Timeline for Invertibility of all left multiplication maps in non-unital rings
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2016 at 16:15 | comment | added | YCor | 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. | |
| Apr 28, 2016 at 16:10 | comment | added | Jeremy Rickard | @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 $R\to R$. | |
| Apr 28, 2016 at 14:20 | comment | added | YCor | 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 $K^n$ $(x,y)\mapsto f(x)g^{-1}(y)$. | |
| Apr 28, 2016 at 13:07 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |