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I'm thinking on the following question: If $A$ is a commutative ring, $M$ is $A$-module if and only if $M$ is $(A,A)$-bimodule?

I know that $M$ is and module-$A$ because $A$ is commutative, but is the question true?

Thanks for your help :)

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    Where are you havin problems checking if this is so or not? – 2017-01-07
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    The problem is that I don't know if it is true. Is it? – 2017-01-07
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    But if you try to prove it, what happeens? – 2017-01-07
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    Which specific part of the definition are you having trouble confirming / disproving? – 2017-01-07

2 Answers 2

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If we define the right $A$-module multiplication on $M$ as the left-module structure, i.e. $$ ma := am, \qquad m \in M, a \in A $$ then this is in fact a right $A$-module structure, due to $A$'s commutativity, we have $$ (ma)b = b(am) = (ba)m = (ab)m = m(ab) $$

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You can easily check that every left $A$-module is in an obvious way a right $A^{\textrm{op}}$-module, where $A^{\textrm{op}}$ is the same ring as $A$ except multiplication $\cdot$ in $A^{\textrm{op}}$ is defined in the opposite way: $a \cdot b$ is defined as $ba$.

If $A$ is commutative, then $A = A^{\textrm{op}}$, so every left $A$-module is also a right $A$-module.