# Theorem of Kaplansky, $R$ is a division ring if every element but one is (right) quasi-invertible.

There is a theorem of Kaplansky that seems to pop up every algebra book. Here rng denotes a ring with possibly no identity. As definition, an element $a$ of a rng $R$ is said to be (right) quasi-invertible if there exists a $b\in R$ such that $a\circ b=a+b-ab=0$.

Kaplansky's theorem states that in a rng $R$ where all but one element is right quasi-invertible, then $R$ is actually a division ring, obviously with identity.

I can't find the origin or proof of this theorem though. Does anyone know a proof, or reference to Kaplansky's proof? Thank you, I would appreciate seeing it.

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If $R$ with its normal operation has identity, then the monoid $(R,\circ)$ is (monoid) isomorphic with $(R,\cdot)$ via the mapping $a\mapsto 1-a$ from $(R,\cdot)$ into $(R,\circ)$. You can easily see that 1 is the only non-RQR element, and that all other elements have right inverses in $(R,\circ)$, and hence everything except 1 has two-sided inverses in $(R,\circ)$. By the isomorphism, everything but $0$ has a twosided inverse in $(R,\cdot)$.

So, all of that follows provided we can show that $(R,\cdot)$ has an identity! We have an obvious candidate: $e$, which is the non-RQR element of $(R,\circ)$. I have arranged the hints below to help you complete this task.

1. Show that $e$ is the identity of $(R,\cdot)$ iff $e$ is two-sided absorbing in $(R,\circ)$, that is, $e\circ a=a\circ e=e$ for all $a\in R$.

2. The fact that $e\circ a=e$ for all $a\in R$ follows from $e$ being the only non RQR element of $(R,\circ)$.

3. (Edit:another, hopefully easier route:) Show that the only idempotents in $(R,\circ)$ are $0$ and $e$. Note $a\circ e$ is idempotent. If $a\circ e=e$, we are done. Examine the case when $a\circ e=0$.

1. The fact that $a\circ e=e$ for all $a\in R$ follows from the fact that $0$ is the unique monoid identity of $(R,\circ)$. That is to say, if $(a\circ e -e)\circ b=0$ for all $b$, you can conclude that $a\circ e -e=0$.

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Thanks rschwieb. I can prove hint 1. I'm having trouble with 2. right now. I know that $e\circ a=e$ iff $ea=a$. I tried finding a contradiction by supposing there exists $a\in R$ such that $e\circ a\neq e$, but I can't construct another element which is not RQR. How can I proceed? – Hailie Mathieson Jul 30 '12 at 17:29
If $e\circ a$ is not $e$, then it is RQR. Hence there exists a $b$ such that $e\circ a\circ b=0$... see a contradiction coming? – rschwieb Jul 30 '12 at 19:40
Oh of course, since $(e\circ a)\circ b=0$, but then by associativity $e\circ(a\circ b)=0$, so $e$ is RQR, contradiction. Thanks! I'll give 3. a try now. – Hailie Mathieson Jul 30 '12 at 20:07
For 3., I calculate $a\circ e-e=a-ae$. Then $$(a\circ e-e)\circ b=(a-ae)+b-(a-ae)b=a-ae+b-ab+aeb=a+b-ae$$ since $eb=b$ by hint 2. So I think $(a\circ e-e)\circ b=0$ iff $b=ae-a$. Have I done something wrong here? – Hailie Mathieson Jul 30 '12 at 20:25
I agree with your computation and am now struggling to see how I overcame that. What I wrote is somewhere buried in my trashcan :( In any case, I was hoping to use the fact that 0 is the twosided monoid identity to get two-sidedness out of $e$. – rschwieb Jul 30 '12 at 21:17