3
$\begingroup$

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.

1 Answers 1

5

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 $a\circ e -e=0$.

  • 0
    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?2012-07-30
  • 0
    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?2012-07-30
  • 0
    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.2012-07-30
  • 0
    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?2012-07-30
  • 0
    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 $e$.2012-07-30
  • 0
    Thanks again! I get that $0$, $e$ and $a\circ e$ are idempotent in $(R,\circ)$. If $a\circ e=0$, I get $ae=a+e$, which implies $aea=a^2+ea$ which implies $0=ea=a$, but plugging back in $0\circ e=e\neq 0$ is a contradiction. The only thing I'm missing is why $0$ and $e$ are the _only_ idempotents in $(R,\circ)$. I see $a\circ a=a$ implies $a^2=a$ in $(R,\cdot)$, but I don't see what goes wrong if $a\neq 0,e$. Do you mind helping me fill in this last detail?2012-07-30
  • 0
    Suppose idempotent $b$ is not $e$. Then it is RQR, so $b\circ c=0$ for some $c$. But then $0=b\circ c=b\circ b\circ c=b\circ 0=b$.2012-07-31
  • 0
    @rschwieb What is "$1$" in the mapping $a\mapsto 1-a$?2014-11-09
  • 0
    Dear @alonsos : sorry, I don't know what you're asking. What is a "1 in a mapping"?2014-11-09
  • 0
    @rschwieb I mean the element of $R$ named "$1$" in the mapping $a\mapsto 1-a$. It seems to me that for the mappping to be an isomorphism from $(R,\cdot)$ to $(R,\circ)$ you must first assume it is the identity in $(R,\cdot)$. Given the latter, then, it seems unnecessary to prove there is an identity element $e$ in $(R,\cdot)$.2014-11-09
  • 0
    @rschwieb Expanding on the above comment: Should $R$ be a ring $1$ would be the multiplicative identity, but that does not hold for $R$ rng, where "$1-a$" ceases to be defined. Also, this problem deals with right quasi-inverses. In Rotman's Introduction to the theory of groups, left quasi inverses are defined, with $a\circ b = a+b-ba$, and there the mapping $1-a$ doesn't seem to work: now matter how I go about it $f(a\cdot b)=1-ab$ while $f(a)\circ f(b)=1-a+1-b-(1-b)(1-a)=1-ba\neq 1-ab$. Is there any workaround?2014-11-09
  • 0
    @alonsos So the structure of the solution is "here is how it works if there is an identity. Now let me show why it has an identity." The latter is certainly necessary for the former to be useful at all. I think you will see what I mean if you read it one more time. Regards.2014-11-09
  • 0
    @alonsos You seem to be over-concerned that the ring does not have an identity by assumption, but that's irrelevant since we can prove it has an identity.2014-11-09
  • 0
    @rschwieb You're absolutely right, I finally see what you mean about the identity. I still can't make the mapping work for left quasi inverses, though.2014-11-09
  • 0
    @alonsos you've got me at a loss for words again :) what does being quasi invertible have to do with the mapping working?2014-11-09
  • 0
    An addendum: we can show $a\circ e$ not RQR a little more directly. Since, if $(a\circ e)\circ b=0$ then $a\circ(e\circ b)=a\circ e=0$.And now as Hailie pointed out this leads to the contradiction $e=0$.2018-02-28