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 10.151.6. Let $R$ be a Noetherian normal domain with fraction field $K$. Then

  1. for any nonzero $a \in R$ the quotient $R/aR$ has no embedded primes, and all its associated primes have height $1$

  2. \[ R = \bigcap \nolimits _{\text{height}(\mathfrak p) = 1} R_{\mathfrak p} \]
  3. For any nonzero $x \in K$ the quotient $R/(R \cap xR)$ has no embedded primes, and all its associates primes have height $1$.

Proof. By Lemma 10.151.4 we see that $R$ has $(S_2)$. Hence for any nonzero element $a \in R$ we see that $R/aR$ has $(S_1)$ (use Lemma 10.71.6 for example) Hence $R/aR$ has no embedded primes (Lemma 10.151.2). We conclude the associated primes of $R/aR$ are exactly the minimal primes $\mathfrak p$ over $(a)$, which have height $1$ as $a$ is not zero (Lemma 10.59.10). This proves (1).

Thus, given $b \in R$ we have $b \in aR$ if and only if $b \in aR_{\mathfrak p}$ for every minimal prime $\mathfrak p$ over $(a)$ (see Lemma 10.62.19). These primes all have height $1$ as seen above so $b/a \in R$ if and only if $b/a \in R_{\mathfrak p}$ for all height 1 primes. Hence (2) holds.

For (3) write $x = a/b$. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be the minimal primes over $(ab)$. These all have height 1 by the above. Then we see that $R \cap xR = \bigcap _{i = 1, \ldots , r} (R \cap xR_{\mathfrak p_ i})$ by part (2) of the lemma. Hence $R/(R \cap xR)$ is a submodule of $\bigoplus R/(R \cap xR_{\mathfrak p_ i})$. As $R_{\mathfrak p_ i}$ is a discrete valuation ring (by property $(R_1)$ for the Noetherian normal domain $R$, see Lemma 10.151.4) we have $xR_{\mathfrak p_ i} = \mathfrak p_ i^{e_ i}R_{\mathfrak p_ i}$ for some $e_ i \in \mathbf{Z}$. Hence the direct sum is equal to $\bigoplus _{e_ i > 0} R/\mathfrak p_ i^{(e_ i)}$, see Definition 10.63.1. By Lemma 10.63.2 the only associated prime of the module $R/\mathfrak p^{(n)}$ is $\mathfrak p$. Hence the set of associate primes of $R/(R \cap xR)$ is a subset of $\{ \mathfrak p_ i\} $ and there are no inclusion relations among them. This proves (3). $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.151: Serre's criterion for normality

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 031T. Beware of the difference between the letter 'O' and the digit '0'.