# Ideals contained in a prime $\mathfrak p$ and containing a $\mathfrak p$-primary ideal

Let $$R$$ be a Noetherian ring and $$\mathfrak q$$ a $$\mathfrak p$$-primary ideal. Is it true that every ideal $$\mathfrak q\subseteq \mathfrak a\subseteq \mathfrak p$$ is $$\mathfrak p$$-primary?

The primary ideals $$\mathfrak q\subseteq \mathfrak a\subseteq \mathfrak p$$ are in bijection with the ideals of $$R_{\mathfrak p}$$ containing $$\mathfrak qR_{\mathfrak p}$$, since the localization induces an order-preserving bijection between primary ideals. Now, take any ideal $$\mathfrak qR_{\mathfrak p}\subseteq \mathfrak b\subseteq \mathfrak pR_{\mathfrak p}$$: since $$R$$ is Noetherian, for some $$n$$ we have $$\mathfrak p^n\subseteq \mathfrak q$$, and so $$\mathfrak p^nR_{\mathfrak p}\subseteq \mathfrak qR_{\mathfrak p}$$; hence $$\mathfrak p^nR_{\mathfrak p}\subseteq \mathfrak b\subseteq \mathfrak pR_{\mathfrak p}$$, and, since the latter is maximal, $$\mathfrak b$$ is $$\mathfrak pR_{\mathfrak p}$$-primary. However I don't see how this implies that every $$\mathfrak q\subseteq \mathfrak a\subseteq \mathfrak p$$ is primary: the bijection, as I said, is not on all the ideals, but just the primary ones. I would rather say that two similar results hold: (1) $$\mathfrak aR_{\mathfrak p}\cap R$$ is always $$\mathfrak p$$-primary and (2) if $$\mathfrak m$$ is a maximal ideal, with $$\mathfrak q$$ being $$\mathfrak m$$-primary, every ideal $$\mathfrak q\subseteq \mathfrak a\subseteq \mathfrak m$$ is $$\mathfrak m$$-primary.

(I found the argument, as a side remark, in a proof that any maximal chain of $$\mathfrak p$$-primary ideals, that contain $$\mathfrak q$$, has the same length. But then, why not say that every chain of arbitrary ideals, that are contained in $$\mathfrak p$$ and contain $$\mathfrak q$$, has the same length?)

• We don't need that every $\mathfrak q\subset\mathfrak a\subset\mathfrak p$ is primary. As you said, there is a bijection between primary ideals. Since every ideal $\mathfrak qR_\mathfrak p\subset\mathfrak b\subset\mathfrak pR_\mathfrak p$ is primary, we show that every maximal chain here has the same length. This then implies every maximal chain of $\mathfrak p$-primary ideals containing $\mathfrak q$ has the same length. This is also Atiyah-MacDonald's Exercise 8.6. Commented Aug 12, 2023 at 13:51

Let $$R=K[X,Y,Z]/(XY-Z^2)$$ and $$\mathfrak p=(x,z)$$. (Small letters denote the residue classes of the indeterminates.) It is well known that $$\mathfrak p^2$$ is not primary. On the other side, the symbolic power $$\mathfrak p^{(3)}$$ is primary, and $$\mathfrak p^{(3)}\subsetneq\mathfrak p^2$$. In order to show this inclusion notice that $$\mathfrak p^{(3)}=(x^2,xz)$$.