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I've looked up numerous definitions, and all of them talk of a new operator that is added to the $R$-module that is the $R$-algebra.

One definition says

An $R$-algebra, where $R$ is a commutative ring, is a ring with identity together with a ring homomorphism $f\colon R \to A$ such that the subring $f(R)$ of $A$ is contained within the center of $A$."

I don't see how the fact that an $R$-algebra is an $R$-module with a bilinear operator follows from this definition. Thanks in advance.

A.P.
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2 Answers2

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An $R$-algebra is an $R$-module, which also has a ring structure. That's probably the definition you have in mind.

The definition above does it the other way round. It says an $R$-algebra is a ring which also has the described additional structure. In fact the morphism $f$ gives $A$ an $R$-module structure:

$A$ is an $R$-module via $ra:=f(r)a$ for $r\in R$ and $a\in A$. Now you want the multiplication in $A$ to be bilinear wrt $R$, which amounts to saying that $f(A)$ lies in the centre.

Conversely if $A$ is an algebra in the original sense, define $f(r)=r\cdot1$, where the multiplication comes from the module structure and $1\in A$ is the identity in the ring $A$.

Edit: Just checked the definition on wikipedia. I understand under an $R$-algebra what Wiki calls associative $R$-algebra. The definition you gave in your question also gives an associative $R$-algebra.

Simon Markett
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    @ayush The centrality condition is basically fulfills the algebra condition that $r(ab)=(ar)b=a(br)$ for $a,b\in R$, $r\in R$. (I guess this could be formulated as part of bilinearity, in some definition.) – rschwieb Apr 03 '13 at 10:49
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So we have the following definitions:

$R$-module: Let $R$ be a ring. Then an abelian additive group $A$ together with a scalar multiplication $\cdot\colon R\times A\to A$ is called an $R$-module iff the following hold:

  1. $r\cdot(a + b) = r\cdot a + r\cdot b$
  2. $(r + s)\cdot a = r\cdot a + s\cdot a$
  3. $(rs)\cdot a = r\cdot(s\cdot a)$
  4. If $R$ has an identity, then we also demand $1_R\cdot a = a$.

$R$-algebra (definition 1): Let $R$ be a ring. Then an $R$-module $A$ together with a multiplication $*\colon A\times A\to A$ is called an $R$-algebra iff the following hold:

  1. $*$ is bilinear, i.e.,
    • $a*(b + c) = a*b + a*c$
    • $(a + b)*c = a*c + b*c$
    • $(r\cdot a)*(s\cdot b) = (rs)\cdot (a*b)$
  2. If $*$ is commutative (respectively associative), we call $A$ a commutative (respectively associative) $R$-algebra.
  3. If $*$ has an identity, then we say that $A$ is an $R$-algebra with identity.

$R$-algebra (definition 2): Let $R$ be a ring. Then a ring $A$ together with a ring homomorphism $f\colon R\to A$ is called an $R$-algebra iff the following hold:

  1. $f(R)$ is contained in $A$'s center.
  2. If $R$ has identity, then we demand that $f(1_R)$ be the identity of $A$.

Now, we can show the following results linking definitions $1$ and $2$:

Theorem 1: Let $R$ be a ring with identity and $A$ be an associative $R$-algebra with identity in sense of definition $1$. Then $A$ together with the homomorphism $r\mapsto r\cdot 1_A$ is an $R$-algebra (with identity) in the sense of definition $2$.

Theorem 2: Let $R$ be a ring and $(A, f)$ be an $R$-algebra in sense of definition $2$. Then the scalar multiplication $\cdot\colon R\times A\to A$ defined by $r\cdot a := f(r)* a$ makes $A$ an associative $R$-algebra.

Note: I don't require homomorphisms to send identity to identity in general.

Atom
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  • in the definition 1.2 it says that we need R to be commutative and unitary, now I'm wondering why do we need that hypothesis! and/or can we omit that or not! i hope I'm not saying something so stupid! – spiritedcaveman Jan 02 '25 at 05:39