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Let $R$ be a ring with unity.

  1. An element $a\in R$ is said to be a unit element if there exists $b\in R$ such that $ab=ba=1$. The ring $R$ is called a division ring if every nonzero element is a unit element.
  2. An element $f\in R$ is said to be a full element if there exists $r,s\in R$ such that $rfs=1$. Every unit element is a full element. If $R$ is commutative, then every full element is a unit element.

I am looking for an example of a ring $R$ whose every nonzero element is a full element but it has at least one nonzero element which is not a unit, i.e., $R$ is not a division ring.

Please suggest me something so that I can find this example.

Jonas Linssen
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Dr. Nirbhay Kumar
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    Could you tell us what you have tried and what examples are you aware of? For example, are you aware of examples of full elements which are not units? – Aryaman Maithani Jul 05 '21 at 06:48

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Let $R$ be the ring of linear transformations of a countably infinite dimensional vector space.

It’s elementary to show that the transformations with infinite dimensional images are full elements, and they remain full when you quotient by the (unique, in this case) maximal ideal $M$ of $R$.

So now it is up to you to verify the hints.

rschwieb
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  • I am not familiar with non-commutative algebra but is it not true that if $M \subset R$ is maximal, then $R/M$ is a division ring? Moreover, the $R$ you mention, does that not already satisfy the conditions that OP wants? Why are we quotient-ing? – Aryaman Maithani Jul 05 '21 at 12:19
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    This is not true in the non commutative case: take $R=M_n(K)$ and $M=(0)$. Then $M$ is maxmal but $R/M$ is not a division ring. – GreginGre Jul 05 '21 at 14:49
  • @AryamanMaithani of course $R$ will not suffice: for any transformation $f$ with finite dimensional image, $RfR\neq R$, so $f$ would not be full. – rschwieb Jul 05 '21 at 15:43
  • Thank you, @GreginGre. Funnily I was aware of that example as an example to see that a ring can have only two ideals without being a division ring but I got confused when I looked up and saw the result being true for maximal one-sided ideals. – Aryaman Maithani Jul 05 '21 at 16:02
  • @rschwieb: Thanks. I misread and somehow considered $R$ to consist of those transformations having infinite-dimensional images. If I understand correctly, here $M$ is the (two-sided) ideal consisting of transformations having finite-dimensional image and then $R/M$ is the example for OP? – Aryaman Maithani Jul 05 '21 at 16:07
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    @AryamanMaithani yes, I believe so! – rschwieb Jul 05 '21 at 19:41
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    @AryamanMaithani of course if I had chosen a vector space of higher dimension, the description of the maximal ideal would have to be changed, but everything else holds. I’ve always thought of von Neumann regular rings being akin to division rings. – rschwieb Jul 05 '21 at 21:09
  • @rschwieb: I see, my guess is that if $\kappa$ is the dimension of the vector space, then $M$ is the maximal ideal consisting of those transformations having image of dimension $< \kappa$. – Aryaman Maithani Jul 05 '21 at 21:26
  • @AryamanMaithani yes, that’s how it works. – rschwieb Jul 05 '21 at 22:30