[Math] Equivalent definitions of an algebra over a ring

abstract-algebramodulesring-theory

I have seen two different definitions from several sources of an algebra over a ring. From Wikipedia:

  • Let R be a commutative ring. An R-algebra is an R-Module A together with a binary operation [·,·]
    [$\cdot$,$\cdot$]: A $\times$ A $\to$ A
    called A-multiplication, which satisfies the following axiom:

Bilinearity:

[$\alpha x + \beta y, z] = \alpha [x, z] + \beta [y, z]$, $\quad [z, \alpha x + \beta y] = \alpha [z, x] + \beta [z, y]$

for all scalars $\alpha, \beta$ in R and all elements x, y, z in A *

I have also frequently seen (this one's from Virginia):

Let R be a commutative ring. An R-algebra is a ring A
which is also an R-module such that the multiplication map A $\times$ A $\to$ A is
R-bilinear, that is,
r $\ast$ (ab) = (r $\ast$ a) b = a $\cdot$ (r $\ast$ b) for any $a, b \in A$ $r \in R$
where $\ast$ denotes the R-action on A.

I'm trying to prove they are equivalent. I am fine with everything apart from proving that, if definition 1 is satisfied, then multiplication in A is associative. Unless this property is what determines whether the algebra is associative or not?

Best Answer

Associativity may or may not be an axiom, depending on your context. Requiring associativity results in a strictly smaller class of objects.

While associative algebras are common, the study of nonassociative algebras is also full of important topics (Lie theory is a good example.)

A basic example to keep in mind is $\Bbb R^3$ with the cross product. That makes an algebra that isn't associative.

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