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:


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

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?

  • $\begingroup$ Associativity is an extra condition that is sometimes omitted from the definition of algebra. $\endgroup$ – Zhen Lin Oct 24 '14 at 11:02
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    $\begingroup$ @MattBurrows I don't see how Definition(2) implies Definition(1). From $r*(ab)=(r*a)b=a(r*b)$ how do you prove$\quad [ \alpha x + \beta y, z] = \alpha [x, z] + \beta [y, z]$ and $\quad [z, \alpha x + \beta y] = \alpha [z, x] + \beta [z, y]$? $\endgroup$ – Babai Apr 24 '16 at 22:02
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    $\begingroup$ @Babai, I think the op assumes in definition 2 that multiplication distributes over addition. $\endgroup$ – R_D May 2 '16 at 16:46
  • $\begingroup$ Another definition is a follows: a ring $A$ is an $R$-algebra, if there exists a homomorphism from $R$ to $A$ which sends the multiplicative identity of $R$ to that of $A$. $\endgroup$ – User0112358 Jul 26 '16 at 9:01

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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