5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ math.stackexchange.com/questions/3698/… Have you looked here? $\endgroup$ – Dylan Yott Nov 14 '13 at 1:09
  • $\begingroup$ 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. $\endgroup$ – Nishant May 24 '14 at 2:32
  • $\begingroup$ 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) $\endgroup$ – Henry Swanson May 24 '14 at 2:53
  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Mar 19 '15 at 16:53