4 ms·
Is there any reason that the definition of "prime" hasn't been changed to "a natural number only divisible by 1 and itself and is non-invertible"?
by saraid216 13y ago
Is there any reason that the definition of "prime" hasn't been changed to "a natural number only divisible by 1 and itself and is non-invertible"?
- wbhart 13y agoIt's kind of murky because in commutative algebra that is actually not the definition of a prime. It just so happens that the prime numbers are precisely the natural numbers with this property. It's fine to define prime numbers that way, but in more general contexts a prime element is actually an element p which cannot divide a product of two elements a.b without dividing one (or both) of them. It just so happens that the non-invertible natural numbers with this property also aren't able to be written as a product p = c.d such that c != 1 and d != 1. In other number rings this isn't necessarily the same thing.
- thaumasiotes 13y agoIs something missing here? As I understand it, that general definition works like this: 6 divides 49, but does not divide either 4 or 9. Therefore, 6 is not prime. But I'm pretty sure 1 cannot divide the product of two integers m, n* while failing to divide m and n individually. So while that general definition does exclude composite numbers, it doesn't seem to exclude 1?
- pavelrub 13y agoThe correct "general" definition also requires the prime element to be non-invertible and non-zero, thus 1 is excluded. It should be noted that this definition excludes "composite numbers" only when the ring is an integral domain. In a general commutative ring this isn't necessarily true.
- pavelrub 13y agoFor this and the "general" definition to coincide, the ring in question has to be a unique factorization domain, which the integers are an example of. In any bigger subset of rings (integral domains or general commutative rings), there are rings for which this doesn't hold.