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

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

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.
$\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (1)

Comment #626 by Wei Xu on

There are also: