this is by the ideal correspondence theorem, there is a bijective correspondence according to this theorem.
Let be a ring homomorphism. Let and . If , then is a prime ideal of , i.e., a point of . Hence induces a map . Show that:
If is surjective, then is a homeomorphism of Y onto the closed subset of .
[ denotes the set of all prime ideals of which contain ].
Here is one way to do it:
By the "Ideal Correspondence Theorem," we know that and induces a bijective map from to . Then I need to show is continuous and is continuous to show it is a homeomorphism.
I don't understand the sentence in red. How do we know induces a bijective map from to ?
Lemma 1. If be a ring homomorphism mapping to and P a prime ideal in B. This induces a map such that .
By lemma 1 and an ideal correspondence theorem, is . The latter corresponds the all prime ideals in B, which is spec(B).
Since the inverse image of a closed subset in Spec(A) is a closed subset in Spec(B), is a continuous map in the Zariski topology.