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Generalized Definition of Prime Numbers


Date: 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/   
    
Associated Topics:
High School Number Theory

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