A division algebra is defined as a (not necessarily finite dimensional, associative, or unital) algebra $A$ over a field, where $\forall a\neq0,b\in A$ the equations $ax=b$ and $ya=b$ have unique solutions. Are all of the following correct?

(1) This definition is equivalent to the maps $L_{a},R_{a}:A\to A$ (defined by left and right multiplication by $a$) having inverse maps.

(2) If $A$ is finite dimensional, then the definition is equivalent to requiring that there be no zero divisors, i.e. $uv=0\Rightarrow u=0$ or $v=0$.

(3) If $A$ is unital, then the definition is equivalent to requiring that unique right and left inverses exist for every non-zero element.

(4) If $A$ is finite dimensional and associative, then $A$ is unital and the left and right inverses are equal, turning the non-zero vectors into a group under multiplication.

The seed of the proof of (2) is here, and I think I understand why the others are true, but all the references I checked mostly state without proof, so any quick proofs would be nice.

  • $\begingroup$ I think you need to be more precise about what you mean by algebras. Algebras over what? Field? Commutative rings? $\endgroup$ – Kimball Mar 29 '18 at 16:58
  • $\begingroup$ Yes, "algebra" is assumed to be over a field; edited the question to clarify, thanks. $\endgroup$ – Adam Marsh Mar 29 '18 at 17:15

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