Strange Integers

Fundamental Theorem of Arithmetic

3 is not prime in Z[√3]. Give it a little thought, and the result is not at all surprising. This is what √3 was invented for - √3 times √3 is 3. 5 is a prime number. But what about 2? Is it a prime? No, it is not. We can write

(1)	2 = (5 + 3√3)(-5 + 3√3),

where none of the factors is a unity (check their N-value.)

In fact, once we found one factorization, we can easily obtain many more. Recollect that U = 2 + √3 as well as U' = 2 - √3 are unities. Multiply the first factor in (1) by U and the second by U'. Apply the same procedure to the result:

(2)

2	= (5 + 3√3)(-5 + 3√3)
	= (19 + 11√3)(-19 + 11√3)
	= (71 + 41√3)(-71 + 41√3)...

Note that the N-value of any factor is 2. This implies that each of them is prime. So we learn not only that, in Z[√3], 2 is not prime, but also that its factorization into prime factors is not unique. Indeed, in the presence of an infinite number of unities, no number may have a unique factorization. However, in (2) factors differ by a multiple of unity. Such primes are called associates. For example, 5 + 3√3 and 71 + 41√3 are associates, for (71 + 41√3) = (5 + 3√3)(7 + 4√3), where N(7 + 4√3) = 1 and 7 + 4√3 is a unity.

By the end of this page I plan to prove the following variant of the Fundamental Theorem of Arithmetic:

In Z[√3], factorization of a number into a product of primes is unique up to the order of associated primes.

First let's establish an analogue of the Euclidean Algorithm:

(3)	For any given pair A and B from Z[√3], B≠0, there exist in Z[√3] numbers Q and R such that. A = BQ + R, and \|N(R)\| < \|N(B)\|

Let A/B = x + y√3, where x and y are rational. Any rational number p lies somewhere between [p] and [p]+1, where [p] is the largest integer smaller than p. Therefore, any rational number is at the distance of at most ½ from an integer. Let r and s be integers such that |x - r| ≤ ½ and |y - s| ≤ ½, respectively.

Let Q = r + s√3 and R = B(x - r) + B(y - s)√3. Then, of course, A = BQ + R, where A, B, Q all belong to Z[√3] and so does R = A - BQ. Let's verify that |N(R)| < |N(B)|.

N(R) = N(B)((x - r)² - 3(y - s)²). By the defintion,

(4)	\|(x - r)² - 3(y - s)²\| ≤ \|x - r\|² + 3\|y - s\|² ≤ (½)² + 3(½)² = 1.

Equality is only achieved when both |x - r| and |y - s| equal ½. But in this case,

|(x - r)² - 3(y - s)²| = |½² - 3½²| = ½ < 1

Therefore always |(x - r)² - 3(y - s)²| < 1, and we have |N(R)| < |N(B)|.

(3) is a very powerful result. Now we may start an analogue of Euclid's Algorithm with confidence, assured that it will terminate in a finite number of steps. When it stops, we shall have found a number D with the property that

D|A and D|B, and
For any C such that C|A and C|B, we also have C|D.

This D is naturally called the greatest common divisor of A and B, and is denoted as D = gcd(A, B). As in N, the algorithm may be extended to yield two numbers S and T such that

AS + BT = gcd(A, B)

The greatest common divisor is obviously defined up to a factor of unity. Two numbers are called coprime (mutually prime) iff their greatest common divisor is a unity. Since 1 associates with every unity,

AS + BT = 1,

for coprime A and B and some S and T; exactly as was the case with integers.

We are in a position to prove the most fundamental property of division:

(5)	Assume that a prime number P divides the product of two numbers A and B: P\|AB. Then either P\|A or P\|B.

In other words, if, for instance, P∤A, then it must be that P|B. As a prime, the only factors of P are its associates and the unities. If any of the associates divided A, so would P, but, by our assumption, this is not the case. Therefore, gcd(P,A) = 1. We thus conclude that there exist two numbers S and T such that PS + AT = 1. Multiply by B: PBS + (AB)T = B. Now, P|PBS and P|AB. Whence P|B.

Uniqueness of the prime factorization follows immediately from (5). Let A = P₁P₂...P_n = Q₁Q₂...Q_m, where all P's and Q's are prime. P₁ divides Q₁Q₂...Q_m. Therefore, by (5), it divides one of the Q's. Both being prime, they are associates of each other and their ratio is a unity. Assume P₁|Q₁ and Q₁/P₁ = U. Replace Q₂ with UQ₂. This will not affect the primality of Q₂. We can chip off factors from one side or another until we get, for example, Q₁Q₂...Q_k = 1, all other factors from the two sides having been matched. The remaining Q's are all unities, hence, not prime. Whence k = 0, n = m, and the proof is completed.

Remark

6 = 3·2 = 3(3-1) = 3² - 3 = (3 + √3)(3 - √3). Since 2|6, we also have that 2|(3 + √3)(3 - √3). However, 2∤(3 + √3) and 2∤(3 - √3), for, say, (3 + √3)/2 = 3/2 + √3/2 which does not belong to Z[√3] at all. Combined with (4), we get another proof that 2 is not prime in Z[√3].

Question

Does or does not the identity 3·2 = (3 + √3)(3 - √3) contradict the fact that the prime factorization in Z[√3] is unique?

Reference

H. M. Stark, An Introduction to Number Theory, The MIT Press, 1995, Ninth printing.

Constructible Numbers, Geometric Construction, Gauss' and Galois' Theories

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