It is a well-known fact that in ZF, the axiom of choice is equivalent to the statement that every commutative ring has a maximal ideal. On the other hand, for Noetherian rings, this is not necessary (either in the sense that it only requires dependent choice or that it requires no choice at all, depending on your definition of Noetherian). It can also be shown (in ZFC) that every Artin ring is Noetherian, and the proof of this fact in Atiyah-MacDonald essentially relies on the following two results:

  1. In an Artin ring, every prime ideal is maximal (proved without choice), and

  2. The nilradical of a ring is the intersection of all of the prime ideals in a ring (proved with choice).

Result 2. is equivalent to the statement that every Artin ring contains a prime ideal. To see this, note simply that given a non-nilpotent element $f$ in a ring $A$, there is a prime ideal of A not containing $f$ if and only if $A_f$ has a prime ideal. According to this answer, the fact that every commutative ring has a prime ideal is equivalent in ZF to the Boolean prime ideal theorem, so at least full choice is not necessary. Can this be weakened for Artin rings? Alternatively, is there another proof that Artin $\Rightarrow$ Noetherian that uses a weaker form of the axiom of choice (or no form of it)? Thank you.


1 Answer 1


Suppose $A$ is a nonzero artinian ring. Then the collection of all nonzero (possibly improper) ideals of $A$ has a minimal element, say $I\subseteq A$. Then $I$ has no nonzero proper submodules, and so is a simple $A$-module. In particular, it must be cyclic (generated by any nonzero element) and the kernel of a surjection $A\to I$ will be a maximal ideal of $A$.

(Assuming "artinian" means "every nonempty collection of ideals has a minimal element", this does not use Choice. Assuming "artinian" means "every descending sequence of ideals stabilizes", it uses only Dependent Choice.)

  • $\begingroup$ Thanks, that's great! Slightly more explicitly, given a minimal $I$ as you suggest, $Ann(I)$ is a maximal ideal of A. That's much simpler than I had expected. $\endgroup$
    – SeanC
    Apr 16, 2017 at 0:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.