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 10.29.3. Let $\varphi : R \to S$ be a ring map. The following are equivalent:

  1. The map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

  2. For any ideal $I \subset R$ the inverse image of $\sqrt{IS}$ in $R$ is equal to $\sqrt{I}$.

  3. For any radical ideal $I \subset R$ the inverse image of $IS$ in $R$ is equal to $I$.

  4. For every prime $\mathfrak p$ of $R$ the inverse image of $\mathfrak p S$ in $R$ is $\mathfrak p$.

In this case the same is true after any base change: Given a ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ has the equivalent properties (1), (2), (3) as well.

Proof. If $J \subset S$ is an ideal, then $\sqrt{\varphi ^{-1}(J)} = \varphi ^{-1}(\sqrt{J})$. This shows that (2) and (3) are equivalent. The implication (3) $\Rightarrow $ (4) is immediate. If $I \subset R$ is a radical ideal, then Lemma 10.16.2 guarantees that $I = \bigcap _{I \subset \mathfrak p} \mathfrak p$. Hence (3) $\Rightarrow $ (2). By Lemma 10.16.9 we have $\mathfrak p = \varphi ^{-1}(\mathfrak p S)$ if and only if $\mathfrak p$ is in the image. Hence (1) $\Leftrightarrow $ (4). Thus (1), (2), (3), and (4) are equivalent.

Assume (1) holds. Let $R \to R'$ be a ring map. Let $\mathfrak p' \subset R'$ be a prime ideal lying over the prime $\mathfrak p$ of $R$. To see that $\mathfrak p'$ is in the image of $\mathop{\mathrm{Spec}}(R' \otimes _ R S) \to \mathop{\mathrm{Spec}}(R')$ we have to show that $(R' \otimes _ R S) \otimes _{R'} \kappa (\mathfrak p')$ is not zero, see Lemma 10.16.9. But we have

\[ (R' \otimes _ R S) \otimes _{R'} \kappa (\mathfrak p') = S \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \]

which is not zero as $S \otimes _ R \kappa (\mathfrak p)$ is not zero by assumption and $\kappa (\mathfrak p) \to \kappa (\mathfrak p')$ is an extension of fields. $\square$


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