Equivalent definitions of an algebra over a ring 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? 

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.
