Lemma 10.16.7. 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 10.16.7. 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

\[ \mathop{\mathrm{Spec}}(R/I) \longrightarrow V(I) \subset \mathop{\mathrm{Spec}}(R). \]

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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## Comments (1)

Comment #626 by Wei Xu on

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