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Section 7: Modular Arithmetic in F[x] and Congruence Classes
HW p. 11 # 1-15 at the end of the notes
In this section, we look at modular arithmetic concerning polynomials in F [x ] , where F
is a field. We will see that many the properties are similar to how modular arithmetic
works over the integers.
Modular Arithmetic in F[x]
Definition 7.1: Let F be a field and f ( x ), g ( x )  F [ x ] with p (x ) non-zero. Then we say
that f (x) is congruent to g (x ) modulo p (x ) , written as f ( x)  g ( x) (mod p( x)) ,
provided that p( x) | ( f ( x)  g ( x)) .
Example 1: In Q[x ] , determine whether x 2  2  2 x  1 (mod x  3)
Solution:
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Example 2: In Z 5 [ x] , determine whether 3x 4  4 x  3x 3  x 2  4 (mod x 2  2)
Solution:
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Example 3: In Z 7 [ x] , determine whether 3x 4  4 x  3x 3  x 2  4 (mod x 2  2)
Solution: In Z 7 [ x] , we want to determine if ( x 2  2) | ((3x 4  4 x)  (3x 3  x 2  4)) or
when simplifying, if ( x 2  2) | (3x 4  4 x 3  6 x 2  4 x  4) . When performing long
division, we see that q( x)  3x 2  4 x and r ( x)  3x  4 . Since the remainder is non-zero,
( x 2  2) | (3x 4  4 x 3  6 x 2  4 x  4) . Hence, 3x 4  4 x  3x 3  x 2  4 (mod x 2  2) .
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Theorem 7.2: Let F be a field and p (x ) be a non-zero polynomial in F [x ] . Then the
relation of congruence modulo p (x ) is an equivalence relation, that is, it satisfies the
following three properties:
(1) Reflexive: f ( x)  f ( x) (mod p( x)) for all f ( x)  F [ x] .
(2) Symmetric: If f ( x)  g ( x) (mod p( x)) , then g ( x)  f ( x) (mod p( x)) .
(3) Transitive: : If f ( x)  g ( x) (mod p( x)) and g ( x)  h( x) (mod p( x)) , then
f ( x)  h( x) (mod p( x)) .
Proof: (1) Reflexive:
(2) Symmetric: Exercise.
(3) Transitive: If f ( x)  g ( x) (mod p( x)) , then p( x) | ( f ( x)  g ( x)) or, by divisibility,
f ( x)  g ( x)  k ( x) p( x) for some k ( x)  F [ x] . Hence
f ( x)  g ( x)  k ( x) p( x)
*
Similarly, if g ( x)  h( x) (mod p( x)) , we can say for some l ( x)  F [ x] that
g ( x )  h( x )  l ( x ) p ( x )
**
Using g (x ) in ** to substitute into * gives
f ( x )  h( x )  l ( x ) p ( x )  k ( x ) p ( x )
or
f ( x)  h( x)  (l ( x)  k ( x)) p( x) .
Hence, p( x) | ( f ( x)  h( x)) and hence by the definition of congruence, we can conclude
that
f ( x)  h( x) (mod p( x)) .
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Theorem 7.3: Let F be a field and p (x ) be a non-zero polynomial in F [x ] . Then if
f ( x)  g ( x) (mod p( x)) and s ( x)  t ( x) (mod p( x)) . Then
(1). f ( x)  s( x)  g ( x)  t ( x) (mod p( x)) .
(2). f ( x) s( x)  g ( x)t ( x) (mod p( x)) .
Proof: (1)
Proof: (2) Exercise.
Fact: If f ( x)  g ( x) (mod p( x)) , the both f (x) and g (x ) have the same remainder r (x )
when divided by the polynomial p (x ) .
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Example 4: In Z 5 [ x] , show that 3x 4  4 x  3x 3  x 2  4 (mod x 2  2) by computing the
integer remainder of 3x 4  4 x and 3x3  x 2  4 when divided by x 2  2 .
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Definition 7.4: Let F be a field and r ( x), p ( x)  F [ x] be polynomials with p ( x)  0 . The
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congruence class (or residue class) of r (x ) modulo p (x ) is denoted by r (x) and is made
up of all polynomials in F [x ] that are congruent to r (x ) modulo p (x ) , that is,
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r ( x)   f ( x)  F[ x], f ( x)  r ( x) (mod p( x))   f ( x)  F[ x], f ( x)  q( x) p( x)  r ( x).
Notes:
1. If deg( r ( x))  deg( p( x)) , then r (x ) would be the polynomial remainder of f (x)
divided by p (x ) .
2. If f ( x)  r ( x) (mod p( x)) , then by definition of congruence, p( x) | ( f ( x)  r ( x)) or
f ( x)  r ( x)  q( x) p( x) for some q ( x)  F [ x] . Hence,
f ( x)  q ( x) p ( x)  r ( x)
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Example 5: In Examples 3 and 4, we showed that 3x 4  4 x  3x 3  x 2  4 (mod x 2  2)
in Z 5 [ x] . Define the congruence class for this case.
Solution: In Example 4, when computing (3x 4  4 x)  ( x 2  2) in Z 5 [ x] , we obtained a
quotient of 3x 2  4 and remainder of 4x  2 . Hence, the division algorithm, we have
(3x 4  4 x)  (3x 2  4)( x 2  2)  4 x  2 .
Similarly, when computing (3x 3  x 2  4)  ( x 2  2) , we obtained a quotient of 3x  1
and a remainder of 4x  2 . Using the division algorithm, this gives
(3x 3  x 2  4)  (3x  1)( x 2  2)  4 x  2
In general, using note 2 given on the last page, in Z 5 [ x] , any polynomial in this
congruence class can be defined by
f ( x)  q( x)( x 2  2)  4 x  2
where f ( x), q( x)  Z 5 [ x] . For example, if q( x)  x 3 , then
f ( x)  x3 ( x 2  2)  4 x  2  x5  2 x3  4 x  2
is a member of the congruence class. Hence, we can say that
x 5  2 x 3  4 x  2  3x 4  4 x  3x 3  x 2  4  4 x  2 (mod x 2  2) .
Hence, using the remainder of the division 4x  2 , we can define these polynomials as
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elements of the set 4 x  2 , where
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

4 x  2  q( x)( x 2  2)  4 x  2, q( x)  Z 5 [ x]
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For the equation f ( x)  q( x) p( x)  r ( x) , where r (x ) is the remainder of the division of
the polynomials f (x) and p (x ) in F [x ] , we can solve for r (x ) and obtain
r ( x)  q ( x) p( x)  f ( x) .
Hence, in Example 5 for instance, we can write the remainder 4x  2 as
4 x  2  (3x 2  4)( x 2  2)  (3x 4  4 x)  (2 x 2  1)( x 2  2)  (3x 4  4 x) .
In general, we could construct other elements g ( x)  F [ x] of the congruence class
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4 x  2 by forming
g ( x)  k ( x)( x 2  2)  (3x 4  4 x)
where k ( x)  F [ x] . For example, the polynomial 3x 3  x 2  4 can be found using the
formula
3x 3  x 2  4  (2 x 2  3x  2)( x 2  2)  (3x 4  4 x)
What this says that any polynomial in a congruence class be used to construct the other
elements in its congruence class. That is, for the polynomials in Z 5 [ x] from Example 5
where we saw
x 5  2 x 3  4 x  2  3x 4  4 x  3x 3  x 2  4  4 x  2 (mod x 2  2) ,
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3
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3
2
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we can say that x  2 x  4 x  2  3x  4 x  3x  x  4  4 x  2 . The fact that
each of these polynomials can be used to represent the same set is constituted by the
following theorem.
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Theorem 7.5: f ( x)  r ( x) (mod p( x)) if and only if f ( x)  r ( x) .
Proof:  Recall that to show that the sets A and B are equal, that is A  B , if A  B
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and B  A. Assume f ( x)  r ( x) (mod p( x)) and suppose that g ( x)  f ( x) . By
Definition 7.4, this says g ( x)  f ( x) (mod p( x)) . Since g ( x)  f ( x) (mod p( x)) and
f ( x)  r ( x) (mod p( x)) , the transitive property of Theorem 7.2 says that
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g ( x)  r ( x) (mod p( x)) . Hence, g ( x)  r ( x) . Thus, f ( x)  r ( x) . Now
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assume h( x)  r ( x) . Then h( x)  r ( x) (mod p( x)) . Since we are given
f ( x)  r ( x) (mod p( x)) , symmetric property of Theorem 7.2 says that
r ( x)  f ( x) (mod p( x)) . Hence, if h( x)  r ( x) (mod p ( x)) and r ( x)  f ( x) (mod p( x)) ,
the transitive property of Theorem 7.2 says that h( x)  f ( x) (mod p( x)) . Hence
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h( x)  f ( x) and r ( x)  f ( x) . Thus, f ( x)  r ( x) .
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 Assume f ( x)  r ( x) . Now, we know f ( x)  f ( x) since the reflexive property of
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Theorem 7.2 says that f ( x)  f ( x) (mod p( x)) . But since f ( x)  r ( x) , this says that
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f ( x)  r ( x) . Thus, by Definition 7.4, f ( x)  r ( x) (mod p( x)) .
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We can now state the following corollary:
Corollary 7.6: Two congruence classes modulo p (x ) are either disjoint (there
intersection is the empty set) or they are identical.
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Proof: Let f (x) and r (x) be two congruence classes modulo p (x ) . If f (x) and r (x)
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are disjoint, than we are done. Suppose they are not and let g ( x)  f ( x) r ( x) . Then
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g ( x)  f ( x) and g ( x)  r ( x) . Then g ( x)  f ( x) (mod p( x)) and g ( x)  r ( x) (mod p( x)) .
Using the symmetric and transitive property of Theorem 7.2, we can say that
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f ( x)  r ( x) (mod p( x)) . Hence, by Theorem 7.5, f ( x)  r ( x) .
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Recall that for the integers Z that the congruence x  b (mod m) partition Z into distinct
congruence classes of the form
_____ 

0, 1, 2, 3,  m  1


Let p( x)  F [ x] be a polynomial of degree m. The division algorithm says that when
dividing any other polynomial in F [x ] by p (x ) , the remainder r (x ) must by zero or of
degree less than m. This leads to this corollary.
Corollary 7.7: Let F be a field and p (x ) a polynomial of degree m in F [x ] . Then each
congruence class modulo p (x ) can be represented by the zero polynomial or some
polynomial of degree less than m in F [x ] . The congruence classes of each of these
polynomials are distinct.
Notation: We denote the set of all congruence classes modulo p (x ) by F [ x] /( p ( x)) ,
which is a notational analogue of the integers modulo m given by Z m .
Example 6: Consider congruence modulo x 2  1 in R [ x ] . Describe the set of congruence
classes R[ x] /( x 2  1).
Solution:
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Example 7: Consider congruence modulo x 2  2 x  2 in Z 3 [ x] . Describe and list the set
of congruence classes for Z 3 [ x] /( x 2  2 x  2).
Solution:
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Example 8: If p( x)  Z m [ x] is a polynomial of degree n , determine the number of distinct
congruence classes in Z m [ x] /( p( x)).
Solution:
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Exercises
1. For the following polynomials, determine whether f ( x)  g ( x) (mod p( x)) by
determining whether p( x) | ( f ( x)  g ( x))
a. 2 x 5  x 4  2 x 3  2 x 2  x  3  x  2 (mod x 2  1) in Q[x ] .
b. 4 x 4  3x 3  x 2  3x  3x 3  x  4 (mod x 2  x  1) in Z 5 [ x] .
c. 4 x 4  3x 3  x 2  3x  3x 3  x  1 (mod x 2  x  1) in Z 5 [ x] .
d. x 5  5x 4  3x 3  4 x 2  6 x  1  6 x 4  x 3  x  3 (mod x 3  4 x  1) in Z 7 [ x] .
2. For the problems in Exercise 1, determine whether f ( x)  g ( x) (mod p( x)) by
determining if f (x) and g (x ) have the same remainder when divided by p (x ) .
3. Prove that for the polynomials f ( x), g ( x), p ( x)  F [ x] , where F is a field, that if
f ( x)  g ( x) (mod p( x)) , then g ( x)  f ( x) (mod p( x)) .
4. Prove that for the polynomials f ( x), g ( x), p ( x)  F [ x] , where F is a field, that if
f ( x)  g ( x) (mod p( x)) and s ( x)  t ( x) (mod p( x)) , then f ( x) s( x)  g ( x)t ( x) (mod p( x)) .
5. Prove that f ( x)  g ( x) (mod p( x)) for the polynomials f ( x), g ( x), p ( x)  F [ x] , where
F is a field, if and only if f (x) and g (x ) have the same remainder when divided by
p (x ) .
6. If p (x ) is a non-zero constant polynomial in F [x ] , show that any two polynomials in
F [x ] are congruent modulo p (x ) .
7. Determine the number and list the congruence classes in Z 3 [ x] /( x 2  1) .
8. Determine the number and list the congruence classes in Z 2 [ x] /( x 3  x  1) .
9. Determine the number and list the congruence classes in Z 2 [ x] /( x 4  x  1) .
10. How many distinct congruence classes are there in Z 5 [ x] /( x 3  2 x  1) ?
11. How many congruence classes are there in R[ x] /( x 4  1) ? List 10 of these
congruence classes.
12. If gcd( p( x), k ( x))  1 and f ( x)k ( x)  g ( x)k ( x) (mod p( x)) , prove that
f ( x)  g ( x) (mod p( x)) .
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13. If p (x ) is irreducible in F [x ] , where F is a field, and f ( x) g ( x)  0 (mod p( x)) ,
prove that f ( x)  0 (mod p( x)) or g ( x)  0 (mod p( x)) .
14. If p (x ) is not irreducible in F [x ] , where F is a field, prove that there exist
f ( x), g ( x)  F [ x] such that f ( x)  0 (mod p ( x)) and g ( x)  0 (mod p( x)) but
f ( x) g ( x)  0 (mod p( x)) .
15. If f ( x), p ( x)  F [ x] and gcd( f ( x), p( x))  1 , where F is a field, prove that there
exists a polynomial u ( x)  F [ x] such that f ( x)u ( x)  1 (mod p( x)) .