Show structure of a commutative ring in a tensor product I need some help with this:
Let $R$ be a commutative ring and $S$ and $T$ be commutative $R$-algebras. Show that $$ S \otimes T
$$ has the structure of a commutative ring with multiplication:  $$ (s \otimes t)(s' \otimes t'):=(ss' \otimes tt')
$$
How do I show here the ring axioms?
 A: It is a common misconception that every element of $S \otimes T$ has the form $s \otimes t$ and that you can define multiplication on $S \otimes T$ via such a formula $(s \otimes t) \cdot (s' \otimes t') = ss' \otimes tt'$ and verify the ring axioms with elements. Rather, one has to use the universal property of the tensor product in order to construct a multiplication map, which is then bilinear by construction. For example, the formula above tells nothing about $(s \otimes t + p \otimes q) \cdot (s' \otimes t')$, but we really (want to) define this as $ss' \otimes tt' + ps' \otimes qt'$. See below for details.
The quickest way of doing this is to realize that an $R$-algebra $S$ is nothing else than an $R$-module equipped with two $R$-linear maps $\eta_S : R \to S$ and $\mu_S : S \otimes_R S \to S$ such that three diagrams commute which state that $\eta_S$ is left and right unital for $\mu$ and that $\mu$ is associative. You can find these diagrams here. The connection to the usual definition is that $\eta_R(1)$ is the unit element for $S$ and that the product of two elements $s_1,s_2 \in S$ is $\mu_S(s_1 \otimes s_2)$.
Now, if $S,T$ are $R$-algebras (not assumed to be commutative), then $P = S \otimes_R T$ becomes an $R$-algebra as follows: We define
$$\eta_P : R \cong R \otimes_R R \xrightarrow{\eta_S \otimes_R \eta_T} S \otimes_R T = P.$$
and
$$\mu_P : P \otimes_R P = (S \otimes_R T) \otimes_R (S \otimes_R T) \cong (S \otimes_R S) \otimes_R (T \otimes_R T) \xrightarrow{\mu_S \otimes_R \mu_T} S \otimes_R T = P.$$
It is then easy to check that the diagrams for $\eta_P$ and $\mu_P$ commute (using those for $\eta_S,\mu_S$ and $\eta_T,\mu_T$), so that $P$ is in fact an $R$-algebra. If $S,T$ are commutative, then $P$ is commutative, too.
Notice that above, I have used some general canonical isomorphisms between (iterated) tensor products. For example, we know that the tensor product is associative and symmetric up to a canonical isomorphisms, and this implies that there is a unique isomorphism of $R$-modules
$$(S \otimes_R T) \otimes_R (S \otimes_R T) \cong (S \otimes_R S) \otimes_R (T \otimes_R T)$$
which maps $(s_1 \otimes t_1) \otimes (s_2 \otimes t_2)$ to $(s_1 \otimes s_2) \otimes (t_1 \otimes t_2)$.
If you want to prove directly, without using the abstract framework above, that $S \otimes_R T$ carries the structure of an $R$-algebra, you basically have to reprove these basic tensor product isomorphisms in a special case. This requires some work which is essentially useless. The unit is easy, it is just $1_S \otimes 1_T$. But you cannot define multiplication simply by $(s \otimes t) \cdot (s' \otimes t') = ss' \otimes tt'$ since not every element of $S \otimes_R T$ has the form $s \otimes t$ and even if this was the case: This representation is not unique and it is hard to decide (in general) when two such tensors are equal.  Every element of $S \otimes_R T$ has the form $\sum_i (s_i \otimes t_i)$ (a finite sum of pure tensors), so that the multiplication may be described as
$$\sum_i (s_i \otimes t_i) \cdot \sum_j (s'_j \otimes t'_j) = \sum_{i,j} (s_i s'_j \otimes t_i t'_j).$$
But now it is even harder (impossible, as far as I know) to show that this is well-defined just using elements. We have to invoke the universal property. One way of doing this is as follows: First, one fixes some pure tensor $s \otimes t$. Then, we look at the map
$$S \times T \to S \otimes_R T,~(s',t') \mapsto ss' \otimes tt'.$$
One checks that it is bilinear, hence extends to a homomorphism $\lambda_{s \otimes t}: S\otimes_R T \to S \otimes_R T$. This is our (what will be the) left multiplication by $s \otimes t$. Now in order to define left multiplication with any element of the tensor product, we can do almost the same trick again: We fix some $y \in S \otimes_R T$ and look at $S \times T \to S \otimes_R T,~(s,t) \mapsto \lambda_{s \otimes t}(y)$. One has to check that this map is bilinear, hence extends to a homomorphism $\rho_y : S \otimes_R T \to S \otimes_R T$. Now, if $x \in S \otimes_R T$, we define $x \cdot y := \rho_y(x)$. Now one has to verify the $R$-algebra axioms. But notice that, for example, the distributivity law in the first variable is given by construction, since $- \cdot y = \rho_y$ is a homomorphism.
Notice how many calculations have to be done with this "concrete" approach (even though I have omitted almost all of them). As I've said, it is so complicated because it ignores basic facts about tensor products and reproves them in a special case. When I started to learn algebra, I had to make so many calculations and nobody and no book told me that these calculations are not necessary at all. Today I'm happy to know that abstraction really helps to simplify or in fact supersede calculations and I hope that I can spread this point of view here.
