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.

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 ⊗ b by^[1]^[2]

(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 1_A ⊗ 1_B.^[3] where 1_A and 1_B 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 of A and B to A ⊗_R B given by^[4]

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:

Hom(A\otimes B,X)\cong \lbrace (f,g)\in Hom(A,X)\times Hom(B,X)\mid \forall a\in A,b\in B:[f(a),g(b)]=0\rbrace

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 constant use in algebraic geometry. For affine schemes X, Y, Z with morphisms from X and Z to Y, so X = Spec(A), Y = Spec(B), and Z = Spec(C) for some commutative rings A,B,C, the fiber product scheme is the affine scheme corresponding to the tensor product of algebras:

X\times _{Y}Z=\operatorname {Spec} (A\otimes _{B}C).

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.
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 $\mathbb {A} _{\mathbb {C} }^{4}$ if f and g are not zero.

Notes

↑ Kassel (1995), p. 32.
↑ Lang 2002, pp. 629-630.
↑ Kassel (1995), p. 32.
↑ Kassel (1995), p. 32.

References

Kassel, Christian (1995), Quantum groups, Graduate texts in mathematics, 155, Springer, ISBN 978-0-387-94370-1 .
Lang, Serge (2002) [first published in 1993]. Algebra. Graduate Texts in Mathematics. 21. Springer. ISBN 0-387-95385-X.

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[1] Kassel (1995), p. 32.

[FOOTNOTELang2002629-630-2] Lang 2002, pp. 629-630.

[3] Kassel (1995), p. 32.

[4] Kassel (1995), p. 32.