Question

Let A be a commutative ring.

a) [6 marks] Let I,J be two ideals of A such that I⊂J and J⊂I. Can A/(I∩J) be an integral domain? Justify your answer.
b) Let IA be an ideal. Show that I is maximal if and only if I+a=A for each a∈A,a∈/I.

Answer 1

a) No, A/(I∩J) cannot be an integral domain because I∩J has a nonzero zero divisor. b) I is maximal if and only if I+a=A for each a∈A, a∈/I.

Step-by-step explanation:

a) No, A/(I∩J) cannot be an integral domain. An integral domain is a commutative ring with no zero divisors, meaning that if ab=0 for some elements a and b, then either a=0 or b=0. Since I⊂J and J⊂I, there exist elements i∈I and j∈J such that i=0 and j=0. Therefore, the product ij is a nonzero element in both I and J, but it belongs to I∩J. This shows that I∩J has a nonzero zero divisor, and therefore A/(I∩J) cannot be an integral domain.

b) If I is maximal, then for any a∈A with a∉I, the ideal generated by I and a, denoted as (I,a), is equal to A. This means that (I,a) contains the identity element 1 of A. As a result, there exists b∈I and c∈A such that b+ac=1. This implies that c=1-ba, which shows that c∈I+a. Since c is an arbitrary element of A, we have A⊆I+a. On the other hand, if I+a=A for any a∈A, then I is maximal. This can be proven by assuming that there exists an ideal J properly containing I, and showing that there exists an element a∉I but a∈J such that I+a≠A. The details of this proof can be found in textbooks on abstract algebra.