division algebra

Let $K$ be a unital ring and $A$ a $K$-algebra. Defining “division”requires special considerations when the algebras are non-associativeso we introduce the definition in stages.

1 Associative division algebras

If $A$ is anassociative algebra then we say $A$ is a division algebraif

1. $A$ is unital with identity $1$. So for all $a\in A$,

   $$a1=1a=a.$$

2. Also every non-zero element of $A$ has an inverse. That is$a\in A$, $a\ne 0$, then there exists a $b\in A$ such that

   $$ab=1=ba.$$

   We denote $b$ by $a^{-1}$ and we may prove $a^{-1}$ is unique to $a$.

The standard examples of associative division algebras are fields, whichare commutative, and the non-split quaternion algebra: $\alpha ,\beta \in K$,

$$\left(\frac{\alpha ,\beta }{K}\right)=\{a_{1}1+a_{2}i+a_{3}j+a_{4}k:i^2=\alpha 1,j^2=\beta 1,k^2=-\alpha \beta 1,ij=k=-ji.\}$$

where $x^2-\alpha$ and $x^2-\beta$ are irreducible over $K$.

2 Non-associative division algebras

For non-associative algebras $A$, the notion of an inverse is not immediate.We use $x.y$ for the product of $x,y\in A$.

Invertible as endomorphisms: Let $a\in A$. Then define $L_a:x\mapsto a.x$and $R_a:x\mapsto x.a$. As the product of $A$ is distributive, both $L_a$ an $R_a$are additive endomorphisms of $A$. If $L_a$ is invertible then we may call $a$“left invertible” and similarly, when $R_a$ is invertible we may call $a$“right invertible” and “invertible” if both $L_a$ and $R_a$ are invertible.

In this model of invertible, $A$ is a division algebra if, and only if,for each non-zero $a\in A$, both $L_a$ and $R_a$ invertible.Equivalently: the equations $a.x=b$ and $y.a=b$ have unique solutionsfor nonzero $a,b\in A$. However, $x$ and $y$ need not be equal.

A common method to produce non-associative division algebras of this sort isthrough Schur’s Lemma.

Invertible in the product:In some instances, the notion of invertible via endomorphisms is notsufficient. Instead, assume $A$ has an identity, that is, an element $1\in A$ suchthat for all $a\in A$,

$$1.a=a=a.1.$$

Next if $a\in A$, we say $a$ is invertible if there exists a $b\in A$such that

$$a.b=1=b.a$$(1)

and furthermore that for all $x\in A$,

$$b.(a.x)=x=(x.a).b.$$(2)

Evidently (1) can be inferred from (2).This added assumption substitutes for the need of associativity in theproofs of uniqueness of inverses and in solving equations with non-associativeproducts.

Proposition 1.

If $A$ is a finite dimensional algebra over a field, theninvertible in this sense forces both $L_a$ and $R_a$ to be invertible as well.

Proof.

Let $x\in A$. Then $x{L}_1=1.x=x=b.(a.x)=x{L}_a{L}_b$. So $L_1=L_a{L}_b$. As$L_1$ is the identity map, $L_a$ is injective and $L_b$ is surjective.As $A$ is finite dimensional, injective and surjective endomorphisms arebijective.∎

In this model, a non-associative algebra is a division algebra $A$ if it isunital and every non-zero element is invertible.

3 Alternative division algebras

The standard examples of non-associative division algebras are actuallyalternative alegbras, specfically, the composition algebras of fields,non-split quaternions and non-split octonions – only the latter areactually not associative. Invertible in the octonions is interpretedin the second stronger form.

Theorem 2 (Bruck-Klienfeld).

Every alternative division algebra is either associative or a non-splitoctonion.

This result is usually followed by two useful results which serve to omitthe need to consider non-associative examples.

Theorem 3 (Artin-Zorn, Wedderburn).

A finite alternative division algebra is associative and commutative, soit is a finite field.

Theorem 4.

An alternative division algebra over an algebraically closed field isthe field itself.