Tensor product of algebras
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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
- [math]\displaystyle{ A \otimes_R B }[/math]
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]
- [math]\displaystyle{ (a_1\otimes b_1)(a_2\otimes b_2) = a_1 a_2\otimes b_1b_2 }[/math]
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 ⊗ 1B.[3] 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.[citation needed]
Further properties
There are natural homomorphisms from A and B to A ⊗R B given by[4]
- [math]\displaystyle{ a\mapsto a\otimes 1_B }[/math]
- [math]\displaystyle{ b\mapsto 1_A\otimes b }[/math]
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:
- [math]\displaystyle{ \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, }[/math]
where [-, -] denotes the commutator. The natural isomorphism is given by identifying a morphism [math]\displaystyle{ \phi:A\otimes B\to X }[/math] on the left hand side with the pair of morphisms [math]\displaystyle{ (f,g) }[/math] on the right hand side where [math]\displaystyle{ f(a):=\phi(a\otimes 1) }[/math] and similarly [math]\displaystyle{ g(b):=\phi(1\otimes b) }[/math].
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:
- [math]\displaystyle{ X\times_Y Z = \operatorname{Spec}(A\otimes_R B). }[/math]
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 [math]\displaystyle{ \mathbb{C}[x,y] }[/math]-algebras [math]\displaystyle{ \mathbb{C}[x,y]/f }[/math], [math]\displaystyle{ \mathbb{C}[x,y]/g }[/math], then their tensor product is [math]\displaystyle{ \mathbb{C}[x,y]/(f) \otimes_{\mathbb{C}[x,y]} \mathbb{C}[x,y]/(g) \cong \mathbb{C}[x,y]/(f,g) }[/math], which describes the intersection of the algebraic curves f = 0 and g = 0 in the affine plane over C.
- More generally, if [math]\displaystyle{ A }[/math] is a commutative ring and [math]\displaystyle{ I,J\subseteq A }[/math] are ideals, then [math]\displaystyle{ \frac{A}{I}\otimes_A\frac{A}{J}\cong \frac{A}{I+J} }[/math], with a unique isomorphism sending [math]\displaystyle{ (a+I)\otimes(b+J) }[/math] to [math]\displaystyle{ (ab+I+J) }[/math].
- Tensor products can be used as a means of changing coefficients. For example, [math]\displaystyle{ \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) }[/math] and [math]\displaystyle{ \mathbb{Z}[x,y]/(f) \otimes_\mathbb{Z} \mathbb{C} \cong \mathbb{C}[x,y]/(f) }[/math].
- Tensor products also can be used for taking products of affine schemes over a field. For example, [math]\displaystyle{ \mathbb{C}[x_1,x_2]/(f(x)) \otimes_\mathbb{C} \mathbb{C}[y_1,y_2]/(g(y)) }[/math] is isomorphic to the algebra [math]\displaystyle{ \mathbb{C}[x_1,x_2,y_1,y_2]/(f(x),g(y)) }[/math] which corresponds to an affine surface in [math]\displaystyle{ \mathbb{A}^4_\mathbb{C} }[/math] if f and g are not zero.
- Given [math]\displaystyle{ R }[/math]-algebras [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] whose underlying rings are graded-commutative rings, the tensor product [math]\displaystyle{ A\otimes_RB }[/math] becomes a graded commutative ring by defining [math]\displaystyle{ (a\otimes b)(a'\otimes b')=(-1)^{|b||a'|}aa'\otimes bb' }[/math] for homogeneous [math]\displaystyle{ a }[/math], [math]\displaystyle{ a' }[/math], [math]\displaystyle{ b }[/math], and [math]\displaystyle{ b' }[/math].
See also
- Extension of scalars
- Tensor product of modules
- Tensor product of fields
- Linearly disjoint
- Multilinear subspace learning
Notes
References
- Kassel, Christian (1995), Quantum groups, Graduate texts in mathematics, 155, Springer, ISBN 978-0-387-94370-1, https://archive.org/details/quantumgroups0000kass.
- Lang, Serge (2002). Algebra. Graduate Texts in Mathematics. 21. Springer. ISBN 0-387-95385-X.
Original source: https://en.wikipedia.org/wiki/Tensor product of algebras.
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