Strange Integers, divisors and primes
As we know, there are many different kinds of numbers:
- N - natural numbers, 1, 2, ... (sometimes are also called counting numbers, and sometimes the set includes 0, too)
- Z - integers ..., -2, -1, 0, 1, 2, ...
- Q - rational numbers in the form p/q, where p and q are integers,
q ≠ 0 - R - real numbers that have an infinite decimal expansion
- C - complex numbers, perfect for solving equations
Although many a student has found it hard to digest the division rule for fractions,
To divide p/q by r/s, turn the divisor r/s upside down to get s/r, and then multiply the result by p/q.
unquestionably the introduction of rational numbers Q has trivialized the operation of division. Division is interesting where it is not commonplace. Among numbers, divisibility problems are characteristic of integers. A number is prime if it's divisible only by itself and 1. And there are no primes among rational, real, or complex numbers unless, of course, they are also integers.
However, as we'll see, on the subset (subring) Z[√3] of real numbers R division gets a second going. Defining primality for Z[√3] makes perfect sense. The definition is more complex than that for N but models after the latter. First we need a definition of a divisor and another of a unity. In the following, unless stated otherwise, capital letters A, B, ... denote elements of Z[√3], small case letters a,b, ... denote integers from Z.
Definition
- A non-zero A divides B iff there exists C such that AC = B. If A divides B, B is said to be divisible by A while A is said to be a divisor of B.
- U is said to be a unity iff it divides any A
- A is said to be prime iff it is divisible only by itself and all available unities.
The definition of primality is very reasonable since, as 1 in N, no divisor is very important if it divides every number. To carry the analogy further, let's prove several simple facts. (Recollect the definition of function N:
- If U is a unity, then N(A) = ±1.
Indeed, assume on the contrary that |N(A)| = a > 1. Find b coprime to a. Then, there exists C such that AC = b and fromN(AB) = N(A)N(B) we geta|N(C)| = b. Since |N(C)| is a regular integer, a|b. A contradiction with our choice of b. - U is a unity iff it divides 1.
If U is a unity, it divides every number from Z[√3]; 1 in particular. Conversely, since 1 divides every number, any of its divisors has the same property. - If N(A) = ±1, then A is a unity.
Indeed, either AA' = 1 or A(-A') = 1. In either case, there exists B such that AB = 1 which exactly means that A is a unity. - Every number A is a product of primes.
The simple formula N(AB) = N(A)N(B) (which, if you write it explicitly, may not look so simple, after all) proved quite useful. We are going to use it once more, and the foregoing discussion is suggestive as to how it may be best used.But note first that by induction we have
N(A_{1}A_{2}...A_{n}) = N(A_{1})N(A_{2})...N(A_{n}) for any number of factors.
Assume A is not prime. Then it necessarily has divisors other than unities: A = A_{1}A_{2}. If both A_{1} and A_{2} are prime, we are finished. Otherwise, at least one of them can be divided further. Changing notations, we now have A as a product of three factors none of which is a unity: A = A_{1}A_{2}A_{3}. We may continue this way. The point of the proof is that we can't continue in this manner indefinitely. The process must stop after a finite number of steps. Indeed, assume that we have A = A_{1}A_{2}...A_{n}, where none of the factors is a unity. Then N(A) = N(A_{1})N(A_{2})...N(A_{n}). But N(A) has only a finite number of factors (different from ±1.) Therefore, the process is bound to stop at some point. When it does, all factors A_{1}, A_{2}, ..., A_{n} are bound to be prime, for, otherwise, we would be able to continue the process further.
- If N(A) is a prime integer then A is also prime.
As one example, we get that √3 is prime in Z[√3] while, consequently, 3 is not.
The number of unities in Z[√3] is infinite. U = 2+√3 is a unity as is U' = 2-√3. By definition, for any n>0, U^{n} is also a unity. Since, U^{-1} = U', the same is true for negative n as well. (All powers of U are different; so that there is indeed an infinite number of distinct unities.)
Prime numbers are defined up to a unity. We know that √3 is prime whereas 2 + √3 is a unity. Whence, √3(2+√3) = 3 + 2√3 is another prime (which, in a sense, is not much different from √3.)
Counterexample
#5 above asserts that if N(A) is a prime integer then A is also prime. The reverse of the statement is not true. There are primes for which N(A) is not a prime integer. 5 is one example. Assume 5 = AB where neither A nor B are unities. Then 25 = N(5) = N(A)N(B). Consequently,
(1) | N(A) = ±5 and N(B) = ±5 |
To show that 5 is a prime number (although N(5) = 25 which is not prime), we ought to prove that equations (1) have no integer solutions. Let a+b√3 be a generic number from
(2) | a^{2} + 2b^{2} = 0 |
Modulo 5, a^{2} may be 0, 1, or 4 whereas 2b^{2} may only be 0, 2, 3. Testing all possible combination (9 of them), we see that the right hand side is 0 only when both a and b are 0 modulo 5. In that case, 5|a and 5|b. Therefore, 25|a^{2} and 25|b^{2}, and also
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