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 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 . This ring is an ''R''-algebra, associative and unital with identity element given by . 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 from ''A'' and ''B'' to given byKassel (1995), p. 32
:$a\mapsto a\otimes 1_B$
:$b\mapsto 1_A\otimes b$
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(A\otimes B,X) \cong \lbrace (f,g)\in \text(A,X)\times \text(B,X) \mid \forall a \in A, b \in B: (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:
:$X\times_Y Z = \operatorname(A\otimes_R B).$
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 ,y/math>-algebras \mathbb ,yf, \mathbb ,yg, then their tensor product is \mathbb ,y(f) \otimes_ \mathbb ,y(g) \cong \mathbb ,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 $\frac\otimes_A\frac\cong \frac$, 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 ,y(x^3 + 5x^2 + x - 1)\otimes_\mathbb \mathbb/5 \cong \mathbb/5 ,y(x^3 + x - 1) and \mathbb ,y(f) \otimes_\mathbb \mathbb \cong \mathbb ,y(f).
* Tensor products also can be used for taking products of affine schemes over a field. For example, \mathbb _1,x_2(f(x)) \otimes_\mathbb \mathbb _1,y_2(g(y)) is isomorphic to the algebra \mathbb _1,x_2,y_1,y_2(f(x),g(y)) which corresponds to an affine surface in $\mathbb^4_\mathbb$ if ''f'' and ''g'' are not zero.

See also

* Extension of scalars
*Tensor product of modules
*Tensor product of fields
*Linearly disjoint
*Multilinear subspace learning

Notes

References

* .
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{{DEFAULTSORT:Tensor Product Of Algebras
Algebras
Ring theory
Commutative algebra
Multilinear algebra