Lemma 10.16.4. Suppose that $\varphi : R \to R'$ is a ring homomorphism. The induced map

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

** Functoriality of the spectrum **

Lemma 10.16.4. 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.
$\square$

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## Comments (1)

Comment #3794 by slogan_bot on

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