# Invertibility of all left multiplication maps in non-unital rings

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 $b=a^{-1}$.

• @YCor: Indeed, I meant to assume that every $\lambda_a$ with $a \neq 0$ is invertible. I've edited the question to fix this. – Tom De Medts Apr 28 '16 at 12:02
• I guess associativity should be some minimal requirement. – Andreas Thom Apr 28 '16 at 12:27
• @AndreasThom: In what sense? – Tom De Medts Apr 28 '16 at 12:29
• In order to expect the possibility of a positive answer. – Andreas Thom Apr 28 '16 at 12:31
• 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. – Andreas Thom Apr 28 '16 at 12:36

## 2 Answers

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.

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

• 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 $K^n$ $(x,y)\mapsto f(x)g^{-1}(y)$. – YCor Apr 28 '16 at 14:20
• @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 $R\to R$. – Jeremy Rickard Apr 28 '16 at 16:10
• 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. – YCor Apr 28 '16 at 16:15