1
Section 8: Congruence Class Arithmetic
HW p. 13 # 1-7 at the end of the notes
In this section, we examine how to combine polynomials in different congruence classes.
We start with the following theorem
Modular Arithmetic in F[x]
Theorem 8.1: Let F be a field and p (x ) a non-zero and non-constant polynomial in
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F [x ] . If f ( x)  g ( x) and h( x)  k ( x) in F [ x] /( p ( x)) , then
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f ( x)  h( x)  g ( x)  k ( x) and f ( x)h( x)  g ( x)k ( x)
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Proof: Since f ( x)  g ( x) and h( x)  k ( x) , Theorem 7.5 from Section 7 says that
f ( x)  g ( x) (mod p( x)) and h( x)  k ( x) (mod p( x)) . Using Theorem 7.3 from Section 7,
we can say that
f ( x)  h( x)  g ( x)  k ( x) (mod p( x)) and f ( x)h( x)  g ( x)k ( x) (mod p( x)) .
Using Theorem 7.5 again, the result follows.
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Example 1: In Z 5 [ x] , it can be shown that 3x 4  4 x  3x 3  x 2  4 (mod x 2  2)
( 3x 4  4 x and 3x 3  x 2  4 give an integer remainder of 4x  2 when divided by x 2  2 ).
Also, 3x 5  3x 3  x 2  x  1  3x 4  3x 2  2 x  3 (mod x 2  2) ( 3x5  3x3  x 2  x  1 and
3x 4  3x 2  2x  3 give an integer remainder of 2x  4 when divided by x 2  2 ). Hence,
we can say
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4
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3
2
3x  4 x  3x  x  4
and
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5
3
2
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4
2
3x  3x  x  x  1  3x  3x  2 x  3 .
Theorem 8.1 says that
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4
5
3
2
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3
2
4
2
(3x  4 x )  (3x  3x  x  x  1  (3x  x  4)  (3 x  3x  2 x  3)
or simplifying both sides modulo 5 says that
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5
4
3
2
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4
3
2
3x  3x  3x  x  1  3x  3x  4 x  2 x  2
This can be verified that when divided by x 2  2 that both 3x5  3x 4  3x3  x 2  1 and
3x 4  3x3  4 x 2  2 x  2 give a remainder of x  1 .
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Theorem 8.1 allows us to perform congruence class arithmetic and get the same result no
matter which polynomial we use to represent the congruence class. We can now give the
following definition to define this arithmetic.
Definition 8.2: Let F be a field and p (x ) a non-constant polynomial in F [x ] . We define
addition and multiplication in F [ x] /( p ( x)) by
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f ( x) g ( x)  f ( x)  g ( x) and f ( x) g ( x)  f ( x) g ( x)
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Example 2: Compute 2 x  1  3x  4 and 2 x  1  3x  4 in Z 5 [ x] /( x 2  2) .
Solution:
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Example 3: Determine the rules for addition and multiplication of the congruence classes
for Q[ x] /( x 2  5) .
4
5
Example 4: Write an addition and multiplication table for the congruence classes of
Z 2 [ x] /( x 2  x  1) .
Solution:
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Theorem 8.3: Let F be a field and p (x ) be a non-constant polynomial in F [x ] . Then the
set F [ x] /( p ( x)) of congruence classes modulo p (x ) is a commutative ring with identity.
Furthermore, F [ x] /( p ( x)) contains a subring F * that is isomorphic to F.
Proof: To show F [ x] /( p ( x)) is a ring, let f ( x), g ( x), h( x)  F [ x] and consider the
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elements f ( x), g ( x) , h( x)  F[ x] /( p( x)) . Note that F [ x] /( p ( x)) is closed under addition
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and multiplication since f ( x)  g ( x)  f ( x)  g ( x)  F[ x] /( p( x)) and
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f ( x) g ( x)  f ( x) g ( x)  F[ x] /( p( x)) . We next show that F [ x] /( p ( x)) is a ring. Recall
that F [x ] itself is a commutative ring with identity. We next verify the ring properties for
F [ x] /( p ( x)) .
1. F [ x] /( p ( x)) is an abelian group under addition since
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a. F [ x] /( p ( x)) is associative under addition since for all f ( x), g ( x) , h( x)  F[ x] /( p( x))
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( f ( x) g ( x) )  h( x)  ( f ( x)  g ( x))  h( x)  f ( x)  g ( x)  h( x)  f ( x) ( g ( x)  h( x))  f ( x) ( g ( x)  h( x) )
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b. 0 is identity element in F [ x] /( p ( x)) since for any f ( x)  F[ x] /( p( x)) ,
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0  f ( x )  0  f ( x )  f ( x )  f ( x)  0  f ( x )  0 .
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c. For f ( x)  F[ x] /( p( x)) ,  f (x) is the additive inverse in F [ x] /( p ( x)) since
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 f ( x) f ( x)   f ( x)  f ( x)  0  f ( x)  ( f ( x))  f ( x)   f ( x) .
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d. F [ x] /( p ( x)) is commutative under addition since for f ( x), g ( x)  F[ x] /( p( x))
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f ( x )  g ( x )  f ( x )  g ( x )  g ( x )  f ( x )  g ( x)  f ( x )
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2. F [ x] /( p ( x)) is associative under multiplication since for all f ( x), g ( x) , h( x)  F[ x] /( p( x))
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( f ( x)  g ( x) )  h( x)  ( f ( x)  g ( x) )  h( x)  f ( x)  g ( x)  h( x)  f ( x)  ( g ( x)  h( x) )  f ( x)  ( g ( x) h( x) )
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3. The distributive laws, since for f ( x), g ( x) , h( x)  F[ x] /( p( x))
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f ( x) ( g ( x)  h( x) )  f ( x) ( g ( x)  h( x) )  f ( x)( g ( x)  h( x))  f ( x) g ( x)  f ( x)h( x)  f ( x) g ( x)  f ( x)h( x)  f ( x) g ( x)  f ( x) h( x)
Similarly,
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( f ( x)  g ( x) ) h( x)  f ( x) h( x)  g ( x) h( x) .
4. F [ x] /( p ( x)) is a commutative ring since F [x ] is commutative and for all
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f ( x), g ( x)  F[ x] /( p( x)) ,
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f ( x ) g ( x)  f ( x ) g ( x )  g ( x ) f ( x)  g ( x ) f ( x ) .
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5.
1 is the multiplicative identity in F [ x] /( p ( x)) since F [x ] has the identity
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polynomial of 1 and for all f ( x)  F[ x] /( p( x)) ,
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f ( x )  1  f ( x)  1  f ( x )  1  f ( x )  1  f ( x ) .
A subring is a subset S of a ring R that is itself a ring under the operations of addition and
multiplication. To prove that a subset of a ring is a subring, one only need to prove that
(i) the subset S is closed under addition and multiplication, that is, if a, b  S , then
a  b  S and ab  S , (ii) the additive identity 0  S , and (iii) each element a  S has
an additive inverse  a  S where a  (a)  (a)  a  0 .
The subset F * of F [ x] /( p ( x)) that is a subring of F [ x] /( p ( x)) is the set of all
congruence classes of all of the constant polynomials of F [ x] /( p ( x)) , that is
F *  {a | a  F}
This can be seen since for all a, b  F , a  b  F and ab  F since F is a field (and
hence a ring) and the following conditions are hence satisfield.
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(i).
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For a , b  F * , a  b  a  b  F * , a  b  a  b  F * .
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(ii). 0  F and hence 0  F * .
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(iii). For all a  F * ,  a  F * and a   a  a  (a)  0  (a)  a   a  a .
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When two rings are isomorphic, they have the same structure in the sense that addition
and multiplication are the same except with the elements of each ring suitably relabeled.
Two rings R and S are isomorphic if there is a function f : R  S such that
f is one-to-one, that is, if f ( x)  f ( y ) for x, y  R , then x  y .
(i)
(ii) f is onto, that is for any element s  S , there exists an x  R where f ( x)  s .
(iii) f is a homomorphism, that is, for all x, y  R , f ( x  y )  f ( x)  f ( y ) and
f ( xy)  f ( x) f ( y ) .
To show F and F * are isomorphic, define the function  : F  F * by
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 (a)  a .
Then,
(i)
 is one-to-one, since for a, b  F , if
 (a)   (b) ,
then
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a  b.
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If a  b , then a  b (mod p( x)) (Theorem 7.5 in Section 7). Hence, this says that
p ( x) | (a  b) . However, p (x ) is assumed to be non-constant and hence
deg p( x)  1 . Since a  b  F is a constant, p ( x) | (a  b) only if a  b  0 . Thus
a  b and  is one-to-one.
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(ii).  is onto by its definition, that is if a  F * , then we have a  F where  (a)  a .
(iii)  is a homomorphism for both addition and multiplication, since if a, b  F , then
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 (a  b)  a  b  a  b   (a)   (b) .
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 (ab)  ab  a b   (a) (b)
Thus, the set of constant polynomials F and the sent of constant polynomial congruences
F * are isomorphic.
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Hence, up to isomorphism, we can say that F [ x] /( p ( x)) is a commutative ring with
identity that contains the field F as a subset. By replacing F * with its isomorphic copy of
F , we have the following theorem.
Theorem 8.4: Let F be a field and p (x ) and a non-constant polynomial in F [x ] . Then
F [ x] /( p ( x)) is a commutative ring with identity that contains F.
Recall that a field F is a commutative ring with identity where every non-zero element in
F has a multiplicative inverse (is a unit). That is, if a  F and a  0 , then there is a
a 1  F where aa 1  1 . Depending on the polynomial p (x ) used for the ring
F [ x] /( p ( x)) , F [ x] /( p ( x)) can also be a field.
Example 5: Consider the multiplication table for the field Z 2 [ x] /( x 2  x  1) .

0
1
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0
1
0
0
0
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x
0
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x 1
0
0
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1
x
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x 1
x 1
x
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x 1
x
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x 1
0
1
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1
x
Find the multiplicative inverses for the non-zero elements of the table.
Solution: Looking at the table, we can see that all of the non-zero elements have
multiplicative inverses. Summarizing, we see that
The multiplicative inverse of 1 is 1 since 1  1  1 .
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The multiplicative inverse of x is x  1 since x  x  1  1 .
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The multiplicative inverse of x  1 is x since x  1  x  1 .
Since all of the non-zero elements of Z 2 [ x] /( x 2  x  1) have multiplicative inverses, we
see that Z 2 [ x] /( x 2  x  1) is a field that contains the field Z 2 .
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Example 6: Consider the multiplication table for the field Z 2 [ x] /( x 3  1) .

0
1
x
x 1
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2
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2
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2
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2
0
0
0
0
0
0
0
0
0
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1
0
1
x
x 1
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2
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2
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2
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2
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2
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2
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x x
1
x 1
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2
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2
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x
0
x
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x
x
x
x 1
x 1
x x
x x
x 1
x  x 1
x  x 1
x  x 1
x 1
x 1
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2
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2
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2
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2
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0
x x
x 1
0
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2
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2
x 1
x
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2
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1
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2
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0
x x
x 1
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2
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2
x 1
x 1
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2
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2
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x 1
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2
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0
x x
x 1
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2
x 1
0
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2
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2
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2
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0
x 1
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2
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2
x x
0
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2
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2
0
0
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2
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x
x x
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2
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x  x 1 0
x x
x
x x
x 1
x x
x 1
x 1
x  x 1 x2  x 1
x 1
x  x 1
0
x 1
x  x 1
x  x 1
Find the multiplicative inverses for the non-zero elements of the table.
Solution: Looking at the table, we see that only some of the elements in Z 2 [ x] /( x 3  1)
have multiplicative inverses:
The multiplicative inverse of 1 is 1 since 1  1  1 .
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2
The multiplicative inverse of x is x
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2
The multiplicative inverse of x
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2
since x  x  1 .
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2
is x since x  x  1 .
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2
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2
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2
The non-zero elements x  1 , x  1 , x  x , and x  x  1 all have no multiplicative
inverse. Thus, Z 2 [ x] /( x 3  1) is not a field. In fact, Z 2 [ x] /( x 3  1) is not even and
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integral domain – the elements x  1 , x 2  1 , x 2  x , and x 2  x  1 are all zero
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2
divisors. For example, x  1  x  x  1  0 .
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Example 7: Show that Q[ x] /( x 2  5) is a field.
Solution:
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Example 8: Show that Q[ x] /( x 2  1) is not a field.
Solution: To be a field, every non-zero element must have a multiplicative inverse. We
will demonstrate an element where the multiplicative inverse does not exist. Consider the
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element x  1 . To find the inverse of this element, we look for a linear congruence of the
form [cx  d ] where
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x  1 cx  d  1 .
This gives
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cx 2  (d  c) x  d  1
We can reduce cx 2  (d  c) x  d to find its equivalent congruence representation by
finding the remainder of the division of cx 2  (d  c) x  d by x 2  1 . When this division is
performed, we obtain a quotient of q( x)  c and remainder of r ( x)  (d  c) x  (c  d ) .
Hence, this gives
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(d  c) x  (c  d )  1
These congruences classes can only be equal if the polynomials that form them are equal.
That is, we want
( d  c ) x  (c  d )  1
or
( d  c ) x  (c  d )  0 x  1 .
Polynomials can be equation only if their corresponding coefficients are equal. This gives
the system of equations
d  c  0 and c  d  1.
Solving the first equation for d gives d  c . Substituting this result into the second
equation gives
c c 1
or
0 1
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Hence, no solution exists and hence the element x  1 in Q[ x] /( x 2  1) has no
multiplicative inverse. Thus, Q[ x] /( x 2  1) is not a field.
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Exercises
1. Perform the following calculations for the following congruence class rings.
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a. 8 x  3  6 x  7 and 8 x  3  6 x  7 in R[ x] /( x 2  1) .
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b.
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x  1  x  2 and x  1  x  2 in Z 2 [ x] /( x 2  1) .
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c. 2 x  1  x  2 and 2 x  1  x  2 in Z 3 [ x] /( x 2  1) .
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d. 4 x  3  2 x  1 and 4 x  3  2 x  1 in Z 5 [ x] /( x 2  2) .
e.
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2
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2
x  x 1 x
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2
2
and x  x  1  x
in Z 2 [ x] /( x 3  x  1) .
2. Write out the addition and multiplication tables for the following congruence class
rings. In each case, is the ring a field? An integral domain?
a. Z 2 [ x] /( x 2  1) .
b. Z 3 [ x] /( x 2  1) .
c. Z 2 [ x] /( x 3  x  1) .
d. Z 5 [ x] /( x 2  1) .
e. Z 5 [ x] /( x 2  2) .
3. In the following problems, each element of the congruence-class ring can be written in
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the form ax  b . Determine the rules for addition and multiplication of congruence.
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classes. That is, if the product ax  b  cx  d is rx  s , describe how to find r and s
from a, b, c, and d.
a. R[ x] /( x 2  1) .
b. Q[ x] /( x 2  2)
c. Q[ x] /( x 2  3)
d. Q[ x] /( x 2 )
4. a. Show R[ x] /( x 2  1) is a field by verifying that every non-zero congruence class
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ax  b is a unit. Hint: Show that the inverse of ax  b is cx  d , where
b
a
and d  2
.
c 2
2
a  b2
a b
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b. Find a polynomial cx  d  R[ x] where 5 x  4  cx  d  1 .
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5. For the congruence class ring Q[ x] /( x 2  2) .
a. Show Q[ x] /( x 2  2) is a field.
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b. Find a polynomial cx  d  Q[x] where 3x  3  cx  d  1 .
6. Show that Q[ x] /( x 2  4) is not a field.
7. Show that Q[ x] /( x 2 ) is not a field. Hint: Attempt to find the multiplicative inverse of
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the element x .