[16] When p, q ≥ 1, for each (k, l) with 1 ≤ k ≤ p, 1 ≤ l ≤ q, there is an R-multilinear map: where ∈ T End : a N M {\displaystyle \{m_{i}\mid i\in I\}} Crossref Hassan Haghighi, Massoud Tousi, Siamak Yassemi, Tensor product of algebras over a field, Commutative Algebra, 10.1007/978-1-4419-6990-3, (181-202), (2011). ⊗ , is exact but not after taking the tensor with E Anyway, the main reason why I asked the connectedness question has just collapsed: the product I was thinking of is not the tensor product. will be a generating set for There is an alternative argument. Let R be a commutative ring and let A and B be R-algebras. J M ϕ r The tensor product turns the category of R-algebras into a symmetric monoidal category. 1. {\displaystyle T\otimes _{R}-} Given a monoidal category and a coalgebra C in ℳ denote by ℳC (resp.Cℳ) the category of right (resp. The notion of extension of scalars has important senses in situations which are qualitatively dierent than complexication of real vector spaces. The term tensor producthas many different but closely related meanings. is exact in both positions, and the two given generating sets are bases, then ⊗ 1 {\displaystyle \{m_{i}\otimes n_{j}\mid i\in I,j\in J\}} Theorem 1.11 gives a necessary condition for a tensor product R⊗S to be a finite PIR, where ϕ y R ( The tensor product is not the coproduct in the category of all R-algebras. {\displaystyle f(a):=\phi (a\otimes 1)} There is always a canonical homomorphism E → E∗∗ from E to its second dual. , , is an exact functor. {\displaystyle {\mathfrak {T}}_{p}^{q}} $\begingroup$ Thanks a lot, though much of this answer is over my head (I'm still not beyound Atiyah/Macdonald in commutative algebra). E N There the coproduct is given by a more general free product of algebras. p ∣ ⊗ A proof is spelled out for instance as (Conrad, theorem 4.1).Related concepts. Sections of the exterior bundle are differential forms on M. One important case when one forms a tensor product over a sheaf of non-commutative rings appears in theory of D-modules; that is, tensor products over the sheaf of differential operators. Unlike the commutative case, in the general case the tensor product is not an R-module, and thus does not support scalar multiplication. ). x If S and T are commutative R-algebras, then S ⊗R T will be a commutative R-algebra as well, with the multiplication map defined by (m1 ⊗ m2) (n1 ⊗ n2) = (m1n1 ⊗ m2n2) and extended by linearity. YCor. E ∣ x {\displaystyle (f,g)} E tensor product of modules. 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: 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: More generally, the fiber product of schemes is defined by gluing together affine fiber products of this form. E Since A and B may both be regarded as R-modules, their tensor product, is also an R-module. Instead, the construction below of the tensor product V RC = complexication of V of a real vectorspace V with C over R is exactly right, as will be discussed later. 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: with the differential given by: for x in Xi and y in Yj, For example, if C is a chain complex of flat abelian groups and if G is an abelian group, then the homology group of It is an isomorphism if E is a free module of finite rank. M ( ∈ where the first map is multiplication by If the tensor products are taken over a field F, we are in the case of vector spaces as above. ⊗ T X → ( commutative ring with identity is a PIR. M ∣ r We denote the natural pairing of its dual E∗ and a right R-module E, or of a left R-module F and its dual F∗ as. j f indeed forms a basis for If E is a finitely generated projective R-module, then one can identify M tensor product of modules. the tensor functor is covariant in both inputs. B is a tensor field of type (p, q). ( The tensor product of two modules A and B over a commutative ring R is defined in exactly the same way as the tensor product of vector spaces over a field: One also has distributivity of the tensor product over direct sums (What else? Then is called an-bilinearfunctionif satisfies the followingproperties: 1. is -biadditive 2. 0 The tensor product of commutative algebras is of constant use in algebraic geometry. Thus, E∗ is the set of all R-linear maps E → R (also called linear forms), with operations. ⊗ j The most classical versions are for vector spaces (modules over a field), more generally modules over a ring, and even more generally algebras over a commutative monad. {\displaystyle g(b):=\phi (1\otimes b)} , q under In mathematics, the tensor product of modules is a construction that allows arguments about bilinear maps to be carried out in terms of linear maps.
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