# Artin's Theorem on Alternative Algebras

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## Theorem

Let $A = \struct {A_R, \oplus}$ be an algebra over the ring $R$ such that $A$ is *not* a boolean algebra.

Then $A$ is alternative if and only if:

- $\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
- $\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$

## Alternative Definition

Some sources take this as a definition of an alternative algebra and from it deduce that $A$ is alternative if and only if:

- for all $a, b \in A_R$, the subalgebra generated by $\set {a, b}$ is an associative algebra.

However, the latter statement is how an alternative algebra is defined on this site, and from it we deduce the statements given in the exposition of this theorem.

The approaches are clearly equivalent.

## Proof

When $A$ is

So suppose that:

- $\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
- $\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$

Then:

- $\sqbrk {a, a, b} = 0$
- $\sqbrk {b, a, a} = 0$

where $\sqbrk {a, a, b}$ denotes the associator of $a, b \in A_R$.

Now let us compute, using the linearity of $\oplus$ and the two suppositions:

\(\displaystyle \paren {a - b} \oplus \paren {\paren {a - b} \oplus a}\) | \(=\) | \(\displaystyle a \oplus \paren {\paren {a - b} \oplus a} - b \oplus \paren {\paren {a - b} \oplus a}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \oplus \paren {a \oplus a} - a \oplus \paren {b \oplus a} - b \oplus \paren {a \oplus a} + b \oplus \paren {b \oplus a}\) | |||||||||||

\(\displaystyle \paren {\paren {a - b} \oplus \paren {a - b} } \oplus a\) | \(=\) | \(\displaystyle \paren {a \oplus a} \oplus a - \paren {a \oplus b} \oplus a - \paren {b \oplus a} \oplus a + \paren {b \oplus b} \oplus a\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \sqbrk {a - b, a - b, a}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sqbrk {a, a, a} - \sqbrk {a, b, a} - \sqbrk {b, a, a} + \sqbrk {b, b, a}\) | Subtract the two expressions | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sqbrk {a, b, a}\) |

$\blacksquare$

## Source of Name

This entry was named for Emil Artin.

## Sources

*This article incorporates material from alternative algebra on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.*