Lemma 10.17.7. Let R be a ring. Let I \subset R be an ideal. The map R \to R/I induces via the functoriality of \mathop{\mathrm{Spec}} a homeomorphism
The inverse is given by \mathfrak p \mapsto \mathfrak p / I.
Lemma 10.17.7. Let R be a ring. Let I \subset R be an ideal. The map R \to R/I induces via the functoriality of \mathop{\mathrm{Spec}} a homeomorphism
The inverse is given by \mathfrak p \mapsto \mathfrak p / I.
Proof. It is immediate that the image is contained in V(I). On the other hand, if \mathfrak p \in V(I) then \mathfrak p \supset I and we may consider the ideal \mathfrak p /I \subset R/I. Using basic notion (51) we see that (R/I)/(\mathfrak p/I) = R/\mathfrak p is a domain and hence \mathfrak p/I is a prime ideal. From this it is immediately clear that the image of D(f + I) is D(f) \cap V(I), and hence the map is a homeomorphism. \square
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