prove that the coproduct of commutative rings with unity is the tensor product of the underlying abelian groups How to prove that the coproduct of commutative rings with unity is the tensor product of the underlying abelian groups? 
I don't know where to start and how to finish. May I get some ideas or the proof.
 A: Let $A$ and $B$ be commutative rings with unity. Then, the tensor product $A \otimes B$ as abelian groups is an abelian group.
To recall, this is the tensor product over the ring $\Bbb Z$. (A $\Bbb Z$-module is the same as an abelian group.)
Moreover, recall the (essence of) the universal property of tensor product: Given a $\Bbb Z$-bilinear map $\varphi: A \times B \to C$, we get a unique well-defined $\Bbb Z$-linear map $\tilde \varphi : A \otimes B \to C$ satisfying $\tilde \varphi(a \otimes b) = \varphi((a, b))$ for all $a \in A$ and $b \in B$.

Now, we must first turn $A \otimes B$ into a ring. This is done by defining $$(a \otimes b) \cdot (a' \otimes b) := (aa') \otimes (bb')$$
and then extending it linearly.
We must check that the above makes sense.
First, for $(a, b) \in A \times B$, define the map $m_{a, b} : A \times B \to A \otimes B$ as $$m_{a, b}(a', b') = (aa') \otimes (bb').$$
For a fixed $(a, b) \in A \times B$, the map $m_{a, b}$ is bilinear. Thus, we get a well-defined linear map $\tilde m_{a, b} : A \otimes B \to A \otimes B$ given by the tensor property.
Now, $M : A \times B \to \operatorname{Hom}(A \otimes B, A \otimes B)$ given by $(a, b) \mapsto \tilde m_{a, b}$ is again bilinear. Thus, we get a well-defined map $\tilde M : A \otimes B \to \operatorname{Hom}(A \otimes B, A \otimes B)$ given by $$\tilde M(a \otimes b) = \tilde m_{a, b}.$$
Thus, our previous definition of the product in $A \otimes B$ makes sense since it is precisely
$$[\tilde M(a \otimes b)](a' \otimes b').$$
Moreover, the distributive property follows since $\tilde M$ and $\tilde m_{a, b}$ are linear maps. Associativity is easy to see. Moreover, the ring has a multiplicative identity, namely $1 \otimes 1$.

Now, for it to be the coproduct, we must give maps $A \xrightarrow{i_1} A \otimes B \xleftarrow{i_2} B$ and show that it satisfies the universal property.
The maps are the obvious ones, namely $a \mapsto a \otimes 1$ and $b \mapsto 1 \otimes b$. These are easily checked to be ring maps.
Now, suppose we are given ring maps $f_1 : A \to C$ and $f_2 : B \to C$.
We wish to define a ring map $g : A \otimes B \to C$ making the desired diagram commute. Moreover, it must be the unique such map.
Firstly, note that such a $g$, if it exists, is forced to satisfy
\begin{align}
g(a \otimes b) &= g((a \otimes 1)(1 \otimes b)) \\
&= g(a \otimes 1) g(1 \otimes b) \\
&= g(i_1(a)) g(i_2(b)) \\
&= f_1(a) f_2(b).
\end{align}
The above proves uniqueness since $g$ is completely determined on the generators. Moreover, it also tells us how we must define $g$.
Now, it is easy to check that $A \times B \to C$ defined by $(a, b) \mapsto f_1(a) f_2(b)$ is bilinear and thus, again, by the universal property of tensor, we get a well-defined map $A \otimes B \to C$ sending $a \otimes b \mapsto f_1(a) f_2(a)$ and we are done. (Why is this a ring map?)
