[proofplan]
We construct the multiplication map on $B \otimes_R C$ as a composition of $R$-linear maps, thereby bypassing any well-definedness issues that would arise from defining a map on pure tensors directly. The key is the canonical reassociation isomorphism $(B \otimes_R C) \otimes_R (B \otimes_R C) \cong (B \otimes_R B) \otimes_R (C \otimes_R C)$, followed by the component-wise multiplication maps $m_B: B \otimes_R B \to B$ and $m_C: C \otimes_R C \to C$. Since each of these maps is $R$-linear and arises from the universal property of the tensor product, the composition is well-defined. The ring axioms are then verified on pure tensors and extended by $R$-linearity.
[/proofplan]
[step:Construct the multiplication maps $m_B$ and $m_C$ via the universal property]
The multiplication in $B$ is the map $B \times B \to B$ given by $(b_1, b_2) \mapsto b_1 b_2$. This map is $R$-bilinear: for all $r \in R$ and $b_1, b_1', b_2 \in B$,
\begin{align*}
(b_1 + b_1') b_2 &= b_1 b_2 + b_1' b_2, \\
(r b_1) b_2 &= r(b_1 b_2) = b_1 (r b_2),
\end{align*}
where the last equality uses the fact that $B$ is a commutative $R$-algebra, so $r$ acts centrally. By the universal property of the tensor product $B \otimes_R B$, this $R$-bilinear map induces a unique $R$-module homomorphism
\begin{align*}
m_B: B \otimes_R B &\to B, \quad b_1 \otimes b_2 \mapsto b_1 b_2.
\end{align*}
By an identical argument, there is an $R$-module homomorphism
\begin{align*}
m_C: C \otimes_R C &\to C, \quad c_1 \otimes c_2 \mapsto c_1 c_2.
\end{align*}
[/step]
[step:Construct the reassociation isomorphism $(B \otimes_R C) \otimes_R (B \otimes_R C) \cong (B \otimes_R B) \otimes_R (C \otimes_R C)$]
There is a canonical $R$-module isomorphism
\begin{align*}
\sigma: (B \otimes_R C) \otimes_R (B \otimes_R C) &\xrightarrow{\;\sim\;} (B \otimes_R B) \otimes_R (C \otimes_R C)
\end{align*}
determined on pure tensors by
\begin{align*}
(b_1 \otimes c_1) \otimes (b_2 \otimes c_2) &\mapsto (b_1 \otimes b_2) \otimes (c_1 \otimes c_2).
\end{align*}
This isomorphism is a standard consequence of the associativity and commutativity isomorphisms for tensor products of $R$-modules: one applies the associativity isomorphism $(B \otimes_R C) \otimes_R (B \otimes_R C) \cong B \otimes_R (C \otimes_R (B \otimes_R C))$, then the commutativity isomorphism $C \otimes_R B \cong B \otimes_R C$ on the inner factor, and then re-associates to obtain $(B \otimes_R B) \otimes_R (C \otimes_R C)$. Each step is a well-defined $R$-module isomorphism arising from the universal property, so $\sigma$ is a well-defined $R$-module isomorphism.
[guided]
Why do we need this reassociation? The multiplication we want to define on $B \otimes_R C$ takes a pair of elements and produces a single element. On pure tensors, the desired formula is $(b_1 \otimes c_1) \cdot (b_2 \otimes c_2) = (b_1 b_2) \otimes (c_1 c_2)$. If we tried to define this directly as a map on $B \otimes_R C \times B \otimes_R C$, we would need to verify it is well-defined with respect to the tensor product relations in each factor -- a messy and error-prone task.
Instead, we exploit the universal property machinery. The product of two elements of $B \otimes_R C$ naturally lives in $(B \otimes_R C) \otimes_R (B \otimes_R C)$, and we want to land in $B \otimes_R C$. The reassociation isomorphism $\sigma$ reorganises the four-fold tensor into $(B \otimes_R B) \otimes_R (C \otimes_R C)$, which separates the $B$-components from the $C$-components. Once separated, we can apply $m_B$ and $m_C$ independently.
The isomorphism $\sigma$ is constructed by composing three standard tensor product isomorphisms. Writing $\alpha$ for associativity and $\tau$ for the swap $C \otimes_R B \cong B \otimes_R C$:
\begin{align*}
(B \otimes_R C) \otimes_R (B \otimes_R C) &\xrightarrow{\alpha} B \otimes_R (C \otimes_R (B \otimes_R C)) \\
&\xrightarrow{\operatorname{id}_B \otimes \alpha^{-1}} B \otimes_R ((C \otimes_R B) \otimes_R C) \\
&\xrightarrow{\operatorname{id}_B \otimes (\tau \otimes \operatorname{id}_C)} B \otimes_R ((B \otimes_R C) \otimes_R C) \\
&\xrightarrow{\alpha^{-1} \otimes \operatorname{id}_C} (B \otimes_R B) \otimes_R (C \otimes_R C).
\end{align*}
Each arrow is a well-defined $R$-module isomorphism, so $\sigma$ is as well.
[/guided]
[/step]
[step:Compose to obtain the multiplication map $\mu: (B \otimes_R C) \otimes_R (B \otimes_R C) \to B \otimes_R C$]
Define the multiplication map as the composition
\begin{align*}
\mu: (B \otimes_R C) \otimes_R (B \otimes_R C) \xrightarrow{\;\sigma\;} (B \otimes_R B) \otimes_R (C \otimes_R C) \xrightarrow{\;m_B \otimes m_C\;} B \otimes_R C.
\end{align*}
Here $m_B \otimes m_C$ denotes the $R$-module homomorphism induced by the $R$-bilinear map $(B \otimes_R B) \times (C \otimes_R C) \to B \otimes_R C$ sending $(x, y) \mapsto m_B(x) \otimes m_C(y)$. On pure tensors:
\begin{align*}
\mu((b_1 \otimes c_1) \otimes (b_2 \otimes c_2)) &= (m_B \otimes m_C)((b_1 \otimes b_2) \otimes (c_1 \otimes c_2)) \\
&= m_B(b_1 \otimes b_2) \otimes m_C(c_1 \otimes c_2) \\
&= (b_1 b_2) \otimes (c_1 c_2).
\end{align*}
Since $\mu$ is a composition of $R$-module homomorphisms, it is a well-defined $R$-module homomorphism. The corresponding bilinear map $(B \otimes_R C) \times (B \otimes_R C) \to B \otimes_R C$ given by $(x, y) \mapsto \mu(x \otimes y)$ is the desired multiplication.
[/step]
[step:Verify the ring axioms on pure tensors and extend by $R$-linearity]
Since $B \otimes_R C$ is spanned as an $R$-module by pure tensors $b \otimes c$, it suffices to check each axiom on pure tensors.
**Associativity.** For pure tensors $b_1 \otimes c_1$, $b_2 \otimes c_2$, $b_3 \otimes c_3$:
\begin{align*}
((b_1 \otimes c_1)(b_2 \otimes c_2))(b_3 \otimes c_3) &= (b_1 b_2 \otimes c_1 c_2)(b_3 \otimes c_3) = (b_1 b_2) b_3 \otimes (c_1 c_2) c_3, \\
(b_1 \otimes c_1)((b_2 \otimes c_2)(b_3 \otimes c_3)) &= (b_1 \otimes c_1)(b_2 b_3 \otimes c_2 c_3) = b_1 (b_2 b_3) \otimes c_1 (c_2 c_3).
\end{align*}
These are equal by associativity of multiplication in $B$ and $C$.
**Commutativity.** For pure tensors $b_1 \otimes c_1$, $b_2 \otimes c_2$:
\begin{align*}
(b_1 \otimes c_1)(b_2 \otimes c_2) = b_1 b_2 \otimes c_1 c_2 = b_2 b_1 \otimes c_2 c_1 = (b_2 \otimes c_2)(b_1 \otimes c_1),
\end{align*}
using commutativity of $B$ and $C$.
**Multiplicative identity.** The element $1_B \otimes 1_C \in B \otimes_R C$ satisfies
\begin{align*}
(1_B \otimes 1_C)(b \otimes c) = 1_B \cdot b \otimes 1_C \cdot c = b \otimes c
\end{align*}
for every pure tensor $b \otimes c$.
**Distributivity.** Since $\mu$ is an $R$-module homomorphism, it is additive in each variable. In particular, for $x, y, z \in B \otimes_R C$:
\begin{align*}
\mu((x + y) \otimes z) = \mu(x \otimes z) + \mu(y \otimes z),
\end{align*}
which gives $(x + y) \cdot z = x \cdot z + y \cdot z$, and similarly on the other side.
Each identity holds on the spanning set of pure tensors and is preserved under $R$-linear combinations (since the multiplication map is $R$-bilinear). Therefore $B \otimes_R C$ with this multiplication is a commutative ring with identity $1_B \otimes 1_C$. The $R$-algebra structure is given by the ring homomorphism $R \to B \otimes_R C$ defined by $r \mapsto f_B(r) \otimes 1_C = 1_B \otimes f_C(r)$ (where $f_B: R \to B$ and $f_C: R \to C$ are the $R$-algebra structure maps), and these two expressions agree because $f_B(r) \otimes 1_C = r \cdot (1_B \otimes 1_C) = 1_B \otimes f_C(r)$ in $B \otimes_R C$.
[/step]
