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Old 04-20-2019, 03:42 PM
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Regarding primes, if R is a ring (let's assume commutative, and containing 1), and like number theorists we consider prime ideals rather than just prime numbers, then what is supposed to happen is that P is prime if and only if R modulo P is an integral domain. The example of the integers shows that (0) is a prime ideal in the ring of integers, for Z = Z/(0) is an integral domain. (Even though 0 is not normally counted as a prime number.) But, similar to the subject of this thread, (1) is not, and Z/(1) = the terminal one-element ring should not be an integral domain!

Relatedly, when people talk about "the field with one element", the thing to keep in mind is that this is just a manner of speaking because there isn't one. As reiterated above the smallest field has 2 elements, namely 0 and 1.
  #52  
Old 04-20-2019, 04:43 PM
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Originally Posted by DPRK View Post
If you count negative integers as numbers, then you have to take them into account and yes, they are; units like 1 and -1 are not counted in the factorization.

It might also be simpler to think of a prime number as one for which whenever p divides ab, then p divides a or p divides b (and p is not zero or a unit).
I don't think so. Saying that -3 is a prime destroys the prime factorization theorem as well as -3 * -3 = 9. If you want to add negative integers, I'd say -1 is a prime and nothing else. -1 has factors of only itself and 1, and the unique prime factorization of -n is -1 * (the unique prime factorization of n)
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Old 04-20-2019, 05:20 PM
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I don't think so. Saying that -3 is a prime destroys the prime factorization theorem as well as -3 * -3 = 9. If you want to add negative integers, I'd say -1 is a prime and nothing else. -1 has factors of only itself and 1, and the unique prime factorization of -n is -1 * (the unique prime factorization of n)
I don't see a problem with the textbook definition that the uniqueness of the factorization into primes is up to units, so when you match up the primes if q = up where u is a unit then everything is OK.

How else would you suggest to generalise it? Take, for example, Z[√2]. Let's factor 7 = (3 + √2)(3 - √2). But also 7 = (2√2 + 1)(2√2 -1). The trouble is, we cannot really regard them as different factorizations, because if p = 3+√2 and q = 2√2-1, then p divides q, since q = (√2-1)p, but also q divides p, since p = (√2+1)q.
  #54  
Old 04-20-2019, 11:13 PM
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Make it even simpler. If -1 is a prime, then 6 = 2*3 = (-1)*(-1)*2*3 , and we've already lost uniqueness. Unless you say that (-1)*(-1) "doesn't count", and at that point, we might as well just say "up to units".
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Old Yesterday, 02:00 AM
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All those statements of theorems that currently start "For any odd prime p" instead of "For primes p > 2" would have to be reworded to the latter (assuming that it fails for p=1, which would be a reasonable assumption if it fails for p=2 but is good for all other primes). I like the former wording because it sounds more clever, and it would be a shame to lose it.
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Old Yesterday, 05:48 AM
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According to Wikipedia, "1 is not prime, as it is specifically excluded in the definition." Why is 1 excluded? What's the reasoning for not including it? Does it lead to some sort of contradiction that I'm not seeing?

Thanks in advance for your answers.
The prime number defines that the number should be exactly divisible by itself by or another positive integer number. But remember, 1 is only divisible by itself therefore, 1 is not a prime number.
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Old Yesterday, 11:44 PM
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Originally Posted by Deltree View Post
According to Wikipedia, "1 is not prime, as it is specifically excluded in the definition." Why is 1 excluded? What's the reasoning for not including it? Does it lead to some sort of contradiction that I'm not seeing?

Thanks in advance for your answers.
I also prefer Wikipedia for a generic definition.
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