Definition 10.64.1. 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 $\mathfrak p^{(n)} = \mathop{\mathrm{Ker}}(R \to R_\mathfrak p/\mathfrak p^ nR_\mathfrak p)$.
10.64 Symbolic powers
Here is the definition.
Note that $\mathfrak p^ n \subset \mathfrak p^{(n)}$ but equality does not always hold.
Lemma 10.64.2. 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 10.64.3. 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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