The Stacks project

Proof. Assume (1) and (2). Let $\mathfrak q, \mathfrak q'$ be primes of $S$ lying over the same prime ideal $\mathfrak p$ of $R$. Suppose $x \in S$ with $x \in \mathfrak q$, $x \not\in \mathfrak q'$. Then $x^ n \in \mathfrak q$ and $x^ n \not\in \mathfrak q'$ for all $n > 0$. If $x^ n = \varphi (y)$ with $y \in R$ for some $n > 0$ then

which is a contradiction. Hence there does not exist an $x$ as above and we conclude that $\mathfrak q = \mathfrak q'$, i.e., the map on spectra is injective. By assumption (2) the kernel $I = \mathop{\mathrm{Ker}}(\varphi )$ is contained in every prime, hence $\mathop{\mathrm{Spec}}(R) = \mathop{\mathrm{Spec}}(R/I)$ as topological spaces. As the induced map $R/I \to S$ is integral by assumption (1) Lemma 10.36.17 shows that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R/I)$ is surjective. Combining the above we see that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is bijective. If $x \in S$ is arbitrary, and we pick $y \in R$ such that $\varphi (y) = x^ n$ for some $n > 0$, then we see that the open $D(x) \subset \mathop{\mathrm{Spec}}(S)$ corresponds to the open $D(y) \subset \mathop{\mathrm{Spec}}(R)$ via the bijection above. Hence we see that the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism.

To see the statement on residue fields, let $\mathfrak q \subset S$ be a prime lying over a prime ideal $\mathfrak p \subset R$. Let $x \in \kappa (\mathfrak q)$. If we think of $\kappa (\mathfrak q)$ as the residue field of the local ring $S_\mathfrak q$, then we see that $x$ is the image of some $y/z \in S_\mathfrak q$ with $y \in S$, $z \in S$, $z \not\in \mathfrak q$. Choose $n, m > 0$ such that $y^ n, z^ m$ are in the image of $\varphi $. Then $x^{nm}$ is the residue of $(y/z)^{nm} = (y^ n)^ m/(z^ m)^ n$ which is in the image of $R_\mathfrak p \to S_\mathfrak q$. Hence $x^{nm}$ is in the image of $\kappa (\mathfrak p) \to \kappa (\mathfrak q)$. $\square$