Download PRIME FACTORIZATION FOR THE INTEGER PERPLEXES

PRIME FACTORIZATION FOR THE INTEGER PERPLEXES
In 1843 Kummer submitted a paper with his proof of the Fermat Theorem.
His proof used factorization in the ring 𝐷𝑛 , which elements are numbers of
the form
𝑎0 + 𝑎1 𝜉 + ⋯ + 𝑎𝑛−2 𝜉 𝑛−2 ,
where 𝑎0 , 𝑎1 , … , 𝑎𝑛−2 are integers and
𝜉 = cos
2𝜋
𝑛
+ 𝑖 𝑠𝑖𝑛
2𝜋
𝑛
th
is the principal 𝑛 root of unity. Dirichlet found a serious gap in Kummer's
proof. He claimed that his assumption about the unique prime factorization
in 𝐷𝑛 was not valid. Trying to restore the unique prime factorization,
Kummer added to the numbers of the ring 𝐷𝑛 some imaginary numbers,
which he called "ideal" numbers. After that, with efforts of many
mathematicians, Fermat Theorem was proved for a wide range of prime
exponents. Still, this range did not cover all possible primes 𝑛 ≥ 3. Only the
development of new branches of modern algebraic number theory led to the
complete proof of Fermat Theorem by Andrew Wiles in 1994.
Meanwhile, there exists another algebraic structure, more convenient then
the ring 𝐷𝑛 , and which was missed by investigators of the Fermat Problem.
It is algebra 𝔸𝑛 , which could be defined as a set of all numbers of the form
𝑡 = 𝑥0 𝑗̂0 + 𝑥1 𝑗̂1 + ⋯ + 𝑥𝑛−1 𝑗̂𝑛−1 ,
where 𝑥0 , 𝑥1 , … , 𝑥𝑛−1 are integers and 𝑗̂𝑖 are bases elements in ℝn. The
products of the bases elements define the bilinear product of two elements in
𝔸𝑛 :
𝑗̂𝑖 ∙ 𝑗̂𝑘 = 𝑗̂𝑖+𝑘 (mod 𝑛) .
There exists a natural homomorphism 𝜑 between 𝔸𝑛 and the ring 𝐷𝑛 :
𝜑(𝑡) = (𝑥0 − 𝑥𝑛−1 ) + (𝑥1 − 𝑥𝑛−1 )𝜉 + ⋯ + (𝑥𝑛−2 − 𝑥𝑛−1 )𝜉 𝑛−2 ,
Still, prime factorizations of elements from 𝔸𝑛 are quite different from the
prime factorizations of their images in 𝐷𝑛 , since each element from 𝐷𝑛 has
infinitely many preimages in 𝔸𝑛 . Nonetheless, the prime factorization in 𝔸𝑛
was not studied yet.
Hypothesis 1 If 𝑛 is a prime number, then any element from 𝔸𝑛 which norm
is not divisible by 𝑛 could be uniquely written as a product of prime factors.
Moreover, it looks like the more strong assumption is correct.
Hypothesis 2 If 𝑛 is a natural number, then any element from 𝔸𝑛 which
norm is mutually prime with 𝑛 could be uniquely written as a product of
prime factors.
Does not matter if these hypotheses are correct for all 𝑛, or only for some of
them, the factorization in algebras 𝔸𝑛 deserved to be studied. The simplest
example of algebras 𝔸𝑛 is the algebra 𝔸2 of the integer perplex (or splitcomplex) numbers. This algebra is a two-dimensional vector space over the
ring of integers:
𝔸2 = {𝑡 = 𝑥 + 𝑦𝑗̂ | 𝑥, 𝑦 ∈ ℤ},
where 𝑗̂ is the imaginary unit, 𝑗̂2 = 1. Split-complex numbers were studied
by many researchers. In this paper we describe prime factorization of integer
perplexes and confirm our hypothesis for algebra 𝔸2 . We prove that:
(1) Every integer perplex with an odd norm can be written as a product of
prime perplexes, and any two its prime factorizations are associative.
(2) Every integer perplex with an even norm can be written as a product of
two integer perplexes. The norm of the first perplex is a power of base 2, and
the norm of the second is an odd number. Any two such factorizations are
associative.
(3) Prime factorization of integer perplexes which norm is a power of base 2
is not unique, but the uniqueness could be restored by extending 𝔸2 . All
possible prime factorization of these integer perplexes are described.
For two perplex numbers 𝑡1 = 𝑥1 + 𝑦1 𝑗̂ and 𝑡2 = 𝑥2 + 𝑦2 𝑗̂, their bilinear
product is determined by the following expression:
𝑡1 ∙ 𝑡2 = (𝑥1 𝑥2 + 𝑦1 𝑦2 ) + (𝑥1 𝑦2 + 𝑦1 𝑥2 )𝑗̂.
The function 𝑁(𝑡) = |𝑥 2 − 𝑦 2 | ≥ 0 defines the norm on the set of
perplexes: 𝑁(𝑡1 ∙ 𝑡2 ) = 𝑁(𝑡1 ) ∙ 𝑁(𝑡2 ). The element 𝑡 ∗ = (𝑥, −𝑦) is called
the conjugate to an element 𝑡 = (𝑥, 𝑦). Obviously, 𝑡 ∙ 𝑡 ∗ = 𝑥 2 − 𝑦 2 . The
perplex conjugate satisfies similar properties to usual complex conjugate. A
perplex 𝑢 ∊ 𝔸2 is called a unit element if there exists 𝑢−1 ∊ 𝔸2 such that
𝑢 ∙ 𝑢−1 = 𝑢−1 ∙ 𝑢 = 1. The only unit elements in 𝔸2 are ±1 and ±𝑗̂.
An element 𝑡 ∊ (𝔸2 )𝑁>1 = {𝑡 ∊ 𝔸2 | 𝑁(𝑡) > 1} is called composite if
there exist two elements 𝑡1 , 𝑡2 ∊ (𝔸2 )𝑁>1 such that 𝑡 = 𝑡1 ∙ 𝑡2 . These
elements are called divisors or factors of 𝑡. An element 𝑡 ∊ (𝔸2 )𝑁>1 that is
not composite is called a prime element. Obviously, if 𝑡 ∊ (𝔸2 )𝑁>1 is a
prime element then 𝑡 ∗ is also a prime element. Using induction by norm, it
is easy to prove that any element 𝑡 ∊ (𝔸2 )𝑁>1 can be written as a product of
prime elements.
An element 𝑡1 ∊ (𝔸2 )𝑁>1 is said to be divisible by an element 𝑡2 ∊
(𝔸2 )𝑁>1 with a remainder, if there exist q ∊ 𝔸2 and r ∊ 𝔸2 such that 𝑡1 =
𝑡2 ∙ 𝑞 + 𝑟 and 𝑟 = (0, 0) or 0 < 𝑁(𝑟) < 𝑁(𝑡2 ). It is not always possible to
divide one integer perplex by another with a remainder.
Example 𝑡1 = (3, 1) is not divisible by 𝑡2 = 2 with a remainder.
Proof Assume opposite: (3, 1) = 2 ∙ 𝑞 + 𝑟 and 𝑟 = (0, 0) or 0 < 𝑁(𝑟) < 4.
Obviously, 𝑟 ≠ (0, 0). Hence, 0 < 𝑁(𝑟) < 4. Let 𝑞 = (𝑎, 𝑏), where 𝑎, 𝑏 ∈
ℤ, then 𝑟 = (3 − 2𝑎, 1 − 2𝑏) and
𝑁(𝑟) = |(3 − 2𝑎)2 − (1 − 2𝑏)2 | = 4 ∙ |(1 − 𝑥 + 𝑦)(2 − 𝑥 − 𝑦)|.
Hence, 𝑁(𝑟) = 0 or 𝑁(𝑟) ≥ 4, which contradicts with 0 < 𝑁(𝑟) < 4.
Still, if the divisor has an odd norm, the division with a remainder is
possible.
Division with Remainder Theorem Let 𝑡1 , 𝑡2 ∊ (𝔸2 )𝑁>1 and 𝑁(𝑡2 ) ≡ 1
(mod 2). Then there exist q ∊ 𝔸2 and r ∊ 𝔸2 , such that 𝑡1 = 𝑡2 ∙ 𝑞 + 𝑟
and 𝑟 = 0 or 0 < 𝑁(𝑟) < 𝑁(𝑡2 ) and 𝑁(𝑟) ≡ 1 (mod 2).
Proof Let 𝜉 = 𝑡1 ∙ 𝑡2−1 = (𝑥, 𝑦), α = 𝑥 − ⌈𝑥 ⌉, β = 𝑦 − ⌈𝑦⌉.
WLOG, we can assume that 0 ≤ α ≤ β < 1 and we will consider several
cases.
Case 1 α = β = 0. In this case, 𝑞 = 𝜉 ∊ 𝔸, 𝑟 = 0.
1
Case 2 0 < α = β ≠ .
2
Let 𝑞 = (⌈𝑥 ⌉, ⌈𝑦⌉ + 1), 𝑟 = 𝑡1 − 𝑡2 ∙ 𝑞.
Then, 0 < 𝑁(𝑟) < 𝑁(𝑡2 ) and 𝑁(𝑟) ≡ 1 (mod 2).
Case 3 0 < α + β < 1 and α < β.
Let 𝑞1 = (⌈𝑥 ⌉, ⌈𝑦⌉), 𝑞2 = (⌈𝑥 ⌉, ⌈𝑦⌉ + 1), 𝑟𝑖 = 𝑡1 − 𝑡2 ∙ 𝑞𝑖 , 𝑖 = 1, 2.
Then, 0 < 𝑁(𝑟𝑖 ) < 𝑁(𝑡2 ) and 𝑁(𝑟1 ) ≡ 1 (mod 2) or 𝑁(𝑟2 ) ≡ 1 (mod 2),
since 𝑟1 − 𝑟2 = 𝑡2 ∙ 𝑗̂. Therefore, either 𝑞1 , 𝑟1 or 𝑞2 , 𝑟2 satisfy the
theorem.
Case 4 α + β = 1 and α < β.
Let 𝑞 = (⌈𝑥 ⌉, ⌈𝑦⌉), 𝑟 = 𝑡1 − 𝑡2 ∙ 𝑞.
Then, 0 < 𝑁(𝑟) < 𝑁(𝑡2 ) and 𝑁(𝑟) ≡ 1 (mod 2).
Case 5 1 < α + β < 2 and α < β.
Let 𝑞1 = (⌈𝑥 ⌉, ⌈𝑦⌉ + 1), 𝑞2 = (⌈𝑥 ⌉ + 1, ⌈𝑦⌉ + 1), 𝑟𝑖 = 𝑡1 − 𝑡2 ∙ 𝑞𝑖 ,
Then, 0 < 𝑁(𝑟𝑖 ) < 𝑁(𝑡2 ) and 𝑁(𝑟1 ) ≡ 1 (mod 2) or 𝑁(𝑟2 ) ≡ 1 (mod 2).
Therefore, either 𝑞1 , 𝑟1 or 𝑞2 , 𝑟2 satisfy the statement of the theorem.
An element δ ∊ (𝔸2 )𝑁≠0 is called the greatest common divisor (gcd) of the
elements 𝑡1 , 𝑡2 ∊ (𝔸2 )𝑁>1 , if δ is the divisor of both 𝑡1 and 𝑡2 , and δ is
divisible by any other common divisor of 𝑡1 and 𝑡2 . Not every pair of
elements from (𝔸2 )𝑁>1 have the greatest common divisor.
Example The greatest common divisor of 8 and (12, 4) does not exist.
Proof Assume that δ is the gcd of 8 and (12, 4). Then 4 and (3, 1) divide δ,
since 8 = (3, 1) ∙ (3, −1). Hence, δ = (3, 1) ∙ δ1 = (3𝑥 + 𝑦, 3𝑦 + 𝑥),
where δ1 = (𝑥, 𝑦), 𝑥, 𝑦 ∈ ℤ. Since 𝑁(δ) ≤ 𝑁(8) = 64 and 𝑁((3, 1)) = 8,
we get: 𝑁(δ1 ) ≤ 8. Since 4 divides δ, we get: 𝑥 + 𝑦 ⋮ 4 and 𝑥 − 𝑦 ⋮ 2.
Hence, 𝑁(δ1 ) ≥ 8. Thus, 𝑁(δ1 ) = 8 and δ1 = (3, 1) ∙ 𝑢 or δ1 = (3, −1) ∙
𝑢, where 𝑢 is a unit element. Therefore, δ = (10, 6) ∙ 𝑢 or δ = 8𝑢. In both
cases, δ does not divide (12, 4), which is a contradiction.
Still, if one of the two integer perplexes has an odd norm, their greatest
common divisor exists.
The GCD Theorem For any 𝑡1 , 𝑡2 ∊ (𝔸2 )𝑁>1 such that 𝑁(𝑡2 ) ≡ 1 (mod 2)
there exist their greatest common divisor δ ∊ (𝔸2 )𝑁≠0 and two integer
perplexes 𝜇, 𝜂 ∊ 𝔸2 such that 𝑡1 𝜇 + 𝑡2 𝜂 = δ.
Proof.
Let
𝑆 = {𝑡 = 𝑡1 𝜇 + 𝑡2 𝜂 | 𝜇, 𝜂 ∊ 𝔸, 𝑁(𝑡) ≢ 0 (mod 2)}.
Obviously, 𝑡2 ∈ 𝑆 and 𝑆 ≠ ∅. Let δ = 𝑡1 𝜇0 + 𝑡2 𝜂0 be an element of the set
S with the smallest norm. Then δ is a common divisor of 𝑡1 and 𝑡2 .
For example, let us prove that δ is a factor of 𝑡1 . Since 𝑁(δ) ≡
1 (mod 2), there exist integer bitenions 𝑞 and 𝑟 such that 𝑡1 = δ𝑞 + 𝑟
and 𝑟 = 0 or 0 < 𝑁(𝑟) < 𝑁(δ), 𝑁(𝑟) ≡ 1 (mod 2).
If 0 < 𝑁(𝑟) < 𝑁(δ) and 𝑁(𝑟) ≡ 1 (mod 2), we get:
𝑟 = 𝑡1 − δ𝑞 = 𝑡1 − (𝑡1 𝜇0 + 𝑡2 𝜂0 )𝑞
= 𝑡1 (1 − 𝜇0 𝑞 ) + 𝑡2 (−𝜂0 𝑞 ) ∊ 𝑆.
Thus, 𝑟 ∊ 𝑆 and 0 < 𝑁(𝑟) < 𝑁(δ) which contradicts that δ is an element
of the set 𝑆 with the smallest norm 𝑁. Therefore, 𝑟 = 0 and δ is a factor
of 𝑡1 . Similarly, we can prove that δ is a factor of 𝑡2 . Therefore, δ is a
common factor of 𝑡1 and 𝑡2 . Obviously, δ = 𝑡1 𝜇0 + 𝑡2 𝜂0 divides any
common factor of 𝑡1 and 𝑡2 . Thus, δ is the greatest common factor of 𝑡1
and 𝑡2 .
Corollary If a prime integer perplex 𝜋 with 𝑁(𝜋) ≡ 1 (mod 2) divides the
product of two integer perplexes 𝑡1 𝑡2 ∊ (𝔸2 )𝑁>1 then 𝜋 divides of one of
these two integer perplexes.
By induction, it is easy to prove that if a prime integer perplex 𝜋 with
𝑁(𝜋) ≡ 1 (mod 2) divides the product 𝑡1 ∙ 𝑡2 ∙ … ∙ 𝑡𝑚 of the integer
perplexes 𝑡1 , 𝑡2 , … , 𝑡𝑚 , then 𝜋 is the factor of one of these integer
perplexes.
Two
elements
𝑡1 , 𝑡2 ∊ (𝔸2 )𝑁≠0 = {𝑡 ∊ 𝔸2 | 𝑁(𝑡) ≠ 0} are
called
associative if there exists a unit element 𝑢 such that 𝑡1 = 𝑢 ∙ 𝑡2 . The
associating relation is the equivalence relation. Hence, (𝔸2 )𝑁≠0 is the union
of disjoint equivalence classes of mutually associative elements. Each class
consists of 4 elements. One of these classes is the set of all unit elements 𝑈.
Two prime factorizations 𝑡 = 𝜋1 ∙ 𝜋2 ∙ … ∙ 𝜋𝑚 and 𝑡 = 𝜋1′ ∙ 𝜋2′ ∙ … ∙ 𝜋𝑘′
of an element 𝑡 ∊ (𝔸2 )𝑁>1 are called associative if 𝑚 = 𝑘 and there exists a
permutation 𝜎 of the integers 1, 2, … 𝑚, such that 𝜋𝑖 and 𝜋𝜎′ (𝑖) are the
associative prime elements for all 𝑖 = 1, 2, … , 𝑚, i.e. 𝜋𝑖 = 𝑢𝑖 ∙ 𝜋𝜎′ (𝑖) , where
𝑢𝑖 ∊ 𝑈 and 𝑢1 ∙ 𝑢2 ∙ … ∙ 𝑢𝑚 = 1.
In algebra 𝔸2 , any odd prime natural number 𝑝 = 2𝑘 + 1 can be factored
onto two prime integer perplexes the following standard way:
𝑝 = 𝜋𝑝 ∙ 𝜋𝑝∗ = (𝑘 + 1, 𝑘) ∙ (𝑘 + 1, −𝑘).
Any prime factor of 𝑝 is associative with either 𝜋𝑝 or 𝜋𝑝∗ , and any prime
factorization of 𝑝 is associative with the standard one.
Lemma The norm 𝑁(𝜋) of any prime integer perplex 𝜋 is either an odd
prime natural number or a power of base 2: 𝑁(𝜋) = 2𝑛 , 𝑛 ≥ 2.
Proof We have two possibilities: either 𝑁(𝜋) =2𝑛 , where 𝑛 ≥ 2, or 𝑁(𝜋) is
divisible by a prime natural number p > 2. Assume that 𝑁(𝜋) is divisible by
a prime natural number p > 2. Let 𝜋1 be a prime divisor of 𝑝, then 𝑁(𝜋1 ) =
𝑝. Since 𝜋 ∙ 𝜋 ∗ = 𝑁(𝜋) is divisible by 𝜋1 , either 𝜋 or 𝜋 ∗ is divisible by 𝜋1 .
Assume that 𝜋 is divisible by 𝜋1 , then 𝜋 = 𝜋1 ∙ 𝑢. Since 𝜋 is a prime
perplex, 𝑢 is a unit element, and, hence, 𝑁(𝜋) = 𝑁(𝜋1 ) = 𝑝. Similarly, if
𝜋 ∗ is divisible by 𝜋1 , then 𝑁(𝜋 ∗ ) = 𝑁(𝜋1 ) = 𝑝. Therefore, 𝑁(𝜋) =
𝑁(𝜋 ∗ ) = 𝑝.
Obviously, 2 = (2, 0) is a prime perplex. Let 𝜈𝑙 = (2𝑙 + 1, 2𝑙 − 1), where
𝑙 ≥ 0, then 𝜈𝑙∗ = (2𝑙 + 1, 1 − 2𝑙 ). It is easy to show that 𝜈𝑙 and 𝜈𝑙∗ , where
𝑙 ≥ 1, are prime perplexes with 𝑁(𝜈𝑙 ) = 𝑁(𝜈𝑙∗ ) = 2𝑙 + 2 . Moreover, if 𝜋 is a
prime perplex with 𝑁(𝜋) = 2𝑙 + 2 , 𝑙 ≥ 1, then 𝜋 = 𝜈𝑙 ∙ 𝑢 or 𝜋 = 𝜈𝑙∗ ∙ 𝑢,
where 𝑢 is a unit element. Every integer perplex 𝑡 = (𝑥, 𝑦) ∊ (𝔸2 )𝑁>1 with
gcd (x, y) = 1 which norm is a power of base 2 and is a prime perplex.
The prime perplexes {𝜈𝑙 } and {𝜈𝑙∗ } satisfy the following equalities:
(1) 𝜈𝑝 ∙ 𝜈𝑞 = 2 ∙ 𝜈𝑝 + 𝑞
(2) 𝜈𝑝∗ ∙ 𝜈𝑞∗ = 2 ∙ 𝜈𝑝∗ + 𝑞
(3) 𝜈𝑝 ∙
𝜈𝑞∗
2𝑞 + 1 ∙ 𝜈𝑝 − 𝑞 , 𝑝 ≥ 𝑞,
= { 𝑝+ 1 ∗
2
∙ 𝜈𝑞 − 𝑝 , 𝑝 < 𝑞.
By induction it is easy to show that 𝜈𝑙 = 2 ∙ 𝛼 𝑙 ,
3 1
3
1
2 2
2
2
𝜈𝑙∗ = 2 ∙ 𝛽 𝑙 , where 𝛼 =
( , ) , 𝛽 = ( , − ), 𝛼 ∙ 𝛽 = 2, 𝑙 ≥ 0.
We say that an element 𝑡 ∊ (𝔸2 )𝑁>1 has a unique prime factorization if any
two prime factorizations of this element are associative. Not every element
𝑡 ∊ (𝔸2 )𝑁>1 has a unique prime factorization. For example, the following
two prime factorizations of 8 are not associative:
8 = (3, 1)(3, −1), 8 = 2 ∙ 2 ∙ 2.
Based on the properties described above, it is easy to get all possible nonassociative prime factorizations of the elements, which norm is a power with
base 2. For example,
(20, 12) = 4 ∙ (5, 3) = 2 ∙ (3, 1)2 = (9, 7) ∙ (3, −1).
The Fundamental theorem of arithmetic (FTA) for algebra 𝔸2 states that:
(1) Every element 𝑡 ∊ (𝔸2 )𝑁>1 such that 𝑁(𝑡 ) ≡ 1 (mod 2), is either prime
itself or is the product of prime elements whose norms are odd prime natural
numbers, and any two prime factorizations of this element are associative.
(2) Every element 𝑡 ∊ (𝔸2 )𝑁>1 such that 𝑁(𝑡) ≡ 0 (mod 2) and 𝑁(𝑡) is not
a power of base 2, is the product of two elements 𝑡 ′ , 𝑡 ′′ ∊ (𝔸2 )𝑁>1 : 𝑡 = 𝑡 ′ ∙
𝑡 ′′ , where 𝑁(𝑡 ′ ) = 2𝑛 , 𝑛 ≥ 2, and 𝑁(𝑡 ′′ ) ≡ 1 (mod 2). Any two such
factorizations of the element 𝑡 are associative.
(3) Every element 𝑡 = (𝑥, 𝑦) ∊ (𝔸2 )𝑁>1 such that 𝑥𝑦 ≠ 0 and 𝑁(𝑡) =
2𝑛 , 𝑛 ≥ 3, can be uniquely written in one of the two forms
𝑡 = 2𝑘 ∙ 𝜈𝑙 ∙ 𝑢 or 𝑡 = 2𝑘 ∙ 𝜈𝑙∗ ∙ 𝑢
where 𝑢 is a unit element, 𝜈𝑙 = (2𝑙 + 1, 2𝑙 − 1) and 𝜈𝑙∗ = (2𝑙 + 1, 1 −
2𝑙 ) are prime elements with norm 𝑁(𝜈𝑙 ) = 𝑁3 (𝜈𝑙∗ ) = 2𝑙 + 2 , 2𝑘 + 𝑙 = 𝑛 −
2, 𝑙 ≥ 1, 𝑘 ≥ 0.
Corollary Any element 𝑡 = (𝑚, 𝑛) ∊ (𝔸2 )𝑁>1 with norm 𝑁(𝑡) ≡ 0 (mod 2)
can be uniquely written as a product 𝑡 = 2𝑘 ∙ 𝑡 ′ ∙ 𝑡 ′′ , where 𝑡 ′ = 𝜈𝑙 or 𝑡 ′ =
𝜈𝑙∗ , 𝑙 , 𝑘 ≥ 0, and 𝑁(𝑡 ′′ ) ≡ 1 (mod 2). Any prime factor
𝜋 of 𝑡 is a factor of 2𝑘 ∙ 𝑡 ′ , if 𝑁(𝜋) is a power with base 2, and is a factor of
𝑡 ′′ , if 𝑁(𝜋) is an odd prime number.
.
Corollary Let 𝑘 be an integer, |𝑘| > 1, and 𝜋 be a prime perplex which
divides 𝑘, 𝜋 ≠ 2 ∙ 𝑢 where 𝑢 is a unit element. Then 𝑁(𝜋) divides 𝑘.
Corollary Let 𝑡 = (𝑚, 𝑛) ∊ (𝔸2 )𝑁>1 and gcd(𝑚, 𝑛) = 1. Then 𝑡 = (𝑚, 𝑛)
and 𝑡 ∗ = (𝑚, −𝑛) are mutually prime integer perplexes.
The other interesting algebra which has very close relation to integer
perplexes, is the algebra 𝔸 = {𝑡 = 𝑥 + 𝑦ω
̂ | 𝑥, 𝑦 ∈ ℤ}, where ω
̂ is an
imaginary unit, such that ω
̂2 = ω
̂ . The bilinear product in this algebra is
determined by the expression
𝑡1 ∙ 𝑡2 = (𝑥1 𝑥2 ) + [(𝑥1 + 𝑦1 )(𝑥2 + 𝑦2 ) − 𝑥1 𝑥2 ]ω
̂,
and the function 𝑁(𝑡) = |𝑥 (𝑥 + 𝑦)| determines the norm. The only units
elements in 𝔸 are ±1 and ±(1 − 2ω
̂ ). The main thing which distinguishes
this algebra from integer perplexes is that algebra 𝔸 is a Euclidean domain,
which means that this algebra allows the division with a remainder. This
implies that algebra 𝔸 is a unique factorization domain. Any prime element
in 𝔸 is a divisor of a prime natural number, and any prime natural number
𝑝 can be factored onto two prime factors the following standard way:
𝑝 = 𝜋̂𝑝 ∙ 𝜋̂𝑝∗ = (1, 𝑝 − 1) ∙ (𝑝, 1 − 𝑝).
Any prime factorization of 𝑝 is associative with the standard one.
The element 𝑡 ∗ = (𝑥 + 𝑦, −𝑦) ∈ 𝔸 is called the conjugate to an element
𝑡 = (𝑥, 𝑦) ∈ 𝔸. If x and y are mutually prime integers then the elements
𝑡 and 𝑡 ∗ are mutually prime.
To restore unique prime factorization in 𝔸2 we extend it to the algebra
𝑥
𝑦
2
2
𝔸′2 = {𝑡 = +
𝑗̂ ∈ 𝔹 | 𝑥, 𝑦 ∈ ℤ, 𝑥 + 𝑦 ≡ 0(mod 2)}.
Algebra 𝔸 is isomorphic to algebra 𝔸′2 . We define the isomorphism φ: 𝔸 →
𝔸′2 the following way:
𝑦
𝑦
2
2
φ(𝑥 + 𝑦ω
̂ ) = (𝑥 + ) +
𝑗̂
Notice that φ maps unit elements of 𝔸 onto unit elements of 𝔸2 .
Let 𝑝 = 2k + 1 be an odd prime natural number, then φ maps two prime
factors of 𝑝 in 𝔸 onto two prime factors of 𝑝 in 𝔸2 :
𝑝 = 𝜋̂𝑝 ∙ 𝜋̂𝑝∗ = (1, 𝑝 − 1) ∙ (𝑝, 1 − 𝑝) in 𝔸,
𝑝 = 𝜋𝑝 ∙ 𝜋𝑝∗ = (𝑘 + 1, 𝑘) ∙ (𝑘 + 1, −𝑘) in 𝔸2 ,
φ(𝜋̂𝑝 ) = 𝜋𝑝 , φ(𝜋̂𝑝∗ ) = 𝜋𝑝∗ .
If 𝑝 = 2 then
φ(𝜋̂2 ) = φ(1 + ω
̂) =
φ(𝜋̂2∗ ) = φ(2 − ω
̂) =
3
2
3
2
1
+ 𝑗̂ = 𝛼,
2
1
− 𝑗̂ = 𝛽.
2
Therefore, prime elements in 𝔸′2 are the same as prime elements in 𝔸2 , but
instead of prime elements which norm is a power of base 2 we have two
prime elements 𝛼 and 𝛽 such that 𝛼 ∙ 𝛽 = 2. The uniqueness of the prime
factorization was restored: every element 𝑡 ∊ (𝔸′2 )𝑁>1 is either prime itself
or is the product of prime elements, and any two prime factorizations of this
element are associative.
Notice also that
φ(2) = 2,
𝜈𝑙 = (2𝑙 + 1) + (2𝑙 − 1)𝑗̂ = φ(2 + 2 ∙ (2𝑙 − 1)ω
̂)
= φ(2) ∙ φ(1 + (2𝑙 − 1)ω
̂ ) = 2φ((1 + ω
̂ )𝑙 )
= 2φ((𝜋̂2 )𝑙 ) = 2[φ(𝜋̂2 )]𝑙 = 2 ∙ 𝛼 𝑙 ,
𝜈𝑙∗ = (2𝑙 + 1) + (1 − 2𝑙 )𝑗̂ = φ(2𝑙 + 1 + 2 ∙ (1 − 2𝑙 )ω
̂)
= φ(2) ∙ φ(2𝑙 + (1 − 2𝑙 )ω
̂ ) = 2φ((2 − ω
̂ )𝑙 )
= 2φ((𝜋̂2∗ )𝑙 ) = 2[φ(𝜋̂2∗ )]𝑙 = 2 ∙ 𝛽 𝑙 .