Tensor product of algebras

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Tensor product of algebras over a field; itself another algebra

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 R B {\displaystyle 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 b 1 ) ( a 2 b 2 ) = a 1 a 2 b 1 b 2 {\displaystyle (a_{1}\otimes b_{1})(a_{2}\otimes b_{2})=a_{1}a_{2}\otimes b_{1}b_{2}}

and then extending by linearity to all of AR 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 ⊗RB given by[4]

a a 1 B {\displaystyle a\mapsto a\otimes 1_{B}}
b 1 A b {\displaystyle 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 B , X ) { ( f , g ) Hom ( A , X ) × Hom ( B , X ) a A , b B : [ f ( a ) , g ( b ) ] = 0 } , {\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 ,}

where [-, -] denotes the commutator. The natural isomorphism is given by identifying a morphism ϕ : A B X {\displaystyle \phi :A\otimes B\to X} on the left hand side with the pair of morphisms ( f , g ) {\displaystyle (f,g)} on the right hand side where f ( a ) := ϕ ( a 1 ) {\displaystyle f(a):=\phi (a\otimes 1)} and similarly g ( b ) := ϕ ( 1 b ) {\displaystyle 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 × Y Z = Spec ( A R B ) . {\displaystyle 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 C [ x , y ] {\displaystyle \mathbb {C} [x,y]} -algebras C [ x , y ] / f {\displaystyle \mathbb {C} [x,y]/f} , C [ x , y ] / g {\displaystyle \mathbb {C} [x,y]/g} , then their tensor product is C [ x , y ] / ( f ) C [ x , y ] C [ x , y ] / ( g ) C [ x , y ] / ( f , g ) {\displaystyle \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 {\displaystyle A} is a commutative ring and I , J A {\displaystyle I,J\subseteq A} are ideals, then A I A A J A I + J {\displaystyle {\frac {A}{I}}\otimes _{A}{\frac {A}{J}}\cong {\frac {A}{I+J}}} , with a unique isomorphism sending ( a + I ) ( b + J ) {\displaystyle (a+I)\otimes (b+J)} to ( a b + I + J ) {\displaystyle (ab+I+J)} .
  • Tensor products can be used as a means of changing coefficients. For example, Z [ x , y ] / ( x 3 + 5 x 2 + x 1 ) Z Z / 5 Z / 5 [ x , y ] / ( x 3 + x 1 ) {\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)} and Z [ x , y ] / ( f ) Z C C [ x , y ] / ( f ) {\displaystyle \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, C [ x 1 , x 2 ] / ( f ( x ) ) C C [ y 1 , y 2 ] / ( g ( y ) ) {\displaystyle \mathbb {C} [x_{1},x_{2}]/(f(x))\otimes _{\mathbb {C} }\mathbb {C} [y_{1},y_{2}]/(g(y))} is isomorphic to the algebra C [ x 1 , x 2 , y 1 , y 2 ] / ( f ( x ) , g ( y ) ) {\displaystyle \mathbb {C} [x_{1},x_{2},y_{1},y_{2}]/(f(x),g(y))} which corresponds to an affine surface in A C 4 {\displaystyle \mathbb {A} _{\mathbb {C} }^{4}} if f and g are not zero.
  • Given R {\displaystyle R} -algebras A {\displaystyle A} and B {\displaystyle B} whose underlying rings are graded-commutative rings, the tensor product A R B {\displaystyle A\otimes _{R}B} becomes a graded commutative ring by defining ( a b ) ( a b ) = ( 1 ) | b | | a | a a b b {\displaystyle (a\otimes b)(a'\otimes b')=(-1)^{|b||a'|}aa'\otimes bb'} for homogeneous a {\displaystyle a} , a {\displaystyle a'} , b {\displaystyle b} , and b {\displaystyle b'} .

See also

Notes

  1. ^ Kassel (1995), p. 32.
  2. ^ Lang 2002, pp. 629–630.
  3. ^ Kassel (1995), p. 32.
  4. ^ Kassel (1995), p. 32.

References

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