1.3 Fermat's and Euler's Theorem:
* Fermat's Theorem :
For any field, the nonzero elements form a group under the field
multiplication.
In particular, for ℤ𝑝 , the elements 1, 2, 3, … , 𝑝 − 1 form group of order
𝑝 − 1 under multiplication modulo 𝑝.
Theorem 3.1: (Little Theorem of Fermat): If 𝑎 ∈ ℤ and 𝑝 is a prime
not dividing 𝑎, then 𝑝 divides 𝑎𝑝−1 − 1 , that is, 𝑎𝑝−1 ≡ 1 (𝑚𝑜𝑑 𝑝) for
𝑎 ≢ 0 (𝑚𝑜𝑑 𝑝) .
Corollary 3.2: If 𝑎 ∈ ℤ , then 𝑎𝑝 ≡ 𝑎 (𝑚𝑜𝑑 𝑝) for any prime 𝑝.
Proof: Follows from Theorem 3.1 if 𝑎 ≢ 0 (𝑚𝑜𝑑 𝑝) . If 𝑎 ≡ 0 (𝑚𝑜𝑑 𝑝),
then both side reduce to 0 𝑚𝑜𝑑𝑢𝑙𝑜 𝑝.
Example: Compute the remainder of 𝟖𝟏𝟎𝟑 when divided by 13 using
Fermat's Theorem.
8103 ≡ (812 )8 (87 ) ≡ (18 )(87 ) ≡ 87 ≡ (−5)7 ≡ (25)3 (−5) ≡
(−1)3 (−5) ≡ 5 (𝑚𝑜𝑑 13) .
* Euler's Generalization :
Theorem 3.3: The set 𝐺𝑛 of nonzero elements of ℤ𝑛 that are not 0
divisors forms a group under multiplications modulo 𝑛.
Proof: First we must show that 𝐺𝑛 is closed under multiplication modulo
𝑛. Let 𝑎, 𝑏 ∈ 𝐺𝑛 . If 𝑎𝑏 ∉ 𝐺𝑛 , then there would exists 𝑐 ≠ 0 in ℤ𝑛 such
that (𝑎𝑏)𝑐 = 0 . So (𝑎𝑏)𝑐 = 0 implies that 𝑎(𝑏𝑐) = 0 . Since 𝑏 ∈ 𝐺𝑛
1
and 𝑐 ≠ 0 , we have 𝑏𝑐 ≠ 0 by definition of 𝐺𝑛 . But then 𝑎(𝑏𝑐) = 0
would imply that 𝑎 ∉ 𝐺𝑛 contrary to assumption.
We now show that 𝐺𝑛 is a group. Multiplication modulo 𝑛 is associative,
and 1 ∈ 𝐺𝑛 . It remains to show that for 𝑎 ∈ 𝐺𝑛 , there is 𝑏 ∈ 𝐺𝑛 such
that 𝑎𝑏 = 1. Let
1, 𝑎1 , … , 𝑎𝑟
be the elements of 𝐺𝑛 . The elements
𝑎1, 𝑎 𝑎1 , … , 𝑎𝑎𝑟
are all different, for if 𝑎𝑎𝑖 = 𝑎𝑎𝑗 , then 𝑎(𝑎𝑖 − 𝑎𝑗 ) = 0, and since 𝑎 ∈ 𝐺𝑛
and thus is not a divisor of 0, we must have 𝑎𝑖 − 𝑎𝑗 = 0 . Therefore, by
counting, we find that either 𝑎1 = 1, or some 𝑎𝑎𝑖 must be 1, so 𝑎 has a
multiplicative inverse.
Definition: Let 𝑛 be a positive integer. 𝜑(𝑛) is the number of positive
integers less than or equal to 𝑛 and relatively prime to 𝑛. Note that
𝜑(1) = 1.
Example: Let 𝑛 = 12. The positive integers less than or equal to 12 and
relatively prime to 12 are 1, 5, 7, and 11. So 𝜑(12) = 4.
Note: By the definition of the characteristic of a ring, 𝜑(𝑛) is the number
of nonzero elements of ℤ𝑛 that are not divisors of 0. This function
𝜑: ℤ+ ⟶ ℤ+ is the Euler phi-function.
Theorem 3.4: ( Euler's Theorem): If 𝑎 is an integer relatively prime to
𝑛, then 𝑎𝜑(𝑛) − 1 is divisible by 𝑛, that is, 𝑎𝜑(𝑛) ≡ 1 (𝑚𝑜𝑑 𝑛).
Proof: If 𝑎 is relatively prime to 𝑛, then the coset 𝑎 + 𝑛ℤ of 𝑛ℤ
containing 𝑎 contains an integer 𝑏 < 𝑛 and relatively prime to 𝑛.
As the multiplication of these cosets by multiplication modulo 𝑛 of
representatives is well-defined, we have
2
𝑎𝜑(𝑛) ≡ 𝑏 𝜑(𝑛) (𝑚𝑜𝑑 𝑛).
But by Theorem 2.1 and Theorem 3.3, 𝑏 can be viewed as an element of
the multiplication group 𝐺𝑛 of order 𝜑(𝑛) consisting of the 𝜑(𝑛)
elements of ℤ𝑛 relatively prime to 𝑛. Thus, 𝑏 𝜑(𝑛) ≡ 1 (𝑚𝑜𝑑 𝑛) and our
theorem follow.
Example: Compute 𝟕𝟒 (𝒎𝒐𝒅 𝟏𝟐) using Euler's Theorem.
Let 𝑛 = 12. From the previous example, we have 𝜑(12) = 4. Thus, for
any integer relatively prime to 12, we have 𝑎4 ≡ 1 (𝑚𝑜𝑑 12). For
example, with 𝑎 = 7, we have
74 = (49)2 = 2,401 = 12(200) + 1
So,
74 ≡ 1 (𝑚𝑜𝑑 12).
The easy way to compute 74 (𝑚𝑜𝑑 12) without using Euler's Theorem is
to compute it in ℤ12 . In ℤ12 , we have 7=-5 so
(74 ) = (72 )2 = ((−5)2 )2 = (25)2 = (1)2 = 1 .
3
1.4 The Field of Quotients of an Integral Domain:
* The Construction:
The steps we need to enlarge an integral domain 𝐷 to a field of quotients
𝐹 are as follow:
1. Define what the elements of 𝐹 are to be.
2. Define the binary operations of addition and multiplication on 𝐹.
3. Check all the field axioms to show that 𝐹 is a field under these
operations.
4. Show that 𝐹 can be viewed as containing 𝐷 as an integral
subdomain.
Step 1: Let 𝐷 be an integral domain, and form the Cartesian product
𝐷 × 𝐷 = {(𝑎, 𝑏)⁄ 𝑎 , 𝑏 ∈ 𝐷}.
The ordered pair (𝑎, 𝑏) is representing a formal quotient 𝑎⁄𝑏 . So, if 𝐷 =
2
ℤ , the pair (2, 3) represent the number . The pair (2, 0) represent no
3
element of ℚ . Thus we need to cut the set 𝐷 × 𝐷 down. Let 𝑆 be the
subset of 𝐷 × 𝐷 given by
𝑆 = {(𝑎, 𝑏)⁄𝑎 , 𝑏 ∈ 𝐷, 𝑏 ≠ 0}.
Now 𝑆 is still not going to be a field because, with 𝐷 = ℤ , different pairs
of integers such as (2, 3) and (4, 6) can represent the same rational
number. So, we will define when two elements of 𝑆 represent the same
element of 𝐹, that is, when two elements of 𝑆 are equivalent.
Definition: Two elements (𝑎, 𝑏) and (𝑐, 𝑑) in 𝑆 are equivalent, denoted
by (𝑎, 𝑏)~(𝑐, 𝑑) iff 𝑎𝑑 = 𝑏𝑐 .
Lemma 4.1: The relation between elements of the set 𝑆 as just described
is an equivalence relation.
Proof: We need to check the three properties of an equivalence relation.
4
Reflexive. (𝑎, 𝑏)~(𝑎, 𝑏) since 𝑎𝑏 = 𝑏𝑎 , for multiplication in 𝐷 is
commutative.
Symmetric. If (𝑎, 𝑏)~(𝑐, 𝑑) , then 𝑎𝑑 = 𝑏𝑐 . Since multiplication in 𝐷 is
commutative, thus 𝑐𝑏 = 𝑑𝑎 . Therefore, (𝑐, 𝑑) = (𝑎, 𝑏) .
Transitive. If (𝑎, 𝑏)~(𝑐, 𝑑) and (𝑐, 𝑑)~(𝑟, 𝑠), then 𝑎𝑑 = 𝑏𝑐 and 𝑐𝑠 = 𝑑𝑟.
As 𝑑 ≠ 0 and 𝐷 is an integral domain, so the cancellation laws hold.
Thus, 𝑎𝑑𝑐𝑠 = 𝑏𝑐𝑑𝑟 and so 𝑎𝑠 = 𝑏𝑟. So (𝑎, 𝑏)~(𝑟, 𝑠) .
Notation: ~ gives a partition of 𝑆 into equivalence classes. So let [(𝑎, 𝑏)]
be the equivalence class of (𝑎, 𝑏) in 𝑆 under the relation ~ . Step 1 is
finished by defining 𝐹 to be the set of all equivalence classes [(𝑎, 𝑏)] for
(𝑎, 𝑏) ∈ 𝑆 .
Step 2: We need the following lemma to define addition and
multiplication in 𝐹. Note that if 𝐷 = ℤ and [(𝑎, 𝑏)] is viewed as (𝑎⁄𝑏) ∈
ℚ , these definition applied to ℚ give the usual operation.
Lemma 4.2: For [(𝑎, 𝑏)] and [(𝑐, 𝑑)] in , the equations
[(𝑎, 𝑏)] + [(𝑐, 𝑑)] = [(𝑎𝑑 + 𝑏𝑐, 𝑏𝑑)]
and [(𝑎, 𝑏)][(𝑐, 𝑑)] = [(𝑎𝑐, 𝑏𝑑)]
give well-defined operations of addition and multiplication on 𝐹 .
Proof: If [(𝑎, 𝑏)] , [(𝑐, 𝑑)] ∈ 𝐹 , then (𝑎, 𝑏), (𝑐, 𝑑) ∈ 𝑆 . So 𝑏 ≠ 0 and ≠
0 . As 𝐷 is an integral domain, 𝑏𝑑 ≠ 0. Hence,
(𝑎𝑑 + 𝑏𝑐, 𝑏𝑑), (𝑎𝑐, 𝑏𝑑) ∈ 𝑆.
Now we need to show that addition and multiplication are well-defined.
So we must show that if different representatives in 𝑆 are chosen, the
same element of 𝐹 will result. Suppose that (𝑎1 , 𝑏1 ) ∈ [(𝑎, 𝑏)] and
(𝑐1 , 𝑑1 ) ∈ [(𝑐, 𝑑)] . We must show that
(𝑎1 𝑑1 + 𝑏1 𝑐1 , 𝑏1 𝑑1 ) ∈ [(𝑎𝑑 + 𝑏𝑐, 𝑏𝑑)]
and (𝑎1 𝑐1 , 𝑏1 𝑑1 ) ∈ [(𝑎𝑐, 𝑏𝑑)] .
5
(𝑎1 , 𝑏1 ) ∈ [(𝑎, 𝑏)] means that (𝑎1 , 𝑏1 )~(𝑎, 𝑏) , that is
𝑎1 𝑏 = 𝑏1 𝑎 . ⟶ (1)
Similarly, (𝑐1 , 𝑑1 ) ∈ [(𝑐, 𝑑)] implies that
𝑐1 𝑑 = 𝑑1 𝑐 . ⟶ (2)
By multiplying (1) by 𝑑1 𝑑 and (2) by 𝑏1 𝑏 and adding the resulting
equation, we have
𝑎1 𝑏𝑑1 𝑑 + 𝑐1 𝑑𝑏1 𝑏 = 𝑏1 𝑎𝑑1 𝑑 + 𝑑1 𝑐𝑏1 𝑏 .
Using axioms of 𝐷, we have
(𝑎1 𝑑1 + 𝑏1 𝑐1 )𝑏𝑑 = 𝑏1 𝑑1 (𝑎𝑑 + 𝑏𝑐) ,
(𝑎1 𝑑1 + 𝑏1 𝑐1 , 𝑏1 𝑑1 )~(𝑎𝑑 + 𝑏𝑐, 𝑏𝑑) .
Thus
(𝑎1 𝑑1 + 𝑏1 𝑐1 , 𝑏1 𝑑1 ) ∈ [(𝑎𝑑 + 𝑏𝑐, 𝑏𝑑)] .
Foe multiplication, we multiply (1) and (2). So
𝑎1 𝑏𝑐1 𝑑 = 𝑏1 𝑎𝑑1 𝑐 ,
Using axioms of 𝐷 we have
𝑎1 𝑐1 𝑏𝑑 = 𝑏1 𝑑1 𝑎𝑐 .
Thus
(𝑎1 𝑐1 , 𝑏1 𝑑1 )~(𝑎𝑐, 𝑏𝑑) .
Hence
(𝑎1 𝑐1 , 𝑏1 𝑑1 ) ∈ [(𝑎𝑐, 𝑏𝑑)] .
Step 3: Now we will check the axioms of the field 𝐹 .
1. Addition in 𝐹 is commutative.
Proof : By definition [(𝑎, 𝑏)] + [(𝑐, 𝑑)] = [(𝑎𝑑 + 𝑏𝑐, 𝑏𝑑)] and
[(𝑐, 𝑑)] + [(𝑎, 𝑏)] = [(𝑐𝑏 + 𝑑𝑎, 𝑑𝑏)] . We must show that (𝑎𝑑 +
𝑏𝑐, 𝑏𝑑)~(𝑐𝑏 + 𝑑𝑎, 𝑑𝑏) . Using the axioms of 𝐷 , we have 𝑎𝑑 +
𝑏𝑐 = 𝑐𝑏 + 𝑑𝑎 and 𝑏𝑑 = 𝑑𝑏 .
2. Addition is associative.
3. [(0, 1)] is an identity element for addition in 𝐹.
4. [(−𝑎, 𝑏)] is an additive inverse for [(𝑎, 𝑏)] in 𝐹.
5. Multiplication in 𝐹 is associative.
6. Multiplication in 𝐹 is commutative.
6
7. The distributive laws hold in 𝐹.
8. [(1, 1)] is a multiplicative identity element in 𝐹.
9. If [(𝑎, 𝑏)] ∈ 𝐹 is not the additive identity element, then 𝑎 ≠ 0 in 𝐷
and [(𝑏, 𝑎)] is a multiplicative inverse for [(𝑎, 𝑏)] .
Proof : Let [(𝑎, 𝑏)] ∈ 𝐹. If 𝑎 = 0, then
𝑎1 = 𝑏0 = 0 ,
So
(𝑎, 𝑏)~(0, 1) ,
That is [(𝑎, 𝑏)] = [(0, 1)] . But [(0, 1)] is the additive identity by
part 3. Thus if [(𝑎, 𝑏)] is not the additive identity in 𝐹, we have
𝑎 ≠ 0 . Now [(𝑎, 𝑏)][(𝑏, 𝑎)] = [(𝑎𝑏, 𝑏𝑎)] . But in 𝐷 we have
𝑎𝑏 = 𝑏𝑎 , so (𝑎𝑏)1 = (𝑏𝑎)1 , and (𝑎𝑏, 𝑏𝑎)~(1, 1) .
Thus [(𝑎, 𝑏)][(𝑏, 𝑎)] = [(1, 1)] , and [(1, 1)] is the multiplicative
identity by part 8.
Step 4: We show that 𝐹 contains 𝐷 . Thus, we must show that there
is an isomorphism 𝑖 of 𝐷 with a subdomain 𝐹. The next lemma
gives us this isomorphism. We will inject 𝐷 into 𝐹.
Lemma 4.3: The map 𝑖: 𝐷 ⟶ 𝐹 given by 𝑖(𝑎) = [(𝑎, 1)] is an
isomorphism of 𝐷 with a subring of 𝐹.
Proof: Let 𝑎, 𝑏 ∈ 𝐷 . Then 𝑖(𝑎 + 𝑏) = [(𝑎 + 𝑏, 1)] .
Also, 𝑖(𝑎) + 𝑖(𝑏) = [(𝑎, 1)] + [(𝑏, 1)] = [(𝑎1 + 1𝑏, 1)] =
[(𝑎 + 𝑏, 1)] .
So, 𝑖(𝑎 + 𝑏) = 𝑖(𝑎) + 𝑖(𝑏) .
Also, 𝑖(𝑎𝑏) = [(𝑎𝑏, 1)],
While 𝑖(𝑎)𝑖(𝑏) = [(𝑎, 1)][(𝑏, 1)] = [(𝑎𝑏, 1)] .
So, 𝑖(𝑎𝑏) = 𝑖(𝑎)𝑖(𝑏) .
Now we show that 𝑖 is one to one map. If 𝑖(𝑎) = 𝑖(𝑏) , then
[(𝑎, 1)] = [(𝑏, 1)] , so (𝑎, 1)~(𝑏, 1) gives 𝑎1 = 1𝑏 , that is 𝑎 = 𝑏.
Thus 𝑖 is an isomorphism of 𝐷 with 𝑖[𝐷] and 𝑖[𝐷] is a subdomain
of 𝐹.
Theorem 4.4: Any integral domain 𝐷 can be enlarge to (or
embedded in) a field 𝐹 such that every element of 𝐹 can be
expressed as a quotient of two elements of . (Such a field 𝐹 is a
field of quotients of 𝐷.)
7
Proof: Since [(𝑎, 𝑏)] = [(𝑎, 1)][(1, 𝑏)] = [(𝑎, 1)]⁄[(𝑏, 1] =
𝑖(𝑎)𝑖(𝑏) which holds in 𝐹
8
1.5 Rings of Polynomial:
If 𝑅 is a ring, then 𝑅[𝑥] is the set of all polynomials with coefficients in
the ring 𝑅. 𝑥 in this case is called an indeterminate rather than variable.
Definition: Let 𝑅 be a ring. A polynomial 𝑓(𝑥) with coefficient in 𝑅 is
an infinite formal sum
∞
∑ 𝑎𝑖 𝑥 𝑖 = 𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 + ⋯
𝑖=0
where 𝑎𝑖 ∈ 𝑅 and 𝑎𝑖 = 0 for all but a finite number of values of 𝑖. The 𝑎𝑖
are coefficient of 𝑓(𝑥). If for some 𝑖 ≥ 0 it is true that 𝑎𝑖 ≠ 0, the largest
such value of 𝑖 is the degree of 𝑓(𝑥). If all 𝑎𝑖 = 0, then the degree of
𝑓(𝑥) is undefined.
Note: * If
𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 + ⋯
and
𝑔(𝑥) = 𝑏0 + 𝑏1 𝑥 + ⋯ + 𝑏𝑛 𝑥 𝑛 + ⋯,
then for polynomial addition, we have
𝑓(𝑥) + 𝑔(𝑥) = 𝑐1 + 𝑐1 𝑥 + ⋯ + 𝑐𝑛 𝑥 𝑛 + ⋯ 𝑤ℎ𝑒𝑟𝑒 𝑐𝑛 = 𝑎𝑛 + 𝑏𝑛 ,
and for polynomial multiplication, we have
𝑛
𝑓(𝑥)𝑔(𝑥) = 𝑑0 + 𝑑1 𝑥 + ⋯ + 𝑑𝑛 𝑥 𝑛 + ⋯ 𝑤ℎ𝑒𝑟𝑒 𝑑𝑛 = ∑ 𝑎𝑖 𝑏𝑛−𝑖
𝑖=0
* 𝑐𝑖 and 𝑑𝑖 are 0 for all but a finite number of values of 𝑖.
* If 𝑅 is not commutative, then
𝑛
∑ 𝑎𝑖 𝑏𝑛−𝑖
𝑖=0
need not equal
9
𝑛
∑ 𝑏𝑖 𝑎𝑛−𝑖
𝑖=0
Theorem 5.1: The set 𝑅[𝑥] of all polynomials in an indeterminate
𝑥 with coefficients in a ring 𝑅 is a ring under polynomial addition
and multiplication. If 𝑅 is commutative, then so is 𝑅[𝑥], and if 𝑅
has unity 1 ≠ 0 ,then 1 is also unity for 𝑅[𝑥].
Example: ℤ[𝑥] is the ring polynomials in 𝑥 with integral coefficients.
ℚ[𝑥] is the ring polynomials in 𝑥 with rational coefficients.
Example: Find the sum and the product of the given polynomials in
in the given polynomial ring:
𝒇(𝒙) = 𝒙 + 𝟏 ,
𝒈(𝒙) = 𝒙 + 𝟏
𝒊𝒏 ℤ𝟐 [𝒙].
𝑓(𝑥) + 𝑔(𝑥) = (𝑥 + 1) + (𝑥 + 1) = 2𝑥 + 2 = 0𝑥 + 0 = 0.
𝑓(𝑥)𝑔(𝑥) = (𝑥 + 1)(𝑥 + 1) = 𝑥 2 + 2𝑥 + 1 = 𝑥 2 + 1.
𝒇(𝒙) = 𝟒𝒙 − 𝟓,
𝒈(𝒙) = 𝟐𝒙𝟐 − 𝟒𝒙 + 𝟐
𝒊𝒏 ℤ𝟖 [𝒙].
In ℤ8 [𝑥], we have
𝑓(𝑥) = 4𝑥 + 3 ,
𝑔(𝑥) = 2𝑥 2 + 4𝑥 + 2
𝑓(𝑥) + 𝑔(𝑥) = 2𝑥 2 + 8𝑥 + 5 = 2𝑥 2 + 0𝑥 + 5 = 2𝑥 2 + 5
𝑓(𝑥)𝑔(𝑥) = 8𝑥 3 + 16𝑥 2 + 8𝑥 + 6𝑥 2 + 12𝑥 + 6 = 6𝑥 2 + 4𝑥 + 6
* The Evaluation homomorphism:
Theorem 5.2: (The Evaluation Homomorphism for Field Theory):
Let 𝐹 be a subfield of a field 𝐸, let 𝛼 be any element of 𝐸, and let 𝑥 be an
indeterminate. The map 𝜙𝛼 : 𝑓[𝑥] ⟶ 𝐸 defined by
𝜙𝛼 (𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ) = 𝑎0 + 𝑎1 𝛼 + ⋯ + 𝑎𝑛 𝛼 𝑛
10
for 𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ) ∈ 𝐹[𝑥] is a homomorphism of 𝐹[𝑥] into 𝐸.
Also, 𝜙𝛼 (𝑥) = 𝛼, and 𝜙𝛼 maps 𝐹 isomorphically by the identity map,
that is, 𝜙𝛼 (𝑎) = 𝑎 for 𝑎 ∈ 𝐹. The homomorphism 𝜙𝛼 is evaluation at 𝛼.
Proof: Insertion or deletion of terms 0𝑥 𝑖 does not affect the value of
𝜙𝛼 (𝑓(𝑥)). If
𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 , 𝑔(𝑥) = 𝑏0 + 𝑏1 𝑥 + ⋯ +
𝑏𝑚 𝑥 𝑚 , 𝑎𝑛𝑑 ℎ(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) = 𝑐0 + 𝑐1 𝑥 + ⋯ + 𝑐𝑟 𝑥 𝑟 ,
then 𝜙𝛼 (𝑓(𝑥) + 𝑔(𝑥)) = 𝜙𝛼 (ℎ(𝑥)) = 𝑐0 + 𝑐1 𝛼 + ⋯ + 𝑐𝑟 𝛼 𝑟 ,
and 𝜙𝛼 (𝑓(𝑥)) + 𝜙𝛼 (𝑔(𝑥)) = (𝑎0 + 𝑎1 𝛼 + ⋯ + 𝑎𝑛 𝛼 𝑛 ) + (𝑏0 + 𝑏1 𝛼 +
⋯ + 𝑏𝑚 𝛼 𝑚 ).
By definition of polynomial addition, we have 𝑐𝑖 = 𝑎𝑖 + 𝑏𝑖 .
Hence, 𝜙𝛼 (𝑓(𝑥) + 𝑔(𝑥)) = 𝜙𝛼 (𝑓(𝑥)) + 𝜙𝛼 (𝑔(𝑥)) .
For multiplication, we have
𝑓(𝑥)𝑔(𝑥) = 𝑑0 + 𝑑1 𝑥 + ⋯ + 𝑑𝑠 𝑥 𝑠 .
So 𝜙𝛼 (𝑓(𝑥)𝑔(𝑥)) = 𝑑0 + 𝑑1 𝛼 + ⋯ + 𝑑𝑠 𝛼 𝑠 ,
Also, [𝜙𝛼 (𝑓(𝑥))][𝜙𝛼 (𝑔(𝑥))] = (𝑎0 + 𝑎1 𝛼 + ⋯ + 𝑎𝑛 𝛼 𝑛 )(𝑏0 + 𝑏1 𝛼 +
⋯ + 𝑏𝑚 𝛼 𝑚 ).
By definition of polynomial multiplication, we have
𝑗
𝑑𝑗 = ∑ 𝑎𝑖 𝑏𝑗−𝑖
𝑖=0
Thus, 𝜙𝛼 (𝑓(𝑥)𝑔(𝑥)) = [𝜙𝛼 (𝑓(𝑥))][𝜙𝛼 (𝑔(𝑥))].
Hence 𝜙𝛼 is a homomorphism.
By the definition of 𝜙𝛼 applied to a constant polynomial 𝑎 ∈ 𝐹[𝑥], where
𝑎 ∈ 𝐹, 𝜙𝛼 (𝑎) = 𝑎. Thus, 𝜙𝛼 maps 𝐹 isomorphically by the identity map.
Again by definition of 𝜙𝛼 , we have 𝜙𝛼 (𝑥) = 𝜙𝛼 (1𝑥) = 1𝛼 = 𝛼.
Example: Let 𝑭 = ℚ, 𝒂𝒏𝒅 𝑬 = ℝ. Compute for the indicated
evaluation 𝝓𝟎 .
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𝜙0 : ℚ[𝑥] ⟶ ℝ.
𝜙0 (𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ) = 𝑎0 + 𝑎1 0 + ⋯ + 𝑎𝑛 0𝑛 = 𝑎0 .
Thus, every polynomial is mapped onto its constant term.
Example: Let 𝑭 = ℚ, 𝒂𝒏𝒅 𝑬 = ℝ. Compute for the indicated
evaluation 𝝓𝟐 (𝒙𝟐 + 𝒙 − 𝟔).
𝜙2 : ℚ[𝑥] ⟶ ℝ.
𝜙2 (𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ) = 𝑎0 + 𝑎1 2 + ⋯ + 𝑎𝑛 2𝑛 .
𝜙2 (𝑥 2 + 𝑥 − 6) = 22 + 2 − 6 = 0 .
Thus, 𝑥 2 + 𝑥 − 6 is in the kernel 𝑁 of 𝜙2 .
Also, 𝑥 2 + 𝑥 − 6 = (𝑥 − 2)(𝑥 + 3) because 𝜙2 (𝑥 2 + 𝑥 − 6) = 0 is that
𝜙2 (𝑥 − 2) = 2 − 2 = 0.
Example: Let 𝑭 = ℚ, 𝒂𝒏𝒅 𝑬 = ℂ. Compute for the indicated
evaluation 𝝓𝒊 (𝒙𝟐 + 𝟏).
𝜙𝑖 : ℚ[𝑥] ⟶ ℂ
𝜙𝑖 (𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ) = 𝑎0 + 𝑎1 𝑖 + ⋯ + 𝑎𝑛 𝑖 𝑛 𝑎𝑛𝑑 𝜙𝑖 (𝑥) = 𝑖.
𝜙𝑖 (𝑥 2 + 1) = 𝑖 2 + 1 = 0.
So 𝑥 2 + 1 is in the kernel 𝑁 of 𝜙𝑖 .
Definition: Let 𝐹 be a subfield of a field 𝐸, and let 𝛼 be an element of 𝐸.
Let 𝑓(𝑥) = 𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 be in 𝐹[𝑥] and let 𝜙𝛼 : 𝐹[𝑥] ⟶ 𝐸 be
the evaluation homomorphism of Theorem 5.2. Let 𝑓(𝛼) denote
𝜙𝛼 (𝑓(𝑥)) = 𝑎0 + 𝑎1 𝛼 + ⋯ + 𝑎𝑛 𝛼 𝑛 .
If 𝑓(𝛼) = 0, then 𝛼 is a zero of 𝑓(𝑥).
Example: Find all real numbers 𝒓 such that 𝒓𝟐 + 𝒓 − 𝟔 = 𝟎.
Let 𝐹 = ℚ and 𝐸 = ℝ, we need to find all 𝛼 ∈ ℝ such that
𝜙𝛼 (𝑥 2 + 𝑥 − 6) = 0,
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that is, finding all zeros of 𝑥 2 + 𝑥 − 6 in ℝ.
{𝛼 ∈ ℝ | 𝜙𝛼 (𝑥 2 + 𝑥 − 6) = 0} = {𝑟 ∈ ℝ | 𝑟 2 + 𝑟 − 6 = 0} = {2, −3}.
Theorem 5.3: The polynomial 𝑥 2 − 2 has no zeros in the rational
numbers. Thus √2 is not a rational number.
Proof: Suppose that 𝑚⁄𝑛 for 𝑚, 𝑛 ∈ ℤ is a rational number such that
(𝑚⁄𝑛)2 = 2. Let the fraction 𝑚⁄𝑛 be in the lowest terms with
gcd(𝑚⁄𝑛) = 1. Then 𝑚2 = 2𝑛2 , where both 𝑚2 and 2𝑛2 are integers.
Since 2 is a factor of 2𝑛2 , 2 must be one of the factors of 𝑚2 . But as a
square, 𝑚2 has as factors of 𝑚 repeated twice. Thus 𝑚2 must have two
factors 2. Hence, 2𝑛2 must have two factors 2, so 𝑛2 must have 2 as
factor, and consequently 𝑛 has 2 as a factor. Thus 𝑚 and 𝑛 must be
divisible by 2 which is contradicts that gcd(𝑚, 𝑛) = 1. So 2 ≠ (𝑚⁄𝑛)2
for any 𝑚, 𝑛 ∈ ℤ.
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1.6 Factorization of Polynomials Over a Field:
Let 𝐸 and 𝐹 be fields, with 𝐹 ≤ 𝐸. Suppose that 𝑓(𝑥) ∈ 𝐹[𝑥] factors in
𝐹[𝑥], so that 𝑓(𝑥) = 𝑔(𝑥)ℎ(𝑥) ∈ 𝐹[𝑥] and let 𝛼 ∈ 𝐸. Now for the
evaluation homomorphism 𝜙𝛼 , we have
𝑓(𝛼) = 𝜙𝛼 (𝑓(𝑥)) = 𝜙𝛼 (𝑔(𝑥)ℎ(𝑥)) = 𝜙𝛼 (𝑔(𝑥))𝜙𝛼 (ℎ(𝑥))
= 𝑔(𝛼)ℎ(𝛼).
Thus, if 𝛼 ∈ 𝐸, then 𝑓(𝛼) = 0 if and only if either 𝑔(𝛼) = 0 𝑜𝑟 ℎ(𝛼) =
0.
*The Division Algorithm in F[x]:
Theorem 6.1 (Division algorithm for F[x]):
Let 𝑓(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎0
and 𝑔(𝑥) = 𝑏𝑚 𝑥 𝑚 + 𝑏𝑚−1 𝑥 𝑚−1 + ⋯ + 𝑏0
be two elements of 𝐹[𝑥], with 𝑎𝑛 and 𝑏𝑚 both nonzero elements of 𝐹 and
𝑚 > 0. Then there are unique polynomials 𝑞(𝑥) and 𝑟(𝑥) in 𝐹[𝑥] such
that 𝑓(𝑥) = 𝑔(𝑥)𝑞(𝑥) + 𝑟(𝑥), where either 𝑟(𝑥) = 0 or the degree of
𝑟(𝑥) is less than the degree 𝑚 of 𝑔(𝑥).
Example: Find 𝒒(𝒙) and 𝒓(𝒙) as described by the division algorithm
theorem.
𝒇(𝒙) = 𝒙𝟒 − 𝟑𝒙𝟑 + 𝟐𝒙𝟐 + 𝟒𝒙 − 𝟏 ,
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𝒈(𝒙) = 𝒙𝟐 − 𝟐𝒙 + 𝟑 𝒊𝒏 ℤ𝟓 [𝒙].
Corollary 6.2: (Factor Theorem):
An element 𝑎 ∈ 𝐹 is a zero of 𝑓(𝑥) ∈ 𝐹[𝑥] if and only if 𝑥 − 𝑎 is a factor
of 𝑓(𝑥) in 𝐹[𝑥].
Proof: Suppose that for 𝑎 ∈ 𝐹 we have 𝑓(𝑎) = 0. By Theorem 6.1. there
exist 𝑞(𝑥), 𝑟(𝑥) ∈ 𝐹[𝑥] such that
𝑓(𝑥) = (𝑥 − 𝑎)𝑞(𝑥) + 𝑐 .
Applying the evaluation homomorphism, 𝜙𝛼 : 𝐹[𝑥] ⟶ 𝐹 of Theorem 5.2,
we find
0 = 𝑓(𝑎) = 0𝑞(𝑎) + 𝑐,
so it must be that 𝑐 = 0. Then 𝑓(𝑥) = (𝑥 − 𝑎)𝑞(𝑥), so 𝑥 − 𝑎 is a factor
of 𝑓(𝑥).
Conversely, if 𝑥 − 𝑎 is a factor of 𝑓(𝑥) in 𝐹[𝑥], where 𝑎 ∈ 𝐹, then
applying the evaluation homomorphism 𝜙𝛼 to 𝑓(𝑥) = (𝑥 − 𝑎)𝑞(𝑥), we
have 𝑓(𝑎) = 0𝑞(𝑎) = 0.
Example: The polynomial (𝒙𝟒 + 𝟑𝒙𝟑 + 𝟐𝒙 + 𝟒) can be factored into
linear factors in ℤ𝟓 [𝒙]. Find this factorization.
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Corollary 6.3: A nonzero polynomial 𝑓(𝑥) ∈ 𝐹[𝑥] of degree 𝑛 can have
at most 𝑛 zeros in a field 𝐹.
Proof: If 𝑎1 ∈ 𝐹 is a zero of 𝑓(𝑥), then
𝑓(𝑥) = (𝑥 − 𝑎1 )𝑞1 (𝑥),
where the degree of 𝑞1 (𝑥) is 𝑛 − 1. A zero 𝑎2 ∈ 𝐹 of 𝑞1 (𝑥) then result in
a factorization
𝑓(𝑥) = (𝑥 − 𝑎1 )(𝑥 − 𝑎2 )𝑞2 (𝑥).
Continuing this process, we have
𝑓(𝑥) = (𝑥 − 𝑎1 ) … (𝑥 − 𝑎𝑟 )𝑞𝑟 (𝑥),
where 𝑞𝑟 (𝑥) has no further zeros in 𝐹. Since the degree of 𝑓(𝑥) is 𝑛, at
most 𝑛 factors (𝑥 − 𝑎𝑖 ) can appear on the right-hand side of the
preceding equation, so 𝑟 ≤ 𝑛. Also, if 𝑏 ≠ 𝑎𝑖 for 𝑖 = 1, … , 𝑟 and 𝑏 ∈ 𝐹,
then
𝑓(𝑏) = (𝑏 − 𝑎1 ) … (𝑏 − 𝑎𝑟 )𝑞𝑟 (𝑏) ≠ 0,
since 𝐹 has no divisors of 0 and none of 𝑏 − 𝑎𝑖 or 𝑞𝑟 (𝑏) are 0 by
construction. Hence, the 𝑎𝑖 for 𝑖 = 1, . . , 𝑟 ≤ 𝑛 are all the zeros in 𝐹 of
𝑓(𝑥).
Corollary 6.4: The multiplicative group of all nonzero elements of a
finite field is cyclic.
* Irreducible Polynomials:
Definition: A nonconstant polynomial 𝑓(𝑥) ∈ 𝐹[𝑥] is irreducible over 𝐹
or is an irreducible polynomial in 𝐹[𝑥] if 𝑓(𝑥) cannot be expressed as a
product 𝑔(𝑥)ℎ(𝑥) of two polynomials 𝑔(𝑥)
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