Proving Continuity of Spec(S) -> Spec(R) Homomorphism

[tom.young84]

f: Spec(S) -> Spec(R)

How do I prove the homomorphism between the two prime spectrums of R and S is continuous?

I have a strategy, but I'm having problems trying to see how to formulate a proof.

My strategy is as follows. I was able to prove earlier in the assignment that a prime ideal P in S will get mapped back to a prime ideal f^-1(P) in R. From this I gather that there will be some ideal A contained in a collection of prime ideals V(B) both in R. Then f^-1(A)=f^-1(V(A)) and that's going to be in P which is in S. I think. As of right now this is my "gut reaction" to seeing the question.

**V(I) is the collection of all prime ideals of R that contain I.

 

[lippyka]

**To prove that the homomorphism between the two prime spectrums of R and S is continuous, we need to show that for any open set U in Spec(R), the preimage f-1(U) is open in Spec(S). To do this, we can take an arbitrary element P of f-1(U). We then need to find an open neighborhood of P that is also a subset of f-1(U).Let P be an arbitrary element of f-1(U). By definition, there exists some B in U such that f-1(B) = P. Then taking the collection of ideals V(B) in R, we have that f-1(V(B)) = P. Since V(B) is a collection of prime ideals in R, this means that P is a collection of prime ideals in S that contain f-1(B). Let Q be any prime ideal in P. Then we know that f-1(B) is contained in Q since Q contains all prime ideals in P. Since U is an open set in R, this means that there exists some C in U such that B is a subset of C. Then we have that f-1(C) is a subset of f-1(B) since f-1 is a homomorphism. This implies that f-1(C) is also contained in Q. We can now take the set P \cap f-1(C) as our open neighborhood of P that is also a subset of f-1(U). Thus, we have shown that f-1(U) is open in Spec(S), which proves that the homomorphism between the two prime spectrums of R and S is continuous.