Let $A$ be a ring $\neq0$. Show that the set of prime ideals of $A$ has minimal elements with respect to inclusion.

I am trying to do this exercise from Atiyah-Macdonald.

Attempt: We should assume that there is no such minimal prime ideal. Then we have a chain $P_1 \supset P_2 \supset P_3 \supset ...$. Then we should set $P= \bigcap_i P_i$, This would be a minimal element but I can't see why it should be prime?

  • $\begingroup$ Do you know Zorn's lemma ? $\endgroup$ Sep 15, 2020 at 13:25
  • $\begingroup$ yes, to use This i would also need that the intersection of prime ideals is a prime ideal $\endgroup$ Sep 15, 2020 at 14:03
  • 4
    $\begingroup$ @Mark Murray Not just arbitrary intersections, but particular intersections of collections of primes which are totally ordered by inclusion (or more generally rendered into downward directed sets by inclusion). $\endgroup$
    – ΑΘΩ
    Sep 15, 2020 at 14:17
  • $\begingroup$ Strongly useful here: math.stackexchange.com/questions/2724350/… It's not that the intersection of prime ideals is prime (which is false; consider $\langle 2 \rangle \cap \langle 3 \rangle = \langle 6 \rangle$ in $\Bbb{Z}$) but it is true if you know one more more fact about the two ideals. $\endgroup$ Sep 15, 2020 at 21:53
  • 1
    $\begingroup$ @Peter LeFanu Lumsdaine The official term for expressing the notion of ''downward cofinal'' being ''coinitial'', just as a tiny remark on terminology. $\endgroup$
    – ΑΘΩ
    Sep 16, 2020 at 6:02

2 Answers 2


Hint: Prove it by contraposition: if neither $x$ nor $y$ belongs to $\mathfrak p$, then $xy$ is not in $\mathfrak p$.

You'll have to show first that, with this hypothesis on $x$ and $y$, there exists a prime ideal $\mathfrak p_i$ in the chain which contains neither $x$ nor $y$.

A last remark: to show the existence of minimal prime ideals, you should consider a totally ordered (by inclusion) family of prime ideals, not a mere sequence. You have no reason to suppose this family is countable.

  • $\begingroup$ Thanks for a good hint, i'll try this and get back to you! $\endgroup$ Sep 15, 2020 at 14:04

The neat way to argue this is to realise that $\mathscr{Spec}(A)$ is inductively ordered by the dual of the inclusion (so that we may apply Zorn's lemma). More generally, consider a subset $\mathscr{M} \subseteq \mathscr{Spec}(A)$ which is upward directed with respect to the dual of inclusion or equivalently downward directed with respect to inclusion itself. Our objective is to prove that $\mathscr{M}$ has an upper bound with respect to the dual of inclusion, which amounts to a lower bound with respect to inclusion itself.

Since $A$ is not a degenerate ring, $\mathscr{Spec}(A) \neq \varnothing$ is nonempty and any prime ideal serves as a lower bound in the particular case $\mathscr{M}=\varnothing$.

When $\mathscr{M} \neq \varnothing$, let us consider $P\colon=\displaystyle\bigcap\mathscr{M}$ and show it is a prime ideal. As it is the nonempty intersection of a collection of proper ideals, it must also be a proper ideal. Assume by contradiction that it were not prime, which would mean the existence of $a, b \in A \setminus P$ such that $ab \in P$. Since neither $a$ are nor $b$ are in the intersection of all members of $\mathscr{M}$, there must exist ideals $Q, R \in \mathscr{M}$ such that $a \notin Q$ and $b \notin R$. Since $\mathscr{M}$ is donward directed with respect to inclusion, there exists $T \in \mathscr{M}$ such that $T \subseteq Q, R$.

Since $ab \in P \subseteq T$ and $T$ is prime, we must have either $a \in T$ or $b \in T$ which lead either to $a \in Q$ or $b \in R$, both of which are contradictions.

Remark: I have not argued at this level of generality above, but the same claim remains valid for arbitrary non-degenerate rings (the assumption of commutativity is not required).

  • $\begingroup$ This looks like an excellent answer to come back to after I have learned about Spec! $\endgroup$ Sep 15, 2020 at 14:03
  • 3
    $\begingroup$ @Mark Murray For all your purposes that are in question here, regarding this problem, you have no need beyond that of merely knowing it is standard notation for the set of all prime ideals of a given ring (it is true that this notation is introduced in the context of defining the Zariski topology which the prime spectrum -- as it is called -- naturally carries, however here we are only concerned with the support set of this topological space). $\endgroup$
    – ΑΘΩ
    Sep 15, 2020 at 14:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .