( x) and we are done. A similar argument
can be made when considering gcd( p( x), f ( x)) .
(2)  (3) If p ( x)  s ( x)t ( x) , then (2) says that p ( x) | s ( x) or p( x) | t ( x) . If, say,
p ( x) | s ( x) , then s ( x)  k ( x) p ( x) and
p( x)  s ( x)t ( x)  k ( x) p( x)t ( x)  k ( x)t ( x) p( x) .
Since F [x ] is an integral domain, we can apply the cancellation laws to p (x ) on both
sides to obtain
k ( x)t ( x)  1 .
Hence, t (x) is a unit and since only constant polynomials can be units, t ( x)  c . A
similar argument for s (x ) holds if we assume p( x) | t ( x) .
(3)  1 Suppose a(x) is any divisor of p (x ) , i.e., suppose p( x)  q( x)a( x) . Then
assertion (3) says that either q(x) or a(x) is a constant polynomial. If q(x) is constant,
say q( x)  c , then p ( x)  ca ( x) or c 1 p( x)  a( x) . Hence, a(x) is an associate of
p (x ) . If q(x) is not constant, then a(x) must be constant. In either case, the divisor
a(x) is a non-zero constant or an associate of p (x ) . Hence, by definition, p (x ) must be
irreducible.
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Corollary 6.5: Let F be a field and p (x ) an irreducible polynomial in F [x ] . If
p( x) | [a1 ( x)a2 ( x) an ( x)] , then p (x ) divides at least one of the ai ( x)' s .
Proof:
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Theorem 6.6: Let F be a field. Then every non-constant polynomial f ( x)  F [ x] is a
product of irreducible polynomials in F [x ] . This factorization is unique in the sense that if
f ( x)  p1 ( x) p2 ( x) pr ( x) and f ( x)  q1 ( x)q2 ( x) q s ( x) ,
where each pi (x) and q j (x) are irreducible, then r  s (the number of irreducible factors
are the same). After the q j (x) are reordered and relabeled, if necessary, then each pi (x)
is an associate for each qi (x) for i  1, 2 r.
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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 f ( x)  R[ x] . An element a  R is said
to be a root of the polynomial f (x) if f (a)  0 .
Example 2: Find the roots of f ( x)  x 2  2 x  3 over Q[x ] .
Solution:
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Example 3: Find the roots of f ( x)  x 2  2 over Q[x ] . Over R[ x ]
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.
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Theorem 6.8: (Factor Theorem): For a field F, an element a  F is a root of
f ( x)  F [ x] if and only if x  a is a factor of f (x ) in F [x ] .
Proof:
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Example 4: Determine if h( x)  x  3 is a factor of f ( x)  x 4  3x 3  x  3 in Q[x ] .
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Example 5: Determine if h( x)  x  1 is a factor of f ( x)  x 3  3x 2  6 x  4 in Z 7 [ x] .
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Corollary 6.9: A non-zero polynomial f ( x)  F [ x] of degree n has at most n zeros in a
field F.
Proof: Let f (x) be a polynomial of degree n is F [x ] . If a1 is a zero of f, the Factor
Theorem says that
f ( x)  q1 ( x)( x  a1 ) , where degree q1 ( x)  n  1 .
If a 2 is a zero of q1 ( x) , then
f ( x)  q2 ( x)( x  a1 )( x  a2 ) , where degree q2 ( x)  n  2 .
Continuing the process until we obtain a quotient with no zeros in F gives
f ( x)  qr ( x)( x  a1 )( x  a2 )( x  ar ) , where degree qr ( x)  n  r .
We claim that a1 , a2 , , ar are the only possible zeros in F for f (x) where r  n
for if b is a zero, b  ai , then
f (b)  qr (b)(b  a1 )(b  a2 )(b  ar )  0
None of the factors of f (b ) 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 ai ' s, i  1, 2, , r  n are the only zeros in F for f (x ).
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Example 6: Factor x 3  3x 2  6 x  4 into linear factors in Z 7 [ x] .
Solution:
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Corollary 4.10: Let F be a field and f ( x)  F [ x] with deg f ( x)  2 .
(1) If f (x) is irreducible in F [x ] , then f (x) has no roots in F.
(2) If f (x) has degree 2 or 3 and has no roots in F, then f (x) is irreducible in F [x ] .
Proof: (1) Since f (x) is irreducible if F [x ] , then f (x) has no linear factors of the form
x  a . Since f (x) has no linear factors, the Factor Theorem (Theorem 4.8) says that f (x)
can have no roots in F.
(2) If f (x) is irreducible, it cannot have a linear factor of degree 1 of the form x  a .
Suppose f ( x)  r ( x) s ( x) . If f (x) is of degree 2 and is reducible, then f (x) must be the
product of two linear one degree polynomials, which is impossible. If f (x) is of degree 3
and is reducible, then f (x) 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 r (x ) or s (x ) must be of degree 0, or must be non-zero constants. Thus, f (x) is
irreducible.
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Note: Part 2 of Corollary 4.10 is not necessarily true if deg f ( x)  4 . For example, the
polynomial
f ( x)  x 4  5x 2  6  ( x 2  2)( x 2  3) .
is reducible in Q[x ] even though there are no roots in Q.
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Example 7: Determine if f ( x)  x 2  2 x  3 is an irreducible polynomial of Z 7 [ x] . If not,
factor the polynomial as a product of irreducible factors.
Solution:
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Example 8: Determine if f ( x)  x 3  2 is an irreducible polynomial of Z 5 [ x] . If not,
factor the polynomial as a product of irreducible factors.
Solution:
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Exercises
1. Find a monic associate of
a. 5x 3  3x 2  4 x  1 in Q[x ] .
b. 2 x 4  3x 3  2 x 2  4 x  1 in Z 5 [ x] .
2. List all of the associates for the following of
a. x 2  x  1 in Z 2 [ x] .
b. x 2  x  1 in Z 5 [ x] .
c. 4 x 3  2 x 2  6 x  5 in Z 7 [ x] .
3. For a prime p, show that a non-zero polynomial in Z p [x] has exactly p  1
associates.
4. Determine if h(x) is factor of f (x) :
a. h( x)  x  2 and f ( x)  x 3  3x 2  4 x  12 in R[ x ] .
1
3
b. h( x)  x  and f ( x)  2 x 4  x 3  x  in Q[x ] .
2
4
5
4
3
c. h( x)  x  2 and f ( x)  3x  4 x  2 x  x 2  2 x  1 in Z 5 [ x] .
d. h( x)  x  3 and f ( x)  x 6  x 3  x  5 in Z 7 [ x] .
5. For what value of k is x  1 a factor of x 4  5 x 3  5 x 2  3x  k in Q[x ] .
6. For what value of k is x  1 a factor of x 4  2 x 3  3x 2  kx  1 in Z 5 [ x] .
7. Show that x  1 divides f ( x)  an x n  an1 x n1    a1 x  a0  F[ x] if and only if
an  an1    a1  a0  0 .
8. Determine if the given polynomials are irreducible.
a. x 2  5 in Q[x ] .
e. x 3  9 in Z11[ x] .
b. x 2  5 in R[ x ] .
c.
x 3  3 in Z 7 [ x] .
d. x 2  2 in Z 5 [ x] .
f. 2 x 3  x 2  2 x  2 in Z 5 [ x] .
g. x 4  x 2  1 in Z 3 [ x] .
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9. The following polynomials can be factored into linear factors over the given field.
Find their polynomials.
a. x 4  4 in Z 5 [ x] .
b. x 3  2 x 2  2 x  1 in Z 7 [ x] .
c. 2 x 3  3x 2  7 x  5 in Z11[ x] .
10. Determine if the following polynomials are irreducible over the given field. If not,
factor the polynomial as a product of irreducible factors.
a. x 3  2 x  3 in Z 5 [ x] .
b. 2 x 3  x 2  2 x  2 in Z 5 [ x] .
c.
x 3  3 in Z11[ x] .
d. x 5  2 x 3  2 x 2  4 in Z11[ x] .
11. Find all irreducible polynomials of
a. degree 2 in Z 2 [ x] .
b. degree 2 in Z 3 [ x] .
c. degree 2 in Z 5 [ x] .
d. degree 3 in Z 2 [ x] .
12. For Exercise 9, list all of the monic irreducible polynomials for each case.
13. If b  F is a non-zero root of an x n  an1 x n1    a1 x  a0  F[ x] , then show b 1
is a root of a0 x n  a1 x n1    an1 x  an .
14. a. If f (x) and g (x ) are associates in F [x ] , show that they have the same
roots in F.
b. If f ( x), g ( x)  F [ x] have the same roots in F, are they associates in F [x ] ?
15. Prove the Remainder Theorem: Let f ( x)  F [ x] , where F is a field, and let   F .
Show that the remainder r (x ) when f (x) is divided by x   , in accordance with
the division algorithm, is f ( ) .
16. Suppose r , s  F are roots of ax 2  bx  c  F[ x] , where a  0 . Use the Factor
Theorem (Theorem 6.8) to show r  s  a 1b and rs  a 1c .
17. Prove that x 2  1 is reducible in Z p [x] , where p is prime, if and only if there exist
integers a and b such that p  a  b and ab  1 (mod p) .