Not-necessarily-unital von Neumann regular local commutative rings: are they fields?

Question

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.

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.