5

Typically, a prime is defined as follows: $p$ is prime iff $(p \mid xy \implies p \mid x$ or $p \mid y)$ and $p$ is not a unit or zero. But for ideals, we say the zero ideal is prime.

There is a strong correspondence between statements about primes and statements about prime ideals:

  • "A non-unit is prime if $p \mid ab \implies p \mid a$ or $p \mid b$" vs. "A proper ideal is prime if $P \ni ab \implies P \ni a$ or $P \ni b$"
  • "All non-zero primes are irreducible" vs. "all non-zero prime ideals are maximal"
  • "Primes are only divisible by themselves and units" vs. "prime ideals are only contained by themselves and the whole ring (generated by a unit)"
  • "Zero is divisible in every ring element vs. "the zero ideal is contained in every ideal"
  • "In a UFD, non-zero elements can be uniquely factored into primes" vs. "in a Dedekind ring, non-zero ideals can be uniquely factored into prime ideals"

So, I feel that zero should be prime iff the zero ideal is a prime ideal. Why is there this discrepancy? I lean towards "zero is not a prime", but what are the consequences of rejecting the zero ideal as prime as well?

MJD
  • 65,394
  • 39
  • 298
  • 580
Henry Swanson
  • 12,972
  • http://math.stackexchange.com/questions/3698/why-doesnt-0-being-a-prime-ideal-in-z-imply-that-0-is-a-prime-number

    Have you looked here?

     – Dylan Yott Nov 14 '13 at 01:09
  • The zero ideal IS considered to be prime in an integral domain.

    Prime ideals are not necessarily maximal; this is true for PIDs, but not in general.

     – Nishant May 24 '14 at 02:32
  • Yeah, that was my point: "we do say the zero ideal is prime (in domains), but we don't say 0 is prime". And now that I actually know some ring theory, I see two of my criteria are just... wrong. (prime ideals can be contained in proper ideals, after all) – Henry Swanson May 24 '14 at 02:53
  • 1
    If you modify the definition so that $,0,$ is prime then you need to modify many theorems, e.g. to exclude the nonunique prime factorization $, 0 = 0^2.\ $ See also the links to factorization theory in non-domains in this answer. – Bill Dubuque Mar 19 '15 at 16:53
  • @Bill Dubuque: Actually, in the paper "Factorization in Commutative Rings with Zero-divisors" by Anderson and Valdes-Leon, where your link leads to (among other things), the element $0$ is considered as having the potential to be prime, that is in the case if and only if $(0)$ is a prime ideal, exactly what the thread opener cares about. – Thrash Mar 10 '21 at 22:34
  • @Bill Dubuque: In my opinion, the factorization of $0$ can be omitted because what matters is the monoid structure of $(R\setminus {0},\cdot)$ because $0$ is an absorbing element. Even if we don't allow $0$ to be prime, then we must exclude it from the prime factorization theorem. Or what does a prime factorization of $0$ look like? So if we allow $0$ to be prime, then the prime factorization theorem can be retained as it is, namely for non-zero elements. – Thrash Mar 10 '21 at 22:47
  • @Thrash I don't know precisely what you mean by writing "actually" above, but it seems you may have read more into "see also" than what was intended. – Bill Dubuque Mar 11 '21 at 06:41
  • @Bill Dubuque: Yes, I'm interested in primality and (different notions of) irreducibility in rings with zero divisors. You can find the authors' statements on page 441, 444 and 450 (Theorem 2.13). But I think it's a matter of taste if you allow the element $0$ to be called prime (or irreducible in any form). – Thrash Mar 12 '21 at 13:50

0 Answers0