Let $R, A, B$ commutative rings with unity. Consider ring homomorphisms $\phi_1:R \to A$ and $\phi_2:R \to B$.
Then can build the $R-$module tensor product $A \otimes_R B$. We know $A \otimes_R B$ is an abelian group with the additive operation in $R$.
Let $\phi: A \times B \to A \otimes_R B$ be the canonical $R-$bilinear map.
Define $\cdot_R: (A \otimes_R B)\times (A \otimes_R B) \to A \otimes_R B$ as $\phi(a, b) \cdot_R \phi(a', b') = \phi(aa', bb')$.
I have two questions:
Can the additive operation in $A \otimes_R B$ be defined in terms of $\phi$? Behind this question, I want to show that $\cdot_R$ distributes the additive operation in $A \otimes_R B$, but I clearly don't understand how to achieve this.
Let $C$ be another commutative ring with unity and $f: A \to C$, $g: B \to C$ two ring homomorphisms. I define $\gamma: A \otimes_R B \to C$ as $\gamma(\phi(a, b)) = f(a)g(b)$. How can I prove that $\gamma$ opens addition? That is
$$\gamma(\phi(a,b) + \phi(a',b')) = \gamma(\phi(a,b)) + \gamma(\phi(a',b'))$$
The root of the problem is that I don't understand how addition is defined in $A \otimes_R B $.
The original problem underlying can be find here:
Let $R \in \textbf{Ring}$. Show that $A_1 \otimes_R A_2 \in \textbf{Ring}$.
Let us use the standard construction of $A\otimes_R B$. I'm going to pretend $A$ and $B$ are both $R$-modules for notational convenience, this can be formalised with the $\phi_{1,2}$. Things like $ra$ for $r\in R$ and $a\in A$ are defined as $\phi_1(r)\cdot a$. The fact all rings involved are commutative is important.
We take the free Abelian group on the set $A\times B$ - let's call this $F$. The elements of the group $F$ are linear combinations of the generators, one for each pair $(a,b)\in A\times B$. Addition in $F$ is purely "formal". In fact, $F$ is more correctly stated as the set of all functions $f:A\times B\to\Bbb Z$ for which $f(a,b)=0$ for all but finitely many pairs $(a,b)$. The addition $f+g$ is defined as the obvious function $(a,b)\mapsto f(a,b)+g(a,b)$. We interpret a function $f$ as a linear combination by considering the ("formal"; this is just notation) sum $\sum_{(a,b)\in A\times B}f(a,b)\cdot(a,b)$ e.g. $(a_1,b_1)+2(a_2,b_2)-3(a_3,b_3)$... but this notation is really just a placeholder for the definition via functions. Then we consider the subgroup $G$ which is generated by elements of form:
And finally construct $A\otimes_RB$ as the Abelian group quotient $F/G$. Notice that the Abelian group "addition" structure on $A\otimes_RB$ is automatically given to us: any quotient group is... well, a group! "Explicitly" the addition works as follows: $$[k_1(a_1,b_1)+k_2(a_2,b_2)+\cdots]+[k'_1(a'_1,b'_1)+\cdots]:=[k_1(a_1,b_1)+\cdots+k'_1(a'_1,b'_1)+\cdots]$$Where $[]$ denotes equivalence classes in the quotient and the $k_\bullet,k'_\bullet$ are any finite selection of integers and the pairs $(a_\bullet,b_\bullet),(a'_\bullet,b'_\bullet)$ are any arbitrary finite selection of elements of $A\times B$. Each $[(a,b)]$ is usually written $a\otimes b$ and is called a "pure tensor".
Now to the issue of multiplication. Your definition of $\cdot_R$ is not really a definition at all, there are two issues. Issue number $(1)$: you have to check it is well defined! For $\phi$ is not injective so $\phi(a,b)=\phi(a',b')$ is quite possible for distinct pairs. Issue number $(2)$ is very similar: $\phi$ is not surjective either (in general)! Not all elements of $A\otimes_RB$ are "pure tensors" (the things usually written as $a\otimes b$, i.e. $\phi(a,b)$).
About the ring structure on $A\otimes_R B$. I object to your "definition" and also to the definition given in your linked question; not because it's wrong per se but because it is not enough - more work needs to be done. We seek a suitable $m:A\otimes_RB\times A\otimes_RB\to A\otimes_RB$ and this is best done by, for each $x\in A\otimes_RB$, defining $m(x,-):A\otimes_RB\to A\otimes_RB$ since maps out of $A\otimes_RB$ are easy to define. We have to find a suitable "$R$-balanced product" $A\times B\to A\otimes_RB$. Let $\sum_{j=1}^nk_j(a_j,b_j)$ be a finite linear combination representing $x$. We can define a map $h:A\times B\to A\otimes_RB$ by $h(a,b):=[\sum_{j=1}^n k_j(a\cdot a_j,b\cdot b_j)]$. We need to check that this is well-defined and that this is an $R$-balanced product. Assuming $h$ is well-defined, the last statement is quite easy to see. The first statement is a bit harder to check; you have to realise that if $x$ is also represented by a different linear combination, then there is a finite sequence of "moves" which takes the first combination to the second; then, you must observe that the multiplication by $(a,b)$ does not affect any of these "moves" in the sense that you stay within the same equivalence class.
Once you've done all that, you have a bona fide group homomorphism $m(x,-):A\otimes_RB\to A\otimes_RB$ uniquely induced from $h$. Now of course you need to show $m$ defines a commutative unital and associative operation which distributes over addition. Some of this is automatic; the unital criterion is easy to check and $m(x,y+z)=m(x,y)+m(x,z)$ is automatic from the fact $m(x,-)$ is a group homomorphism. To get the other claims, it is best to first show each $m(-,y)$ is also a group homomorphism (i.e. it comes from an $R$-balanced product) and therefore it suffices to check all further conditions on the pure tensors only, which is easier.
A quick internet search suggests that $A\otimes_RB$ is not in general a ring. We can only expect it to be an $R$-module; you need to be searching for a multiplication map $m:R\times A\otimes_RB\to A\otimes_RB$ which satisfies some axioms. One does this as follows; for each fixed $r\in R$, we have a homomorphism $F\to F$ which is defined as the linear extension of $(a,b)\mapsto(ra,b)$. This map clearly descends to a homomorphism $m(r,\bullet):A\otimes_RB\to A\otimes_RB$ by the universal property of a quotient group. Now, your task is to check that $m(r,-)+m(r',-)$ and $m(r+r',-)$ are the same homomorphism and that $m(rr',-)$ and $m(r,m(r',-))$ are the same homomorphism. This is best done by using the universal property again (well, in this direction we just use surjectivity of the quotient map $F\twoheadrightarrow F/G=A\otimes_RB$). I guess I should leave this as an exercise, now that you know what you need to show.As for point $(2)$, you make the same (conceptual) mistake when you say "that is": $\phi$ isn't in general surjective, so checking $\gamma\circ\phi$ respects addition isn't immediately enough... but in a sense it is enough. Sorry. The real thing you need to do is check that you have a bilinear "$R$-balanced product" $A\times B\to C$. Then (extremely important but straightforward exercise!) you use the definition of $A\otimes_RB$ as a quotient group to prove there is a genuine group homomorphism $\gamma:A\otimes_RB\to C$ such that $\gamma(a\otimes b)=f(a)g(b)$ on pure tensors.
More abstractly, this property that guarantees existence of $\gamma$ is the real reason we care about the tensor product. It's common to define the tensor product by this property, and then consider the construction as an important but secondary detail.