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Suppose that $R$ is a ring, not necessarily commutative nor associative. Assume that for every non-zero $a \in R$, the left multiplication map $$ \lambda_a \colon R \to R \colon x \mapsto ax $$ is invertible. (We do not assume that its inverse is again a left multiplication map $\lambda_b$ for some $b \in R$.)

Is such a ring $R$ necessarily unital (i.e., does it have a unit $1 \in R$), and is it then a division ring (i.e., does every element $a$ have a two-sided inverse)?

Note: if $R$ is unital and associative and every $\lambda_a$ ($a \neq 0$) is invertible, then it is indeed a division ring, and the inverse of $\lambda_a$ is equal to $\lambda_b$ where $b=a^{-1}$.

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  • $\begingroup$ @YCor: Indeed, I meant to assume that every $\lambda_a$ with $a \neq 0$ is invertible. I've edited the question to fix this. $\endgroup$
    – Tom De Medts
    Commented Apr 28, 2016 at 12:02
  • $\begingroup$ I guess associativity should be some minimal requirement. $\endgroup$
    – Andreas Thom
    Commented Apr 28, 2016 at 12:27
  • $\begingroup$ @AndreasThom: In what sense? $\endgroup$
    – Tom De Medts
    Commented Apr 28, 2016 at 12:29
  • $\begingroup$ In order to expect the possibility of a positive answer. $\endgroup$
    – Andreas Thom
    Commented Apr 28, 2016 at 12:31
  • $\begingroup$ I would have thought that maybe in the associative case there must be a unit, or at least the counterexample would be interesting. Anyway, I do not know many positive results about non-associative rings. $\endgroup$
    – Andreas Thom
    Commented Apr 28, 2016 at 12:36

2 Answers 2

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What if your ring is the $\mathbf{R}$-algebra $\mathbf{R}^2$ with the bilinear law $$(x,y)(z,t)=\begin{pmatrix}2x & -y \\ y & x\end{pmatrix}\begin{pmatrix}z \\ t\end{pmatrix}=(2xz-yt,yz+xt)\quad?$$ It's even commutative.

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Another example. Let $R=\mathbb{C}$ as an additive group, with multiplication $$(w,z)\mapsto\overline{wz}.$$

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    $\begingroup$ well, it's a variant of mine. More generally, you can take any linear map $f:K^n\to M_n(K)$ such that $f(x)$ is invertible for all $x\neq 0$, any linear automorphism $g$ of $K^n$ not in the range of $f$, and define the multiplication on $K^n$ $(x,y)\mapsto f(x)g^{-1}(y)$. $\endgroup$
    – YCor
    Commented Apr 28, 2016 at 14:20
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    $\begingroup$ @YCor Ah, OK. The way I was thinking of my example was to take an associative unital multiplication $R\otimes|_\mathbb{Z}R\to R$ and compose with a random group automorphism $R\to R$. $\endgroup$
    – Jeremy Rickard
    Commented Apr 28, 2016 at 16:10
  • $\begingroup$ Ah OK. Indeed bilinear laws can be freely twisted on the right by automorphisms, this affects associativity, left/right units, but not commutativity or invertibility of left/right translations. $\endgroup$
    – YCor
    Commented Apr 28, 2016 at 16:15

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