In this question all rings are commutative, but don't necessarily have a multiplicative identity (so: commutative rngs). On Wikipedia there is the unsourced claim:

It is well known that a local ring which is also a von Neumann regular ring is a division ring.

On the page this appears, von Neumann regular rings are allowed to be non-unital. However on the page dedicated to von Neumann regular rings, the definition states up front the assumption the ring is unital. So I am willing to accept (though a reference would be nice) that a unital commutative von Neumann regular local ring is a field. But I can define a local ring to be a ring with a unique maximal ideal, with no reference to the multiplicative identity. So my question is:

Is a commutative von Neumann regular ring with a unique maximal ideal necessarily a field, even if I don't assume the ring unital a priori?

On the one hand, this seems rather strong: where does the unit come from? But having a unique maximal ideal is rather strong, so perhaps this is enough.

  • 1
    $\begingroup$ Heh, I saw you make that edit. The proof that a local VNR with identity is a division ring is pretty trivial, even for an undergraduate course. I'm not sure where one would find a citation to suit that. One can just note that if $0\neq a=axa$, then $ax$ is an idempotent, and hence has to be $1$ (since the ring is local). So all nonzero elements are units. $\endgroup$
    – rschwieb
    Jul 6, 2020 at 17:17
  • $\begingroup$ I have no idea about the case for commutative local VNR rngs though. There are a lot of questions defining a local ring without identity. Does the maximal ideal have to be modular? Does it have to contain all other proper ideals? All of those things are taken care of when you have an identity. $\endgroup$
    – rschwieb
    Jul 6, 2020 at 17:19

1 Answer 1


Exercise: a local von Neumann regular ring with identity is a division ring.

Proof: If $a$ is nonzero, $axa=a$ for some element $x\in R$, and $ax$ and $xa$ are nonzero idempotents. Since local rings only have trivial idemoptents, $ax=1=xa$. Thus every nonzero element is invertible.

For the purposes of the question, I suppose that "local" for a ring without identity means "has one proper ideal containing all proper ideals." Along with von Neumann regularity, this is enough to prove $R$ has an identity.

Suppose $R$ is VNR and $M$ is the unique maximal ideal. Select $a\in R\setminus M$. Then $a=axa$ and $ax=e$ is an idempotent. Clearly $ax\notin M$, for if it were, $axa=a\in M$.

Then $(e)$ is some ideal of $R$. Suppose $(e)\neq R$: then by our supposition of what "local" means above, $(e)\subseteq M$, contradicting $e\notin M$.

Therefore the only possibility is that $(e)=eR=R$, but then it is easy to see that $e$ is the multiplicative identity of $R$. At that point, the exercise above shows $R$ is a field.

It works with minor modifications in the noncommutative case. Now we suppose there is a proper right ideal containing all proper right ideals, and a proper left ideal containing all proper left ideals.

Use $e=ax$ and $e'=xa$ to argue the same way, and you wind up with $eR=R=Re'$ to get that $e$ is a left identity and $e'$ is a right identity, therefore $e=e'$ is the identity for the ring. The exercise above indicates $R$ is a division ring.

  • $\begingroup$ Aside from making this answer self-contained, I think that's a good an answer I'm going to get. You can rephrase the hypothesis on the ideal as just containing all other proper ideals, since I think that should make it maximal. $\endgroup$ Jul 6, 2020 at 23:29
  • $\begingroup$ I think this argument is simple enough to go on Wikipedia. I will add it myself, if no one gets there before me. $\endgroup$ Jul 6, 2020 at 23:52
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    $\begingroup$ @DavidRoberts I went ahead and made it self-contained with the comment's content. $\endgroup$
    – rschwieb
    Jul 7, 2020 at 14:33

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