Rings, Prime Ideal Correspondence

Prime Ideal Correspondence

This is similar to the ideal correspondence theorem, but now we are dealing with prime ideals. We will see that prime ideals containing the kernel correspond 1-1 with prime ideals in the image.

Let f be a ring epimorphism from R onto S, with kernel K. We know that ideals in R containing K correspond to ideals in S. Verify that the xRy test is satisfied in R iff it is satisfied in S. Therefore an ideal is prime iff its image or preimage is prime.

Note that ideal correspondence, and prime ideal correspondence, hold, even if R does not contain 1.

When ideals don't contain the kernel, all bets are off. For instance, the multiples of 6 are not prime in Z, but they map to Z₂ in Z₄, a maximal/prime ideal.

For the converse, let K be a field and let R be K[x,y]/xy. These are the polynomials in x and y with no mixed terms. Let P be the ideal generated by y. Since R/P is K[x], an integral domain, P is prime. Now consider R mod x². These are polynomials in y, with one possible linear term in x. The ideal P, generated by y, is no longer prime, since it contains x×x, and x is not in P.