Why is $B\otimes_A C$ a ring, where $B,C$ are $A$-algebra Given $B,C$ be two $A$-algebra, it is said that $B\otimes_AC$ has a ring structure, with the multiplication being defined as:
$$
(b_1\otimes c_1)\cdot(b_2\otimes c_2):=(b_1b_2)\otimes (c_1c_2).
$$
However I don't see an easy way to check it is well-defined.
For, given another pair of representatives:
$$
(b_1'\otimes c_1')\cdot(b_2'\otimes c_2'):=(b_1'b_2')\otimes (c_1'c_2')
$$
where
$$
b_1\otimes c_1=b_1'\otimes c_1'\quad\text{and}\quad b_2\otimes c_2=b_2'\otimes c_2'.
$$
How to verify that
$$
(b_1b_2)\otimes (c_1c_2)=(b_1'b_2')\otimes (c_1'c_2')?
$$
 A: In general, you should always be trying to formulate properties of tensor products in terms of the universal property which it satisfies, i.e., if you ever find yourself in the situation of trying to prove two elements of a tensor product are equivalent using "tensor product relations" or something like that, you're probably on the wrong track.
Recall that the tensor product $B\otimes_A C$ (viewing $B$ and $C$ here only as $A$-modules for now) is uniquely characterized up to unique isomorphism by the following universal property. Denote by $\phi:B\times C\to B\otimes_A C$ the canonical $A$-bilinear homomorphism, defined by $\phi(b, c) = b\otimes c$. For any $A$-bilinear homomorphism $\psi:B\times C \to M$ to another $A$-module $M$, there exists a unique $A$-linear homomorphism $\widetilde{\psi}:B\otimes_A C\to M$ such that $\psi = \widetilde{\psi}\circ \phi$, i.e., $\psi(b, c) = \widetilde\psi(b\otimes c)$.
The key here is to recognize that "multiplication by an element" is a bilinear map. In other words, given $b_1\in B$, $c_1\in C$, consider the homomorphism $\psi_{(b_1, c_1)}:B\times C\to B\otimes_A C$ defined by $$\psi_{(b_1, c_1)}(b_2, c_2) = (b_1b_2)\otimes (c_1c_2).$$ Check that this is an $A$-bilinear homomorphism. By the universal property, therefore, there exists a unique $A$-linear homomorphism $\widetilde{\psi}_{(b_1, c_1)}:B\otimes_A C\to B\otimes_A C$ satisfying
  $$\widetilde{\psi}_{(b_1, c_1)}\circ \phi = {\psi}_{(b_1, c_1)}.$$
Note that saying $b_2\otimes c_2 = b_2'\otimes c_2'$ is equivalent to saying that $\phi(b_2, c_2) = \phi(b_2', c_2')$, so by using the above, we get
  \begin{multline}  
    (b_1b_2)\otimes (c_1c_2) = \psi_{(b_1, c_1)}(b_2, c_2) = \widetilde{\psi}_{(b_1, c_1)}(\phi(b_2, c_2)) =\\
    \widetilde{\psi}_{(b_1, c_1)}(\phi(b_2', c_2')) = \psi_{(b_1, c_1)}(b_2', c_2') = (b_1b_2')\otimes (c_1c_2').
  \end{multline}
Using a symmetric argument for $b_1'\otimes c_1'$ gives $(b_1b_2)\otimes (c_1c_2) = (b_1'b_2')\otimes (c_1'c_2')$.
A: That $B$ is an $A$-algebra means that it comes equipped with maps $u_B : A \to B$ (unit) and $m_B : B \otimes_A B \to B$ (multiplication) of $A$-modules such that certain diagrams commute (write them down!). Similarly for $C$. But then $B \otimes_A C$ has the following $A$-algebra structure:
The unit is $A \cong A \otimes_A A \xrightarrow{u_B \otimes u_C} B \otimes_A C$. The multiplication is
$(B \otimes_A C) \otimes_A (B \otimes_A C) \cong (B \otimes_A B) \otimes_A (C \otimes_A C) \xrightarrow{m_B \otimes m_C} B \otimes_A C.$
So it is just induced by functoriality of the tensor product, as well as the usual associativity and symmetry isomorphisms. In fact, this procedure works in any symmetric monoidal category. No need to calculate with elements here.
