This is exercise 8 on page 11 of Atiyah-MacDonald.

The question is:

Let $A$ be a ring $\neq 0$. Show that the set of prime ideals of $A$ has minimal elements with respect to inclusion.

I'm wondering whether there's a constructive proof of this fact that gives me a minimal prime element in some direct way.

This is my actual, non-constructive proof. My attempt to begin a constructive one is below.

I can show this using Zorn's Lemma.

First, I note that every nonzero ring contains a maximal ideal and that maximal ideals are prime ideals.

Let $C$ be a chain of prime ideals. $I = \cap C$ contains $0$, and furthermore is closed under addition and multiplication by an arbitrary element of $A$. So $I$ is an ideal.

Additionally, the complement of $I$ is closed under multiplication, since it is the union of a chain of sets that are closed under multiplication $(\cap C)^c = \left(\bigcup_{x \in C} \, x^c\right)$.

Since every chain contains a lower bound, it follows then by Zorn's Lemma that there exists a minimal element among the prime ideals of $A$ ordered by inclusion.

This is my attempt at a constructive proof.

The very first thing that I tried though in order to solve this problem was to actually find the ideal.

  1. If $A$ is finite, we walk the prime ideals and pick a minimal one.

  2. If $A$ is an infinite integral domain, we pick $(0)$ as our prime ideal.

  3. If $A$ is an infinite ring that is not an integral domain where all zero divisors are nilpotent, then the set of all nilpotent elements $N$ is a minimal prime ideal. So element of $x \in N$ can be in the complement of any ideal $J$ because then $x^n = 0$ would also be in the multiplicative closure of $J^c$, which would mean that $J$ is not prime. So we have an explicit minimal prime ideal.

In the remaining case, $A$ is infinite, not an integral domain, and has at least one zero divisor that is not nilpotent. I'm stuck on how to construct an explicit minimal prime ideal here, assuming that it's even possible in general.


1 Answer 1


It's consistent with ZF that there exists a non-zero ring with no prime ideals at all, let alone minimal prime ideals. I take this to mean that there can be no "constructive" proof of the existence of minimal prime ideals.

Indeed, if $R$ is a Boolean ring, then every prime ideal is maximal, and hence maximal ideals, prime ideals, and minimal prime ideals are all the same. Furthermore, for any prime ideal $I$, the set $\{1+r\mid r\in I\}$ is a ultrafilter on the corresponding Boolean algebra.

For an explicit example, it is consistent with ZF that $\mathcal{P}(\mathbb{N})/I$ has no prime ideals. Here $\mathcal{P}(\mathbb{N})$ is the power set of $\mathbb{N}$, with its Boolean ring structure, and $I$ is the ideal of finite sets. Indeed, a prime ideal on $\mathcal{P}(\mathbb{N})/I$ would pull back to a prime ideal on $\mathcal{P}(\mathbb{N})$ containing all the finite sets, which would correspond to a non-principal ultrafilter on the Boolean algebra $\mathcal{P}(\mathbb{N})$ (i.e., on the set $\mathbb{N}$). And it is consistent with ZF that there is no such thing.

  • $\begingroup$ An interesting harder question would be if in ZF you can have a ring that has a prime ideal but no minimal prime ideals. (For what it's worth, in the absence of AC, the "right" definition of minimal prime is probably one that becomes the nilradical when you localize at it. With this definition, I suspect a ring like $C([0,1])$ can probably consistently have no minimal primes.) $\endgroup$ Commented Feb 27, 2023 at 3:06
  • $\begingroup$ (Oh, actually, if you use that definition, then I have already answered that question here.) $\endgroup$ Commented Feb 27, 2023 at 3:19
  • $\begingroup$ @EricWofsey Yes, thanks for the correction. And for the interesting follow-up question and answer! $\endgroup$ Commented Feb 27, 2023 at 14:16

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