Thread: Ideals

Let I and J be nonzero ideals, in a commutative ring R. If R is a domain, Prove that the intersection of I and J does not equal {0}.


I have that I and J are nonzero so I has a 0 elements and elements a,b such that a+b is also in I. The same goes for J. R is a domain so it has a 1 different from 0. Does the intersection of I and J must contain 1? If that is the case then does that mean that the intersection is nonempty?

if an ideal A contains 1, then A = R, so that's not the right avenue to pursue. what you know is that there is at least one non-zero element a in I, and one non-zero element b in J (we cannot, in general, assume two such elements of I or J. for example, R might be Z4, and I might be {0,2}).

what i suggest, is to consider the element ab of R, and use the fact that R is a domain.

This is what I have. Since I and J are nonzero ideals there exist an a in J and there exist a b in J. Now both a and b have to be in R. Now R is a domain so nonzero ab has to be in there. Now a is in I and ab in R therefore a(ab) has to be in I
Now b is in J and ab in R therefore b(ab) is in J. By associativity there are the same that means they have an element in common.....

I'm I missing anything?

well, yes, you are missing something. there is no reason to suppose that a(ab) = b(ab). even with associativity, these are not necessarily the same

(one is a^2b, the other is ab^2, given that R is commutative. for example, in the domain Z, with I = (3) and J = (5), 45 is not the same element as 75).

since a is in I, and b is in J (and thus R), ab is in I (since I is a (two-sided) ideal (since R is commutative, left ideals are also right-ideals)).

since b is in J, and a is in I (and thus R), ab is in J (same as above).

so ab is in I and J, and thus is in their intersection. by assumption, both a and b are non-zero (since I and J are asssumed to be non-zero ideals).

since R is a domain, ab is non-zero, and is thus the desired element of I∩J.