Equivalent definitions of an algebra over a ring

Question

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.

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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?

Answer 1

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.