Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

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 $a\circ e -e=0$.

share|improve this answer
    
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? –  hmIII 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. –  hmIII 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? –  hmIII 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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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