Tensor product of algebras

Remarks

The multiplication is well defined because the formula is $R$-bilinear in each pair of arguments. The maps $a \mapsto a \otimes 1$ and $b \mapsto 1 \otimes b$ are $R$-algebra homomorphisms, and their images commute. If $A$ and $B$ are commutative, then $A \otimes_R B$ is commutative, and it has a universal property. For every commutative $R$-algebra $C$ and $R$-algebra maps $f \colon A \to C$, $g \colon B \to C$, there is exactly one $R$-algebra map $h \colon A \otimes_R B \to C$ with $h(a \otimes b) = f(a) g(b)$.

Example: $R[x] \otimes_R R[y] \cong R[x, y]$, with $x^i \otimes y^j \mapsto x^i y^j$; both sides are free $R$-modules with these bases, and the map respects products. Another example is $\C \otimes_\R \C \cong \C[t]/(t^2 + 1) \cong \C \times \C$, by the Chinese remainder theorem, since $t^2 + 1 = (t - i)(t + i)$ over $\C$.

Non-example: the direct product $A \times B$ with componentwise operations is in general not $A \otimes_R B$. For $A = B = \Q[x]$ over $R = \Q$, the product $\Q[x] \times \Q[x]$ has zero divisors, such as $(1, 0)(0, 1) = 0$, while $\Q[x] \otimes_\Q \Q[x] \cong \Q[x, y]$ is an integral domain.