up to speed on primes
 

Hi,

I'm trying to get a precise definition of 'prime' in sets of numbers beyond the set of natural numbers.

In my open university book it says :

Let \(R\) be a commutative ring. A non-zero, non-unit element \(p\) of \(R\) is prime if whenever \(p\mid ab\) then either \(p\mid a\) or \(p \mid b\).

When I see what this means for the set of integers I get to a conclusion I know must be wrong :

\(2 \mid 4\times8\) but 2 divides both 4 and 8. It can't be said that 2 divides either 4 or 8.

In other definitions of prime in rings and fields, there's no either in the wording. This makes more sense but is

Let \(R\) be a commutative ring. A non-zero, non-unit element \(p\) of \(R\) is prime if whenever \(p\mid ab\) then \(p\mid a\) or \(p \mid b\).

the right definition of prime?

Yes, you're absolutely right!
The point is that, in the integers for example, 4 divides 6x10 but 4 does not divide 6 or 10.
This is only possible because 4 is not a prime number.

Strike out "either." It should be the "inclusive or." If p is prime (a "prime element"), then p|a*b implies p|a or p|b (or both). The condition that p is [i]not a unit[/i], i.e. p does not divide 1, is to avoid trivialities. Of course, 0 is excluded from this definition since division by zero is undefined.

There is however, a special (and important) type of ring called an [i]integral domain[/i], in which a*b = 0 implies a = 0 or b = 0.

The ring of integers Z is an integral domain. The ring of 2x2 matrices with integer entries is [i]not[/i] an integral domain. The ring of the integers mod 4 is not an integral domain, since 2*2 = 0.

Thanks to both posters. Very helpful.

My example of the ring of 2x2 matrices was inapt. Yes, it has plenty of "divisors of zero," but it's not an integral domain for a much more basic reason.

Its multiplications isn't commutative!
:blush:

One can take commutative subrings, though. For example, the ring generated by (integer multiples of) the 2x2 identity I2 and M = [0,1;1,0] (which obviously commute) has

M^2 = I2, so that

(M - I2)*(M + I2) = 0.

I guess M = [0,1;1,0], means

0 1
1 0

[QUOTE=wildrabbitt;500288]I guess M = [0,1;1,0], means

0 1
1 0[/QUOTE]

Yes (in PARI/GP notation).

I know a weird thing called "ideal", it might be the definition of 'prime' in sets of numbers beyond the set of natural numbers.


Since it is quite difficult to define a "prime" beyond the set of natural numbers.


For example, in [TEX]\mathbb Z[\sqrt{-1}][/TEX], the defination of "prime" will be what you describe, but in [TEX]\mathbb Z[\sqrt{-5}][/TEX], since [TEX]9=3*3=(2+\sqrt{-5})(2-\sqrt{-5})[/TEX], the "prime" you define will be quite different than what you are thinking about.


Actually both [TEX]3|(2+\sqrt{-5})[/TEX] and [TEX]3|(2-\sqrt{-5})[/TEX]could not be true, but [TEX]3|(2+\sqrt{-5})(2-\sqrt{-5})=9[/TEX]

In our Basic Number Theory course, we look at a more general definition of primes here:
[URL]https://www.mersenneforum.org/showthread.php?t=22479[/URL]