Artin's Theorem on Alternative Algebras
Contents |
Theorem
Let $A = \left({A_R, \oplus}\right)$ be an algebra over the ring $R$ such that $A$ is not a boolean algebra.
Then $A$ is alternative iff:
- $\forall a, b \in A: \left({a \oplus a}\right) \oplus b = a \oplus \left({a \oplus b}\right)$
- $\forall a, b \in A: \left({b \oplus a}\right) \oplus a = b \oplus \left({a \oplus a}\right)$
Alternative Definition
Some sources take this as a definition of an alternative algebra and from it deduce that $A$ is alternative iff:
- for all $a, b \in A_R$, the subalgebra generated by $\left\{{a, b}\right\}$ 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:
- $\forall a, b \in A: \left({a \oplus a}\right) \oplus b = a \oplus \left({a \oplus b}\right)$
- $\forall a, b \in A: \left({b \oplus a}\right) \oplus a = b \oplus \left({a \oplus a}\right)$
Then:
- $\left[{a, a, b}\right] = 0$
- $\left[{b, a, a}\right] = 0$
where $\left[{a, a, b}\right]$ denotes the associator of $a, b \in A_R$.
Now let us compute, using the linearity of $\oplus$ and the two suppositions:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({a - b}\right) \oplus \left({ \left({a - b}\right) \oplus a }\right)\) | \(=\) | \(\displaystyle a \oplus \left({ \left({a - b}\right) \oplus a }\right) - b \oplus \left({ \left({a - b}\right) \oplus a }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle a \oplus \left({a \oplus a}\right) - a \oplus \left({b \oplus a}\right) - b \oplus \left({a \oplus a}\right) + b \oplus \left({b \oplus a}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({\left({a - b}\right) \oplus \left({a - b}\right)}\right) \oplus a\) | \(=\) | \(\displaystyle \left({a \oplus a}\right) \oplus a - \left({a \oplus b}\right) \oplus a - \left({b \oplus a}\right) \oplus a + \left({b \oplus b}\right) \oplus a\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \left[{a - b, a - b, a}\right]\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left[{a, a, a}\right] - \left[{a, b, a}\right] - \left[{b, a, a}\right] + \left[{b, b, a}\right]\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Subtract the two expressions | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left[{a, b, a}\right]\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
This suffices, but why it does could do with explanation (although trivial). Also, where do I use that $A$ isn't Boolean? Maybe better use talk page
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$\blacksquare$
Source of Name
This entry was named for Emil Artin.
Sources
- John C. Baez: The Octonions (2002): 1.1 Preliminaries
- This article incorporates material from alternative algebra on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
