Let $A$ be an algebra over a commutative ring $k$ and $M$ and $N$ be modules over $A.$ Is there any natural way to define tensor product of $M$ and $N$ over the algebra $A\ $?
My idea is that since $A$ is an algebra it already has the ring structure. So we can consider the tensor product of $M$ and $N$ when $A$ is regarded as a ring. But when $A$ is a Hopf algebra the monoidal structure is defined in terms of the comultiplication. Namely, given $A$-modules $M$ and $N,$ the $A$-module structure on $M \otimes N$ is defined as follows $:$
$$a \cdot (m \otimes n) : = \Delta (a) \cdot (m \otimes n) = \sum\limits_{a} a_{(1)} m_1 \otimes a_{(2)} m_2$$
by using Sweedler's notation. But why do we need to use the comultiplication here? Can't we simply use the bilinearity of the tensor products? I am pretty confused at this stage. Any suggestion in this regard would be greatly appreciated.
Thanks for your time.
Source $:$ Quantum Groups $:$ A Survey of Definitions, Motivations and Results by Arun Ram.
There are (unique) $A$-module homomorphisms $\alpha_m:A\to M$, $\alpha_n:A\to N$ with $\alpha_m(1)=m$, $\alpha_n(1)=n$. If one turns somehow $A\otimes A$ and $M\otimes N$ into $A$-modules and wants $\alpha_m\otimes\alpha_n:A\otimes A\to M\otimes N$ to be an $A$-module homomorphism too, then one must have$$a\cdot(m\otimes n)=a\cdot((\alpha_m\otimes\alpha_n)(1\otimes1))=(\alpha_m\otimes\alpha_n)(a\cdot(1\otimes1))$$which is the expression with $\Delta$ that you have.