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

The second bullet point from this nLab page 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 this MSE post 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?

PrincessEev
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Sasha
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  • What does "power-associative" mean? Maybe the second answer wouldn't be true without that assumption. – coffeemath Dec 12 '18 at 05:32
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    @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 $a^i$). – Stahl Dec 12 '18 at 05:47
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    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 – Jyrki Lahtonen Dec 12 '18 at 07:30
  • Caveat: I have never seriously studied non-associative division algebras, so take my comments with a grain of salt. – Jyrki Lahtonen Dec 12 '18 at 07:35
  • Inverses are two sided – R.C.Cowsik Dec 12 '18 at 09:37

1 Answers1

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In finite-dimensional unital algebras, being free of zero divisors is equivalent to being a division algebra.

It's not possible if the algebra under consideration is power-associative or flexible, for instance. Because it's a division algebra it must have even dimension bounded by 8. You ask for the right-inverse and left-inverse to be the same, that is the reason you can construct such algebra, we only need to build a division algebra with at least one element whose right and left inverses do not coincide.

In this article it's shown a general method to obtain division algebras using the Cayley-Dickson construction.

For an example, follow this answer and consider $\mathrm{Cay}(\mathbb{H},i)$ given by the vector space of pairs of Quaternions (the usual, generated by $1,i,j,k$) and with multiplication given by:

$(u,v)\cdot(u',v') = (u\cdot u'+ i(\bar{v'}v), v'u + v\bar{u'})$

It's a particular case of the construction above so the article deals with the affirmation "it is a division algebra".

Now take $v=(0,1+i$). The right-inverse for $v$ must satisfy: $(0,1+i)\cdot(x,y) =(i(\bar{y}(1+i)),(1+i)\bar{x})=(1,0)$ implying $x=0$ and $y=\frac{-1+i}{2}$. It can not be a left-inverse because $(0,\frac{-1+i}{2})\cdot(0,1+i)=(-1,0)$.

AHandsomeAlien
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