It is clear that every prime ideal in a commutative ring is primary. The converse is false; for example, in the ring $\mathbb{Z}$ the ideal $(p^2)$ is an example of a primary ideal that is not prime (where $p$ is a prime number). So my question is when does the converse hold:

1) Is there any characterization of commutative rings in which every primary ideal is prime?

One class of rings satisfying the above condition are absolutely flat rings. In fact, it is known that in every absolutely flat ring every primary ideal is maximal (see, for example, exercise 3 in page 55 of "Introduction to Commutative Algebra" by Atiyah & Macdonald). Here is my second question, that is more specific:

2) What would be an example of an ring in which every primary ideal is prime, but not every primary ideal is maximal?

An answer for question 1) would probably solve 2) easily.


  • 2
    $\begingroup$ The powers of maximal ideals are primary and therefore a necessary condition is $m=m^2$ for any maximal ideal. $\endgroup$
    – user26857
    May 12, 2013 at 0:22
  • $\begingroup$ Building on @YACP's comment: Then the Krull Intersection Theorem shows that if $R$ has this property, is Noetherian and either a domain or local, then it must be a field. I think though that there are non-Noetherian domains which have the property: e.g. a valuation ring with value group $\mathbb{Q}$. $\endgroup$ May 12, 2013 at 1:31
  • 1
    $\begingroup$ Well, a valuation ring with value group $\mathbb{Q}$ satisfies YACP's necessary condition at least. Its only nonzero prime ideal is the maximal ideal, which does satisfy $\mathfrak{m} = \mathfrak{m}^2$. I don't see any non-prime primary ideals at the moment, but I'm not yet prepared to claim that there are none. (If this works it answers 2), because the zero ideal is primary but not maximal in this ring.) $\endgroup$ May 12, 2013 at 1:38
  • $\begingroup$ @PeteL.Clark The rings with the property that primary ideals are maximal are von Neumann regular; see here. $\endgroup$
    – user26857
    May 12, 2013 at 6:15
  • $\begingroup$ @YACP: Interesting (though inconclusive). Thanks. $\endgroup$ May 12, 2013 at 16:48

1 Answer 1


Claim: $A$ is a ring such that every primary ideal is prime if and only if $A$ is absolutely flat.

Let us say a ring is a PP(primary is prime) ring if every primary ideal is prime.

Suppose $A$ is absolutely flat, then $A$ is PP. (exercise 3 in page 55 of "Introduction to Commutative Algebra" by Atiyah & Macdonald)

We only need to show that if $A$ is PP then $A$ is absolutely flat.

Notice that every primary ideal of $S^{-1}A$ is of the form $S^{-1}\mathfrak{q}$ where $\mathfrak{q}$ is a $\mathfrak{p}$-primary ideal such that $\mathfrak{p}\cap S=\emptyset$. And if $J/I$ (here $J\supset I$) is a primary ideal of $A/I$ then every zerodivisor in $A/J=(A/I)/(J/I)$ is nilpotent, so $J$ is is primary ideal of $A$.

Now it is easy to show that, if $A$ is PP, then $S^{-1}A$ and $A/I$ is PP for a multiplicatively closed subset $S$ and an ideal $I$ and $\mathfrak{m}^2=\mathfrak{m}$ for every maximal ideal $\mathfrak{m}$.

Based on these properties we are able to prove that PP rings are absolutely flat.

We first show every prime ideal in $A$ is maximal. Suppose not, there are two distinct prime ideals $\mathfrak{p}\subset \mathfrak{m}$ where $\mathfrak{m}$ is maximal. Now $B=(A/\mathfrak{p})_{\mathfrak{m}}$ is a PP, local domain(but not a field) we also use $\mathfrak{m}$ to denote the maximal ideal of $B$. Let $0\neq b\in \mathfrak{m}$, suppose $\mathfrak{q}$ is a minimal prime containing $b$, then $C=B_{\mathfrak{q}}$ is a PP, local domain(not a field) and the only maximal ideal of $C$ is minimal over the ideal $(b)=bC$ so $(b)$ is primary and hence maximal, $(bC)^2=bC$, thus $bC=0$. It is impossible. We have proved that every prime ideal in $A$ is maximal.

Let $\mathfrak{m}$ be any prime ideal in $A$, then $A_{\mathfrak{m}}$ is PP. If $A_{\mathfrak{m}}$ is not a field, pick any nonezero $x\in \mathfrak{m}A_{\mathfrak{m}}$, then $(x)$ is primary hence equals $\mathfrak{m}A_{\mathfrak{m}}$, so $\mathfrak{m}A_m=0$, contradiction.

Hence every prime ideal $\mathfrak{m}$ in $A$ is maximal and $A_{\mathfrak{m}}$ is a field, so $A$ is absolutely flat. We are done.

  • $\begingroup$ Thanks very much! +1 Allow me some time to absorb this argument :) $\endgroup$
    – Prism
    Nov 17, 2013 at 7:54

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