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A ring R is prime if and only if the zero ideal {0} is a prime ideal in the noncommutative sense. This being the case, the equivalent conditions for prime ideals yield the following equivalent conditions for R to be a prime ring: For any two ideals A and B of R, AB = {0} implies A = {0} or B = {0}.
The question of when this happens is rather subtle: for example, for the localization of k[x, y, z]/(x 2 + y 3 + z 5) at the prime ideal (x, y, z), both the local ring and its completion are UFDs, but in the apparently similar example of the localization of k[x, y, z]/(x 2 + y 3 + z 7) at the prime ideal (x, y, z) the local ring is a UFD but ...
The ring of 2×2 matrices with integer entries does not satisfy the zero-product property: if = and = (), then = () = =, yet neither nor is zero. The ring of all functions: [,], from the unit interval to the real numbers, has nontrivial zero divisors: there are pairs of functions which are not identically equal to zero yet whose product is the ...
The prime ideals of the localized ring correspond exactly to those prime ideals of with =. [2] As every non-zero commutative ring has a maximal ideal, which is prime, every non-nilpotent x {\displaystyle x} is not contained in some prime ideal.
The ring = of algebraic integers in a number field K is Noetherian, integrally closed, and of dimension one: to see the last property, observe that for any nonzero prime ideal I of R, R/I is a finite set, and recall that a finite integral domain is a field; so by (DD4) R is a Dedekind domain. As above, this includes all the examples considered ...
In a commutative ring R with at least two elements, if every proper ideal is prime, then the ring is a field. (If the ideal (0) is prime, then the ring R is an integral domain. If q is any non-zero element of R and the ideal (q 2) is prime, then it contains q and then q is invertible.)