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?)

  • $\begingroup$ 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. $\endgroup$ Commented Aug 12, 2023 at 13:51

1 Answer 1


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 $\mathfrak p^{(3)}=(x^2,xz)$.


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