The Stacks project

Functoriality of the spectrum

Lemma 10.17.4. Suppose that $\varphi : R \to R'$ is a ring homomorphism. The induced map

\[ \mathop{\mathrm{Spec}}(\varphi ) : \mathop{\mathrm{Spec}}(R') \longrightarrow \mathop{\mathrm{Spec}}(R), \quad \mathfrak p' \longmapsto \varphi ^{-1}(\mathfrak p') \]

is continuous for the Zariski topologies. In fact, for any element $f \in R$ we have $\mathop{\mathrm{Spec}}(\varphi )^{-1}(D(f)) = D(\varphi (f))$.

Proof. It is basic notion (41) that $\mathfrak p := \varphi ^{-1}(\mathfrak p')$ is indeed a prime ideal of $R$. The last assertion of the lemma follows directly from the definitions, and implies the first. $\square$

Comments (1)

Comment #3794 by slogan_bot on

Suggested slogan: "Pullback of primes gives a continuous map on spectra"

There are also:

  • 4 comment(s) on Section 10.17: The spectrum of a ring

Post a comment

Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00E2. Beware of the difference between the letter 'O' and the digit '0'.