Rings, The Union of Prime Ideals

The Union of Prime Ideals

Let P₁ P₂ … P_n be a set of prime ideals in a commutative ring. Let H be a set, closed under addition and multiplication, such that H is contained in the union of the prime ideals. We aren't talking about the ideal generated by the prime ideals, just their union, as sets. We will show that H is contained in one of the prime ideals.

If there are only two ideals, they don't even have to be prime. If H contains x from P₁-P₂ and y from P₂-P₁, H must contain x+y, which is not contained in either ideal. Yet all the members of H come from some ideal, hence H belongs to one of the two ideals.

For more than two ideals we need primality. Proceed by induction on the number of prime ideals.

Suppose we have a minimal counter example; the smallest number of prime ideals that refutes this theorem. If H is contained in some of the prime ideals, but not all, it is contained in one of them by induction. Thus a piece of H is in every prime ideal, and outside of the others.

Let x_i be an element in H that comes from P_i, and is not present in the other prime ideals. For each i, let y_i be the product of x_j, where j is not equal to i. In other words, y_i is the product of the other x values. Let z be the sum of y_i. For every prime ideal P_i, all the terms of z are contained in P_i, except for y_i. Thus z is in P_i iff y_i is in P_i. Now z is in H by closure, so let it lie in the i^th prime ideal. Thus y_i is in P_i, and y_i is a product, and represents the product of principal ideals in our commutative ring. Thus some x_j from that product is in P_i, and that is a contradiction. Therefore the set H is always contained in one of the prime ideals.