Prove Prime Ideal Problem: I/J ⊆ P

[mathmajor2013]

Let R be a ring with ideals I, J, and P. Prove that if P is a prime ideal and I intersect J is a subset of P, then I is a subset of P or J is a subset of P.

 

[mathmajor2013]

If i like ab in I intersect J, then ab is in P. Therefore a in P or b in P since P is prime. Neither a or b need be in I intersect J though.

 

[disregardthat]

If neither I nor J is contained in P, some element a in I and b in J are neither in P, as you say. But this is a contradiction, since ab is in P and P is prime, hence one of the ideals is contained in P.

Follow-up: Show this for an arbitrary finite number of ideals.

 

[Deveno]

obviously if either I or J is a subset of P, there is nothing to prove. so to negate that, we need some a in I with a NOT in P, AND b in J with b NOT in P.

but since I is an ideal, ab is in I. since J is an ideal ab is in J. therefore ab is in I∩J, and thus in P. since P is prime, either a is in P, contradicting our choice of a, or b is in P, contradicting our choice of b.

the only conclusion is that we cannot pick such a in I AND b in J outside of P, either I or J must lie within P.

 

[disregardthat]

Exactly. So if we can pick an a in I not in P, we cannot pick a b in J not in P. Hence if I is not contained in P, J must be contained in P. Oppositely, if J is not contained in P, I must be contained in P.