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Recently my definition of a bimodule over a $k$-algebra has been challenged and I believe both definitions to be equivalent, am I wrong?


Notation: $k$ is a commutative ring and $A$ is a (unital associative) $k$-algebra.


Definition 1:

An $(A,A)$-bimodule $M$ is an abelian group together with a left and right $A$-module structure such that: \begin{equation} (\forall m \in M) (\forall a,b \in A) (a \cdot m) \cdot b = a \cdot (m \cdot b) \end{equation}


Definition 2:

An $(A,A)$-bimodule $M$ is an $k$-module together with a left and right $A$-module structure such that: \begin{equation} (\forall m \in M) (\forall a,b \in A) (a \cdot m) \cdot b = a \cdot (m \cdot b) \end{equation}


Reasoning

$A$ is a $k$-algebra then the inclusion of $k$ into $A$ induces an action of $k$ on $M$; whence $M$ must also be a $k$-module if it satisfies definition $1$, conversely if $M$ satisfies definition $2$ then it satisfies definition 1 by definition of a $k$-module.


Apparently my logic is flawed but I don't see why...

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    $\begingroup$ When you say the $A$-bimodule $M$ is a $k$-module, do you mean that with $fm=mf$ for all $f\in A$ and $m\in M$? This seems to be included in the second but not in the first definition. I personally include this in my definition, and then an $A$-bimodule is just an $A\otimes_k A^{\mathrm{op}}$-leftmodule. $\endgroup$
    – Ben
    Commented Aug 6, 2014 at 15:59
  • $\begingroup$ @BenA. I think you mean $f\in k$ rather than $f\in A$? $\endgroup$ Commented Aug 6, 2014 at 16:34
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    $\begingroup$ I agree with you. I think the datum of a left $A$-module structure on $M$ includes the datum of a $k$-module structure on $M$, this structure being compatible with the $k$-module structure on $A$. $\endgroup$ Commented Aug 6, 2014 at 17:22
  • $\begingroup$ Of course @JeremyRickard,thank you! $\endgroup$
    – Ben
    Commented Aug 6, 2014 at 20:39

3 Answers 3

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The first definition is what you might call a "ring" bimodule. It only asks for $A$ to be a ring, and it doesn't mind if $A$ has any $k$ algebra structure.

I would call your second definition an algebra-bimodule, but I think you're missing an axiom. Not only does $M$ have a $k$-module structure, but this structure is compatible with $A$'s $k$-structure! You should also insist that $(ka)\cdot m=a\cdot (km)=k(a\cdot m)$ and $m\cdot(kb)=(km)\cdot b=k(m\cdot b)$ for all choices of $k,a,b,m$. This makes sure that the $k$-module action on $M$ is in sync with the $k$-structure on $A$.

In the first definition, simply no comment is made about the $k$-algebra structure on $A$ in relationship to $M$. Certainly you can induce a $k$ module structure on $M$ using the $A$ action on $M$, but this is just not mentioned.

Finally, you could also remember that all rings are $\Bbb Z$ algebras, and if you're working with unital modules (as most people do) then you are automatically guaranteeing that every ring-bimodule is also a $\Bbb Z$-algebra bimodule.

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In Definition 2 the action of $k$ on $M$ induced by the action of $k$ on $A$ can be different from the given $k$-action on $M$.

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As you say in your reasoning, the $A$-module structure on $M$ induces a $k$-module structure. So does the $B$-module structure. But without added conditions these two $k$-module structures could be different.

For example, take $A=B=M=\mathbb{C}$ with the bimodule structure $$a\cdot m\cdot b=am\overline{b},$$ where $\overline{b}$ is the complex conjugate of $b$.

This is a perfectly good bimodule structure if you're dealing with abstract rings (or $\mathbb{R}$-algebras), but working over $\mathbb{C}$, the standard definition would require that (in your first definition) the $k$-module structures on $M$ induced by the $A$-module and $B$-module structures are the same, or equivalently that (in your second definition) the $A$-module and $B$-module structures are both $k$-bilinear.

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