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*}

10.63 Symbolic powers

Here is the definition.

Definition 10.63.1. Let $R$ be a ring. Let $\mathfrak p$ be a prime ideal. For $n \geq 0$ the $n$th symbolic power of $\mathfrak p$ is the ideal $\mathfrak p^{(n)} = \mathop{\mathrm{Ker}}(R \to R_\mathfrak p/\mathfrak p^ nR_\mathfrak p)$.

Note that $\mathfrak p^ n \subset \mathfrak p^{(n)}$ but equality does not always hold.

Lemma 10.63.2. Let $R$ be a Noetherian ring. Let $\mathfrak p$ be a prime ideal. Let $n > 0$. Then $\text{Ass}(R/\mathfrak p^{(n)}) = \{ \mathfrak p\} $.

Proof. If $\mathfrak q$ is an associated prime of $R/\mathfrak p^{(n)}$ then clearly $\mathfrak p \subset \mathfrak q$. On the other hand, any element $x \in R$, $x \not\in \mathfrak p$ is a nonzerodivisor on $R/\mathfrak p^{(n)}$. Namely, if $y \in R$ and $xy \in \mathfrak p^{(n)} = R \cap \mathfrak p^ nR_{\mathfrak p}$ then $y \in \mathfrak p^ nR_{\mathfrak p}$, hence $y \in \mathfrak p^{(n)}$. Hence the lemma follows. $\square$

Lemma 10.63.3. Let $R \to S$ be flat ring map. Let $\mathfrak p \subset R$ be a prime such that $\mathfrak q = \mathfrak p S$ is a prime of $S$. Then $\mathfrak p^{(n)} S = \mathfrak q^{(n)}$.

Proof. Since $\mathfrak p^{(n)} = \mathop{\mathrm{Ker}}(R \to R_\mathfrak p/\mathfrak p^ nR_\mathfrak p)$ we see using flatness that $\mathfrak p^{(n)} S$ is the kernel of the map $S \to S_\mathfrak p/\mathfrak p^ nS_\mathfrak p$. On the other hand $\mathfrak q^{(n)}$ is the kernel of the map $S \to S_\mathfrak q/\mathfrak q^ nS_\mathfrak q = S_\mathfrak q/\mathfrak p^ nS_\mathfrak q$. Hence it suffices to show that

\[ S_\mathfrak p/\mathfrak p^ nS_\mathfrak p \longrightarrow S_\mathfrak q/\mathfrak p^ nS_\mathfrak q \]

is injective. Observe that the right hand module is the localization of the left hand module by elements $f \in S$, $f \not\in \mathfrak q$. Thus it suffices to show these elements are nonzerodivisors on $S_\mathfrak p/\mathfrak p^ nS_\mathfrak p$. By flatness, the module $S_\mathfrak p/\mathfrak p^ nS_\mathfrak p$ has a finite filtration whose subquotients are

\[ \mathfrak p^ iS_\mathfrak p/\mathfrak p^{i + 1}S_\mathfrak p \cong \mathfrak p^ iR_\mathfrak p/\mathfrak p^{i + 1}R_\mathfrak p \otimes _{R_\mathfrak p} S_\mathfrak p \cong V \otimes _{\kappa (\mathfrak p)} (S/\mathfrak q)_\mathfrak p \]

where $V$ is a $\kappa (\mathfrak p)$ vector space. Thus $f$ acts invertibly as desired. $\square$


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