Question

Let R be a commutative ring with identity and suppose that the ideal A of R is contained in a finite union of prime ideals P1∪…∪Pn. Show that A⊂Pi

Answer 1

To show that A is a subset of Pi, assume that A is not contained in any of the prime ideals Pi and derive a contradiction.

Step-by-step explanation:

Let's assume that the ideal A is not a subset of any of the prime ideals P1, P2, ..., Pn. Since A is not contained in any of these prime ideals, it means that for each prime ideal Pi, there exists an element ai in A that is not in Pi.

Now, consider the product of all these elements: a = a1a2...an. Since all the prime ideals are prime, each of the elements ai is not in any of the prime ideals Pi, so their product, a, is also not in any of the prime ideals Pi.

But the product a is an element of A, and it is not in any of the prime ideals Pi, contradicting our assumption. Therefore, A must be a subset of one of the prime ideals Pi.