Multiplication map for algebras in the sense of Hopf algebra

Question

Let $R$ be a commutative ring with unity and $A$ a ring with unity. In the definition of algebras, we require a multiplication map $\mu: A\otimes A \rightarrow A$ satisfying certain property. Indeed, we can view $\mu$ just as a usual multiplication in ring $A$. My question is why we want the map $\mu$ to define on the tensor product $A \otimes A$ rather than the cartesian product $A \times A$ as we did in the definition of general rings.

Answer 1

The definitions are equivalent by the universal property of the tensor product. What you gain with the tensor product is that the multiplication becomes an $R$-module homomorphism, which is often more useful than a bilinear map.

Answer 2

From the definition of the tensor product of two $R$-modules we get that: For any $R$-module $A$, there exists an $R$-module $A\otimes A$ and an $R$-bilinear map $\otimes:A\times A\rightarrow A\otimes A$, such that

the following universal property is satisfied:

For any $R$-bilinear map: $\ \ \cdot:A\times A\rightarrow A$, there exists a unique $R$-linear map (i.e. an $R$-module homomorphism) $\mu:A\otimes A\rightarrow A$ such that: $$\mu\circ\otimes=\cdot$$ or equivalently: $$\mu(a\otimes b)=a\cdot b$$ The converse is obvious by the definition of the tensor product: for any $R$-linear map $\mu:A\otimes A\rightarrow A$ the map $\mu\circ\otimes:A\times A\rightarrow A$ is $R$-bilinear.

The above imply that the $R$-algebra multiplication can be equivalently viewed either as an $R$-bilinear map: $ \ \ \cdot:A\times A\rightarrow A$ or as an $R$-linear map: $\mu:A\otimes A\rightarrow A$