Question

Let f∶ R → S be a ring homomorphism,with R and S commutative.If P is a prime ideal of S,show that the preimage f-¹ (P) is a prime ideal of R

Answer 1

1. Final answer:
2. To show that the preimage f^(-1)(P) is a prime ideal of R, we need to prove two properties: f^(-1)(P) is an ideal of R, and if ab is in f^(-1)(P), then either a or b is in f^(-1)(P).
4. Step-by-step explanation:
5. To show that the preimage f-1(P) is a prime ideal of R, we need to prove two properties:
6. f-1(P) is an ideal of R
7. If ab is in f-1(P), then either a or b is in f-1(P)
8. Property 1: f-1(P) is an ideal of R
9. Let a and b be in f-1(P). Since f is a ring homomorphism, we have f(a+b) = f(a) + f(b). Since a and b are in f-1(P), we have f(a) and f(b) in P. Since P is an ideal of S and f is a ring homomorphism, f(a) + f(b) = f(a+b) is in P. Therefore, a+b is in f-1(P). Similarly, we can prove that -a and ra are in f-1(P), where r is any element in R.
10. Property 2: If ab is in f-1(P), then either a or b is in f-1(P)

Assume that ab is in f-1(P). Then f(ab) = f(a)f(b) is in P since P is an ideal of S. Since S is commutative, we have f(b)f(a) = f(ab). Since P is a prime ideal of S, either f(a) or f(b) is in P. Therefore, either a or b is in f-1(P).