Tensor product of algebras

In mathematics, the tensor product of two algebras over a commutative ring R is also an R-algebra. This gives the tensor product of algebras. When the ring is a field, the most common application of such products is to describe the product of algebra representations.

Definition
Let R be a commutative ring and let A and B be R-algebras. Since A and B may both be regarded as R-modules, their tensor product
 * $$A \otimes_R B$$

is also an R-module. The tensor product can be given the structure of a ring by defining the product on elements of the form a&thinsp;⊗&thinsp;b by
 * $$(a_1\otimes b_1)(a_2\otimes b_2) = a_1 a_2\otimes b_1b_2$$

and then extending by linearity to all of A ⊗R B. This ring is an R-algebra, associative and unital with identity element given by 1A&thinsp;⊗&thinsp;1B. where 1A and 1B are the identity elements of A and B. If A and B are commutative, then the tensor product is commutative as well.

The tensor product turns the category of R-algebras into a symmetric monoidal category.

Further properties
There are natural homomorphisms from A and B to A&thinsp;⊗R&thinsp;B given by
 * $$a\mapsto a\otimes 1_B$$
 * $$b\mapsto 1_A\otimes b$$

These maps make the tensor product the coproduct in the category of commutative R-algebras. The tensor product is not the coproduct in the category of all R-algebras. There the coproduct is given by a more general free product of algebras. Nevertheless, the tensor product of non-commutative algebras can be described by a universal property similar to that of the coproduct:
 * $$\text{Hom}(A\otimes B,X) \cong \lbrace (f,g)\in \text{Hom}(A,X)\times \text{Hom}(B,X) \mid \forall a \in A, b \in B: [f(a), g(b)] = 0\rbrace,$$

where [-, -] denotes the commutator. The natural isomorphism is given by identifying a morphism $$\phi:A\otimes B\to X$$ on the left hand side with the pair of morphisms $$(f,g)$$ on the right hand side where $$f(a):=\phi(a\otimes 1)$$ and similarly $$g(b):=\phi(1\otimes b)$$.

Applications
The tensor product of commutative algebras is of frequent use in algebraic geometry. For affine schemes X, Y, Z with morphisms from X and Z to Y, so X = Spec(A), Y = Spec(R), and Z = Spec(B) for some commutative rings A, R, B, the fiber product scheme is the affine scheme corresponding to the tensor product of algebras:
 * $$X\times_Y Z = \operatorname{Spec}(A\otimes_R B).$$

More generally, the fiber product of schemes is defined by gluing together affine fiber products of this form.

Examples

 * The tensor product can be used as a means of taking intersections of two subschemes in a scheme: consider the $$\mathbb{C}[x,y]$$-algebras $$\mathbb{C}[x,y]/f$$, $$\mathbb{C}[x,y]/g$$, then their tensor product is $$\mathbb{C}[x,y]/(f) \otimes_{\mathbb{C}[x,y]} \mathbb{C}[x,y]/(g) \cong \mathbb{C}[x,y]/(f,g)$$, which describes the intersection of the algebraic curves f = 0 and g = 0 in the affine plane over C.
 * More generally, if $$A$$ is a commutative ring and $$I,J\subseteq A$$ are ideals, then $$\frac{A}{I}\otimes_A\frac{A}{J}\cong \frac{A}{I+J}$$, with a unique isomorphism sending $$(a+I)\otimes(b+J)$$ to $$(ab+I+J)$$.
 * Tensor products can be used as a means of changing coefficients. For example, $$\mathbb{Z}[x,y]/(x^3 + 5x^2 + x - 1)\otimes_\mathbb{Z} \mathbb{Z}/5 \cong \mathbb{Z}/5[x,y]/(x^3 + x - 1)$$ and $$\mathbb{Z}[x,y]/(f) \otimes_\mathbb{Z} \mathbb{C} \cong \mathbb{C}[x,y]/(f)$$.
 * Tensor products also can be used for taking products of affine schemes over a field. For example, $$\mathbb{C}[x_1,x_2]/(f(x)) \otimes_\mathbb{C} \mathbb{C}[y_1,y_2]/(g(y))$$ is isomorphic to the algebra $$\mathbb{C}[x_1,x_2,y_1,y_2]/(f(x),g(y))$$ which corresponds to an affine surface in $$\mathbb{A}^4_\mathbb{C}$$ if f and g are not zero.
 * Given $$R$$-algebras $$A$$ and $$B$$ whose underlying rings are graded-commutative rings, the tensor product $$A\otimes_RB$$ becomes a graded commutative ring by defining $$(a\otimes b)(a'\otimes b')=(-1)^{|b||a'|}aa'\otimes bb'$$ for homogeneous $$a$$, $$a'$$, $$b$$, and $$b'$$.