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 10.59.10. Let $R$ be a Noetherian ring. Let $x \in R$.

  1. If $\mathfrak p$ is minimal over $(x)$ then the height of $\mathfrak p$ is $0$ or $1$.

  2. If $\mathfrak p, \mathfrak q \in \mathop{\mathrm{Spec}}(R)$ and $\mathfrak q$ is minimal over $(\mathfrak p, x)$, then there is no prime strictly between $\mathfrak p$ and $\mathfrak q$.

Proof. Proof of (1). If $\mathfrak p$ is minimal over $x$, then the only prime ideal of $R_\mathfrak p$ containing $x$ is the maximal ideal $\mathfrak p R_\mathfrak p$. This is true because the primes of $R_\mathfrak p$ correspond $1$-to-$1$ with the primes of $R$ contained in $\mathfrak p$, see Lemma 10.16.5. Hence Lemma 10.59.7 shows $\dim (R_\mathfrak p) = 1$ if $x$ is not nilpotent in $R_\mathfrak p$. Of course, if $x$ is nilpotent in $R_\mathfrak p$ the argument gives that $\mathfrak pR_\mathfrak p$ is the only prime ideal and we see that the height is $0$.

Proof of (2). By part (1) we see that $\mathfrak q/\mathfrak p$ is a prime of height $1$ or $0$ in $R/\mathfrak p$. This immediately implies there cannot be a prime strictly between $\mathfrak p$ and $\mathfrak q$. $\square$


Comments (2)

Comment #3527 by Jonas Ehrhard on

I think in the proof of part (2) we want to mod out instead of , because , so .

The same holds for the proof of 0BBZ which is essentially the same.


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