The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.45.3. Let $\varphi : R \to S$ be a ring map. If

  1. for any $x \in S$ there exists $n > 0$ such that $x^ n$ is in the image of $\varphi $, and

  2. $\mathop{\mathrm{Ker}}(\varphi )$ is locally nilpotent,

then $\varphi $ induces a homeomorphism on spectra and induces residue field extensions satisfying the equivalent conditions of Lemma 10.45.2.

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

\[ x^ n \in \mathfrak q \Rightarrow y \in \mathfrak p \Rightarrow x^ n \in \mathfrak q' \]

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 10.35.17 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 $\kappa (\mathfrak p) \to \kappa (\mathfrak q)$. $\square$


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