# Associated prime $\mathfrak{p}$ of $I$ written as $(I:x)$ for $x \notin \mathfrak{p}$.

Let $R$ be a Noetherian ring, let $I$ be an ideal of $R$ and let $\mathfrak{p}$ be a prime ideal associated to $I$. Then it is known from the standard theory that $\mathfrak{p} = (I:x)$ for some $x \in R$.

Is it also true that $\mathfrak{p} = (I:y)$ for some $y \in R\setminus \mathfrak{p}$?

Background: This argument appears in the proof of the Artin Approximation theorem in the book "Local Analytic Geometry" by Pfister/DeJong (in the situation of the proof $\mathfrak{p}$ is minimal over $I$, but this is not explicitely mentioned). Usually the authors' arguments are quite detailed and at this step of the proof nothing more then the Noetherness is mentioned.

The existence of such $y$ implies that $I_p = p_p$: Suppose there exists such $y$. Then $p_p = (I :_R y)_p = (I_p :_{R_p} y_p) = I_p$ since $y_p$ is a unit in $R_p$.
Therefore, you can find an example where such $y$ doesn't exists. More specifically take an ideal $I$ which has embedded associated primes and take $p$ as one of those embedded ones.
• Actually in the proof one has that $I_{\mathfrak{p}} = \mathfrak{p}_{\mathfrak{p}}$. Do you know whether under this assumption the statement holds? (I.e., is the converse to what you showed also true? – Sebastian Oct 10 '13 at 18:54
• If $p$ appears in the primary decomposition and is minimal over $I$, I believe it is true (if an ideal is self-radical, i.e., reduced, then any $p$ associated to $I$ satisfies the condition). However, it is not true in general. Take an Artinian local ring. Then the maximal ideal is associated to any ideal in the ring, but unless it is a field there exist ideals which are not the maximal ideal. In particular, one can take $(0)$. – Youngsu Oct 11 '13 at 3:40
• If we have that $I_{\mathfrak{p}} = \mathfrak{p}$, then $\mathfrak{p}$ is part of any primary decomposition of $I$. Because if $\mathfrak{q_{1}}, \dots, \mathfrak{q_{k}}$ is the part of an irredundant primary decomposition of $I$ whose associated prime is $\mathfrak{p}$, then the localization $\mathfrak{q_{1}} R_{\mathfrak{p}}, \dots, \mathfrak{q_{k}} R_{\mathfrak{p}}$ is an irredundant decomoposition of $I R_{\mathfrak{p}} = \mathfrak{p} R_{\mathfrak{p}}$. So $k=1$, $\mathfrak{q_{1}} R_{\mathfrak{p}} = \mathfrak{p} R_{p}$. – Sebastian Oct 11 '13 at 7:18
• And since $(\mathfrak{q} R_{\mathfrak{p}}) \cap R = \mathfrak{q}$ for any primary ideal $\mathfrak{q} \subset \mathfrak{p}$, we have that $\mathfrak{q} = \mathfrak{p}$, which yields that $\mathfrak{p}$ appears in the primary decomposition. – Sebastian Oct 11 '13 at 7:20
If $I_{\mathfrak{p}} = \mathfrak{p}$, then the assertion is true. As discussed in the comment above, this assumption yields that $\mathfrak{p}$ is a prime appearing in a (irredundant) decomposition of $I$. Further there exists an $h \in R\setminus{p}$ so that $h\cdot \mathfrak{p} \subset I$. Let $I$ be generated by $(f_{1}, \dots, f_{m})$ and let $\mathfrak{p}$ be generated by $(g_{1}, \dots, g_{n})$. Then we can find $s_{i,j} \in R\setminus \mathfrak{p}$ and $r_{i,j} \in R$ so that $g_{i} = \sum_{j=1}^{m} \frac{r_{i,j}}{s_{i,j}} f_{i,j}$. Setting $h:= \prod_{i,j} s_{i,j}$ yields then $h \cdot \mathfrak{p} \subset I$, so $\mathfrak{p} \subset (I:h)$. This implies that $(I:h)$ is prime, for if $ab \in (I:h)$, then $h \cdot ab \in I \subset \mathfrak{p}$, and since $h \notin \mathfrak{p}$ we either have $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$, yielding $a \in (I:h)$ or $b \in (I:h)$, so either $(I:h)$ is prime or all of $R$. But the later is impossible, because this would mean $h \cdot R \subset I$ and consequently $R_{\mathfrak{p}} \subset I_{\mathfrak{p}} = \mathfrak{p} R_{\mathfrak{p}}$. So $(I:h) \in Ass(I)$, and since $I$ has no embedded primes over $\mathfrak{p}$, we conclude that $(I:h) = \mathfrak{p}$.