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Section 9: The Structure of F[x]/p(x) when p(x) is Irreducible
HW p. 9 # 1-6 at the end of the notes
In Example 5 in Section 8, we saw by using the multiplication table of Z 2 [ x] /( x 2  x  1)
that every non-zero element in Z 2 [ x] /( x 2  x  1) had a multiplicative inverse and hence
Z 2 [ x] /( x 2  x  1) was a field. On closer examination of the polynomial x 2  x  1 , it
should be noted by the Factor Theorem that x 2  x  1 is irreducible since neither of the
elements in Z 2  {0, 1} is a root of x 2  x  1 .
In contrast, we saw in Example 6 in Section 8 that Z 2 [ x] /( x 3  1) had elements that in
Z 2 [ x] /( x 3  1) that did not have multiplicative inverses. Hence, Z 2 [ x] /( x 3  1) was not a
field. In addition, Z 2 [ x] /( x 3  1) was not an integral domain because it contained zero
divisors. When examining the polynomial, we should note that x 3  1 is a reducible since
1  Z 2 is a root of x 3  1 giving x  1 as a factor.
This relationship of when F [ x] /( p ( x)) is a field (and hence an integral domain) with
whether p (x ) is irreducible or not is not a coincidence. We summarize this relationship
in the following theorem.
Theorem 9.1: Let F be a field and p (x ) non-zero be a non-constant polynomial in F [x ] .
Then the following statements are equivalent.
(1) p (x ) is irreducible in F [x ] .
(2) F [ x] /( p ( x)) is a field.
(3) F [ x] /( p ( x)) is an integral domain.
Proof: (1)  (2) Suppose p (x ) is irreducible in F [x ] . To show F [ x] /( p ( x)) is a field,
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we must show any non-zero congruence class a( x)  F[ x] /( p( x)) , that is, any
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congruence class where a( x)  0 , has a multiplicative inverse. If a( x)  0 , Theorem
7.5 in Section 7 says that this implies that a( x)  0 (mod p( x)) . Thus, p( x) | (a( x)  0)
or p( x) | a ( x) . Hence, since p (x ) is irreducible, the gcd( a( x), p( x))  1 . Hence,
Theorem 5.3 in Section 5 says there exists polynomials u ( x), v( x)  F [ x] such that
a( x)u ( x)  p( x)v( x)  gcd( a( x), p( x))  1 .
Hence, we can write
a( x)u ( x)  1   p( x)v( x)
2
or
a( x)u ( x)  1  p( x)( v( x))
Thus, p( x) | (a( x)u ( x))  1) or a( x)u ( x)  1 (mod p( x)) . Hence, u (x) is the
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multiplicative inverse of a(x) in F [x ] and hence u(x) is the multiplicative inverse of
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a(x) in F [ x] /( p ( x)) since
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a ( x )  u ( x)  a ( x )  u ( x )  1 .
Thus, since every non-zero element has a multiplicative inverse, F [ x] /( p ( x)) is a field.
(2)  (3) If F [ x] /( p ( x)) is a field, then F [ x] /( p ( x)) must be an integral domain by
Theorem 3.5 in Section 3.
(3)  (1) Assume F [ x] /( p ( x)) is an integral domain. Suppose p( x)  a( x)b( x) . Hence,
a( x)b( x)  0 (mod p( x)) . Thus
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a( x)  b( x)  a( x)  b( x)  0 .
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Since F [ x] /( p ( x)) is an integral domain, this means that a( x)  0 or b( x)  0 . If
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a( x)  0 , then a( x)  0 (mod p( x)) , p( x) | a( x) and hence a( x)  t ( x) p( x) for some
t ( x)  F [ x] . Thus,
p( x)  a( x)b( x)  p( x)t ( x)b( x)
The cancellation property for integral domains implies that
t ( x)b( x)  1
Because deg(1)  0 , deg( b( x))  0 . Hence, b(x ) must be a non-zero constant, that is
b( x)  c for some c  F where c  0 . Thus,
p ( x)  a( x)b( x)  ca ( x)
and hence a(x) is an associate of p (x ) . A similar argument can be obtained for b(x ) if
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we assume b( x)  0 . Definition 6.3 in Section 6, p (x ) is irreducible.
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We can use Theorem 9.1 to quickly determine whether the rings we considered in Section
8 are fields, must faster than by we did in the Examples in that section.
Example 1: Is Z 2 [ x] /( x 2  x  1) a field?
Solution:
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Example 2: Is Z 2 [ x] /( x 3  1) a field?
Solution:
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Example 3: Is Q[ x] /( x 2  5) a field?
Solution:
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Example 4: Is Q[ x] /( x 2  4) a field?
Solution:
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As we will see later, if p is prime and p( x)  Z p [ x] is irreducible polynomial of degree k,
then Z p [ x ] / p ( x) is a finite field of p k elements. This can be seen using the fact that each
element in Z p [ x] / p ( x) can be represented by a polynomial for the form
ak 1 x k 1  ak 2 x k 2   a1 x  a0
where each ai  Z p . Since Z p has p elements, there are p choices for each ai , where
i  0, 1, , k  1 . Multiplying the p choices for the k coefficients gives p k elements.
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Let F be a field and p (x ) be an irreducible polynomial in F [x ] . Then recall that
K  F [ x] / p( x) is a field that contains the base field F. The field K  F [ x] / p( x) is said to
be an extension field of F.
Question: If p (x ) is irreducible, then p (x ) has no roots in the field F. Does p (x ) have
roots in the extension field K  F [ x] / p( x) ?
Example 1: Consider the addition and multiplication tables for the field Z 2 [ x] /( x 2  x  1) .
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
0
0
1
0
1
1
0
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x
x
x 1
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x 1

0
x
0
1
0
0
0
1
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x
0
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x
x 1
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x
0
1
1
0
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x 1
x
0
0
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1
x
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x 1
x 1
x
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x 1
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x 1
1
x 1
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x
1
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x 1
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x 1
0
x
1
Does the irreducible polynomial p( x)  x 2  x  1 have roots in the extension field
Z 2 [ x] /( x 2  x  1) ?
Solution: Since p (x ) is of degree 2, it will have at most two roots in Z 2 [ x] /( x 2  x  1) .
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Setting x  x , we see that x is a root since
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p ( x )  ( x ) 2  x  1  x  x  x  1  x  1 x  1  2 x  2  0 x  0  0 .
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In addition, x  x  1 is root since
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p( x  1)  ( x  1) 2  x  1  1  x  1  x  1  x  1 1  x  x  1 1  1  1  0
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Notes
1. In the previous example, it is not necessary to use a table to determine the roots.
Using Definition 8.2 in Section 8, we can write for the polynomial p( x)  x 2  x  1
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2
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p( x )  ( x )  x  1  x  x  1  0
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2
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2
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2
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2
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p( x  1)  ( x  1)  x  1  1  ( x  1)  x  1  1  x  2 x  1  x  2  x  0  1  x  0  x  x  1  0
2
2. To use notation we are more familiar with, sometimes we will make a variable
substitution so that the root of a polynomial in K  F [ x] / p( x) will not appear as a
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congruence class. For example, if we set   x in the previous example, we can say
that for   F [ x] / p( x) , that for p( x)  x 2  x  1 ,
p( )   2    1  0 .
The next theorem generalizes the ideas given in the last example.
Theorem 9.2: Let F be a field and p (x ) be an irreducible polynomial in F [x ] . Then
F [ x] / p ( x) is an extension field of F that contains a root of p (x ) .
Proof: Let p( x)  an x n  an1 x n1    a1 x  a0 be irreducible in F [x ] , where ai  F .
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Since F [ x] / p ( x) is an extension field of F, each ai  F[ x] /( p( x)) . Setting x  x in
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F [ x] / p ( x) , x is a root since
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__ n
__ n 1
p( x )  a n x  a n1 x
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   a1 x  a0 
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a n x n  a n1 x n1    a1 x  a0  p( x)
 0.
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Example 2: For the irreducible polynomial p( x)  x 2  1 in R[ x ] , what is a root of
p (x ) in the extension field R[ x] /( x 2  1) . Describe the elements in R[ x] /( x 2  1) .
Solution:
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Corollary 9.3: Let F be a field and f (x) a non-constant polynomial in F [x ] . Then there
is an extension field K of F that contains a root of f (x) .
Proof: By Theorem 6.6 in Section, the polynomial f (x) is guaranteed to have an
irreducible factor p (x ) in F [x ] , that is f ( x)  t ( x) p( x) for some t ( x)  F [ x] . By
Theorem 9.2, K  F [ x] /( p( x)) is an extension field of F that contains a root, say
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  x  F[ x] /( p( x)) , that is root of p (x ) . That is, p ( )  0 . Thus,  is also a root of
f (x ) since
f ( )  t ( ) p( )  t ( )(0)  0 .
Hence,  is a root of f ( x ) in K  F [ x] /( p( x)) .
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Example 3: Show that Z 3 [ x] /( x 2  2 x  2) is a field. Find the roots for the polynomial
p( x)  x 2  2 x  2 in the extension field Z 3 [ x] /( x 2  2 x  2) .
Proof:
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Exercises
1. Determine which of the rings are fields.
a. Q[ x] /( x 2  7) .
b. Q[ x] /( x 2  9) .
c. R[ x] /( x 2  x  1) .
d
Z 2 [ x] /( x 3  x  1) .
e. Z 3 [ x] /( x 3  1) .
f.
Z 3 [ x] /( x 3  2 x  1) .
2. Show that Z 3 [ x] /( 2 x 2  2 x  1) is a field. Find the roots for the polynomial
p( x)  2 x 2  2 x  1 in the extension field Z 3 [ x] /( 2 x 2  2 x  1) .
3. Show that Z 2 [ x] /( x 3  x  1) is a field. Show that all three roots of the polynomial
p( x)  x 3  x  1 are contained in the extension field Z 2 [ x] /( x 3  x  1) . Hint: The
addition and multiplication tables found in Exercise 2c in Section 8 for
Z 2 [ x] /( x 3  x  1) will be helpful.
4. Show that Z 2 [ x] /( x 3  1) is not a field. Find the roots of the polynomial p( x)  x 3  1 .
5. If p (x ) is an irreducible quadratic polynomial in F [x ] , show that F [ x] /( p ( x))
contains all of the roots of p (x ) .
6. If f ( x)  F [ x] has degree n, prove that there exists and extension field K of F such that
f ( x)  c0 ( x  c1 )( x  c2 )( x  cn ) for some (not necessarily distinct) ci  K . That is,
K contains all of the roots of f (x ) .