Generalized Definition of Prime NumbersDate: 02/01/2001 at 04:02:33 From: Lakshmanan Narayanan Subject: Prime Numbers I was actually attending a course on set and group theory when the definitions of identity, prime, and composite were given. Identity (e) for an operation * for a, any set A, is defined as e * a = a * e = a for any a, an element of A. If the above is satisfied, then e is the identity on A for the operation *. But later the definition of prime and composite were given with respect to the set of natural numbers. A number is said to be prime if it is only divisible by 1 and itself. Supposing this definition of prime is extended to any set A and for any operation *: For a set A, if e (an element of A) is the identity for the operation *, then an element p of A is prime with respect to the operation * if there exist no two elements q and r of A that are not equal to either of e or p, such that the relation p = q * r holds good. For the set of Natural Numbers and the operation multiplication, this reduces to the conventional definition of a prime number; that is, a natural number p is said to be prime if there exist no two numbers q and r that are natural numbers, and neither of which equal 1 or p such that the relation p = qr holds good. Now the real question is, are there any other operations or set pairs apart from multiplication, Natural Numbers for which the above extended definition of prime holds good? Date: 02/01/2001 at 14:44:57 From: Doctor Rob Subject: Re: Prime Numbers Thanks for writing to Ask Dr. Math, Lakshmanan. Your guess at the generalized definition of a prime number is a good one, but not the right one. For one thing, you have to worry about sets A and operations * such that there exist elements e' other than e that have inverses in A; that is, there exists e" such that e'*e" = e. Elements in A whose inverses are in A are called the units of A. If there were units in A other than e itself, then you could write x = e'*(e"*x) for any x in A, and so there would be no primes at all in A, according to your definition. In the Natural Numbers, 1 is the only unit, and that is a special property of the Natural Numbers. As an example you could have A = Z - {0}, the nonzero integers, with the usual multiplication. Then 1 and -1 are both units, and you could write x = (-1)*(-x) for any x, so there would be no primes in this system. You want 2 to be a prime, however, so this is bad. You could get around that by defining a prime to be an element such that if you wrote p = q*r for q and r in A, then either q or r would have to be a unit. Unfortunately, this is not quite right either. Second of all, there are sets A and operations * for which it happens that p is prime by even this revised definition, and p is a divisor of a*b, but p is not a divisor of either a or b. That is very bad, because the Unique Factorization Theorem fails in this situation. One classic example of this is the set of complex numbers A = {a + b*sqrt(-5): a, b integers, not both zero}, with * being the usual multiplication. The only units in this set are 1 and -1. Here you have the situation 3*7 = 21 = (1 + 2*sqrt(-5))*(1 - 2*sqrt(-5)). 3 is a prime according to the revised definition, and 3 divides the product on the right, but it doesn't divide either factor (and likewise for 7). The correct generalized definition of a prime element p of A is that p is not a unit, and if p is a divisor of a*b for some a and b in A, then either p is a divisor of a or else p is a divisor of b. You can easily see that this is also equivalent to the definition you gave above when A is the Natural Numbers. Elements p such that if you wrote p = q*r for q and r in A, then either q or r would have to be a unit, are called IRREDUCIBLE elements. It is true that prime elements are irreducible, but there may be irreducibles that are not prime, like 3 in the above example. Occasionally one encounters sets in which every irreducible is also prime. An example of this is the set A = {a + b*sqrt(-2): a, b integers, not both zero}, with the usual multiplication. Now there are still only two units 1 and -1 in this set, but every irreducible in this set is also prime. The differences between the two examples are pretty subtle, but they are there. You are investigating some very deep and interesting ideas. Keep asking good questions! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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