A map \beta \colon M \times N \to P of R-modules is bilinear if it is linear in each variable. The tensor product M \otimes_R N is a module with a bilinear map (m, n) \mapsto m \otimes n such that every bilinear \beta factors as \beta(m, n) = \varphi(m \otimes n) for exactly one linear map \varphi [1,2]. It is built as the free module on the pairs (m, n) modulo the relations of bilinearity, and the universal property determines it up to isomorphism.
Its elements are finite sums \sum m_i \otimes n_i, and those of the form m \otimes n are called pure tensors. For free modules, R^m \otimes_R R^n \cong R^{mn} with basis e_i \otimes f_j, and the pure tensor v \otimes w has the coordinates v_i w_j, an m \times n matrix of rank at most 1.
The figure draws v \otimes w for v, w \in \mathbb{R}^5 as a 5 \times 5 grid whose squares have area proportional to |v_i w_j|. Every row is a multiple of w. Adding a second pure tensor p \otimes q gives a grid whose rows are no longer proportional, so the sum is not a pure tensor: most elements of a tensor product are genuine sums.
Torsion can make tensors vanish. In \mathbb{Z}/2 \otimes_{\mathbb{Z}} \mathbb{Z}/3 we have 1 \otimes 1 = 3(1 \otimes 1) - 2(1 \otimes 1) = (1 \otimes 3) - (2 \otimes 1) = 0, because 3 = 0 in \mathbb{Z}/3 and 2 = 0 in \mathbb{Z}/2, so the whole tensor product is 0. In general \mathbb{Z}/m \otimes_{\mathbb{Z}} \mathbb{Z}/n \cong \mathbb{Z}/d with d = \operatorname{gcd}(m, n) [2].
Quotients and localizations of modules are tensor products. For an ideal I, the map \bar{r} \otimes m \mapsto rm gives R/I \otimes_R M \cong M/IM, and for a multiplicative set S, the map (r/s) \otimes m \mapsto rm/s gives S^{-1}R \otimes_R M \cong S^{-1}M [1]. So both operations are cases of tensoring with a fixed module.
Tensoring is right exact: if M_1 \to M_2 \to M_3 \to 0 is exact, so is M_1 \otimes N \to M_2 \otimes N \to M_3 \otimes N \to 0 [1]. The reason is that the cokernel of the first map has the universal property of M_3 \otimes N: a bilinear map on M_3 \times N is the same as a bilinear map on M_2 \times N that vanishes on the image of M_1.
Injectivity is not preserved. Multiplication by 2 on \mathbb{Z} is injective, but after tensoring with \mathbb{Z}/2 it becomes multiplication by 2 on \mathbb{Z} \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2, which is 0. In the figure, the integers of both rows fold into the classes \bar{0} and \bar{1}, and every arrow n \mapsto 2n lands on \bar{0}; in particular \bar{1} goes to \bar{0}.
A module N for which tensoring preserves injective maps is called flat. Free modules are flat, and so is every localization S^{-1}R, which is why localization preserves exact sequences. Flatness has its own article later in this collection.
If A and B are R-algebras, then A \otimes_R B is a ring with (a_1 \otimes b_1)(a_2 \otimes b_2) = a_1a_2 \otimes b_1b_2, and a \mapsto a \otimes 1 and b \mapsto 1 \otimes b are ring homomorphisms [1]. A pair of R-algebra homomorphisms A \to C and B \to C is the same as one R-algebra homomorphism A \otimes_R B \to C.
For example, R[x] \otimes_R R[y] \cong R[x, y], and B \otimes_R R[x]/(f) \cong B[x]/(f): tensoring with B keeps the generators and the relations and changes the coefficients. If A and B are the coordinate rings of X and Y over a field k, then A \otimes_k B is the coordinate ring of X \times Y; in general \operatorname{Spec}(A \otimes_R B) is the fiber product of \operatorname{Spec} A and \operatorname{Spec} B over \operatorname{Spec} R.
Even when A and B are fields, A \otimes_R B need not be one. Since \mathbb{C} = \mathbb{R}[x]/(x^2 + 1), we get \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[x]/(x^2 + 1), and over \mathbb{C} the factors x - i and x + i are coprime, so the Chinese remainder theorem gives \mathbb{C} \times \mathbb{C}. The images of x - i and x + i are nonzero, but their product is 0.
The figure shows the surface |z^2 + 1| over the complex plane. Over the real axis its height x^2 + 1 never reaches 0, which is why \mathbb{R}[x]/(x^2 + 1) is the field \mathbb{C}. Over \mathbb{C} the surface touches 0 at i and -i, and these two points give the two factors of \mathbb{C} \times \mathbb{C}.