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#51




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 oneelement 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




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#53




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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√21, then p divides q, since q = (√21)p, but also q divides p, since p = (√2+1)q. 
#54




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".



#55




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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