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?)
Best Answer
The answer to your main question is negative. The classical example of prime ideal whose second power is not primary can provide a counterexample.
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 $\mathfrak p^{(3)}=(x^2,xz)$.