Editing Power associativity (section)

== Examples and properties ==
Every [[associative algebra]] is power-associative, but so are all other [[alternative algebra]]s (like the [[octonion]]s, which are non-associative) and even non-alternative [[flexible algebra]]s like the [[sedenion]]s, [[trigintaduonion]]s, and [[Okubo algebra]]s.  Any algebra whose elements are [[idempotent]] is also power-associative.

[[Exponentiation]] to the power of any [[positive integer]] can be defined consistently whenever multiplication is power-associative. For example, there is no need to distinguish whether ''x''<sup>3</sup> should be defined as (''xx'')''x'' or as ''x''(''xx''), since these are equal. Exponentiation to the power of zero can also be defined if the operation has an [[identity element]], so the existence of identity elements is useful in power-associative contexts.

Over a [[field (mathematics)|field]] of [[Characteristic (algebra)|characteristic]] 0, an algebra is power-associative if and only if it satisfies <math>[x,x,x]=0</math> and <math>[x^2,x,x]=0</math>, where <math>[x,y,z]:=(xy)z-x(yz)</math> is the [[associator]] (Albert 1948).

Over an infinite field of [[prime number|prime]] characteristic <math>p>0</math> there is no finite set of identities that characterizes power-associativity, but there are infinite independent sets, as described by Gainov (1970):

* For <math>p=2</math>: <math>[x,x^2,x]=0</math> and <math>[x^{n-2},x,x]=0</math> for <math>n=3,2^k</math> (<math>k=2,3...)</math>
* For <math>p=3</math>: <math>[x^{n-2},x,x]=0</math> for <math>n=4,5,3^k</math> (<math>k=1,2...)</math>
* For <math>p=5</math>: <math>[x^{n-2},x,x]=0</math> for <math>n=3,4,6,5^k</math> (<math>k=1,2...)</math>
* For <math>p>5</math>: <math>[x^{n-2},x,x]=0</math> for <math>n=3,4,p^k</math> (<math>k=1,2...)</math>

A substitution law holds for [[real number|real]] power-associative algebras with unit, which basically asserts that multiplication of [[polynomial]]s works as expected. For ''f'' a real polynomial in ''x'', and for any ''a'' in such an algebra define ''f''(''a'') to be the element of the algebra resulting from the obvious substitution of ''a'' into ''f''. Then for any two such polynomials ''f'' and ''g'', we have that {{nowrap|1=(''fg'')(''a'') = ''f''(''a'')''g''(''a'')}}.