This has been bugging me for a while any help would be appreciated.

The second bullet point from here says:

Let A be a non-associative unital algebra with finite dimension, then it's possible to find a case (over R) where A has no zero divisors, but there exists a non-zero element in A that has no inverse (i.e. nonzero x, where xa = ax = 1).

However Prove that any non-zero-divisor of a finite dimensional algebra has an inverse says:

Let A be a non-associative (although power-associative) unital algebra with finite dimension Then if A has no divisors implies every nonzero element in A has an inverse (particularly looking at the proof by Robert Lewis).

Is there a contradiction somewhere, or am I overlooking a use of associativity or technicality?

  • $\begingroup$ What does "power-associative" mean? Maybe the second answer wouldn't be true without that assumption. $\endgroup$ – coffeemath Dec 12 '18 at 5:32
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    $\begingroup$ @coffeemath power-associative means that an expression like $a^3$ is well defined: i.e., that $a\cdot(a\cdot a) = (a\cdot a)\cdot a.$ Robert Lewis' answer definitely makes use of this assumption (you can see this because of the terms $a^i$). $\endgroup$ – Stahl Dec 12 '18 at 5:47
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    $\begingroup$ That nLab page uses some non-standard terminology. For example, in math a division ring is always a ring. And a ring, by definition, is always associative, whereas a division algebra by convention is not necessarily associative. I don't have the time to trace back all the other claims on the nLab page affected by this (if any). Also, their definition of a division algebra simply assumes lack of zero divisors. Not sure whether that is standard or not. I would have thought that a division algebra requires inverses, but I sort of see the potential of just assuming surjectivity of multiplication $\endgroup$ – Jyrki Lahtonen Dec 12 '18 at 7:30
  • $\begingroup$ Caveat: I have never seriously studied non-associative division algebras, so take my comments with a grain of salt. $\endgroup$ – Jyrki Lahtonen Dec 12 '18 at 7:35
  • $\begingroup$ Inverses are two sided $\endgroup$ – R.C.Cowsik Dec 12 '18 at 9:37

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