Tensor products allow us to build "linear" objects from "multilinear" ones. It can refer to: basic ones from linear algebra/module theory, or more sophisticated versions from differential/algebraic geometry (bundles/sheaves), functional analysis (Hilbert/Banach/locally convex spaces), or in their ...

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Is it true that $M\otimes_A F\simeq M^{(I)}$?

Let $A$ be an $R$-algebra ($R$ is a commutative ring with identity $1_R$) and suppose $F$ is a left free module over $A$. Is it true that $$M\otimes_A F\simeq M^{(I)}$$ for any right module $M$ over ...
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1answer
31 views

Another approach to $A/I\otimes_A A/J\simeq A/(I+J)$?

The same question appears here $A/ I \otimes_A A/J \cong A/(I+J)$ however I'm looking for a different approach. Let $A$ be an algebra, $I$ a right ideal and $J$ a left ideal. I'd like to show ...
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1answer
61 views

When is $M\otimes N$ a module?

So suppose we have a $R$-$S$-bimodule $M$, and an $S$-$T$ bimodule $N$, then we can construct the abelian group $M\otimes_S N$. Under what conditions could we make this a module? I would expect this ...
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34 views

Revert problem in Kronecker tensor product

I have a question related to reverting kron tensor product. As We can see the in below example: ...
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31 views

Ring homomorphism of tensor product of algebras

Let $B, C$ be two $A$-algebras, $f:A \to B, g: A\to C$ the corresponding ring homomorphisms. From this we can construct an $A$-algebra $B \otimes _A C$ and the mapping $ a \mapsto f(a) \otimes ...
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80 views

Determinant of the Transpose of an Operator.

Let $V$ be a vector space over a field $F$ of characteristic $0$. A linear operator $T$ on $V$ induces a linear operator $\Lambda^k T:\Lambda^k V\to \Lambda^k V$ such that $\Lambda^k T(v_1\wedge ...
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30 views

Canonical linear mapping is bijective

Let $V$ be a $K$-vector space with finite dimension. Proof that mapping: $V^* \otimes V \rightarrow {\rm End}_K(V), \ h\otimes a\mapsto (x\mapsto h(x)a)$ is bijective. So we have one mapping, which is ...
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1answer
56 views

Kernel of the Symmetrizing Map $Sym:\bigotimes^k V\to \bigotimes^k V$

$\DeclareMathOperator{\sym}{Sym}$ Let $V$ be a finite dimensional vector space over a field of characterisitc $0$ and $\sym:\bigotimes^k V\to \bigotimes^k V$ be the map given by $$ ...
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30 views

Tensor Product of Vectors

let $S,T$ be respectively $k-, n- $ tensors; $k,n>0$. Then we define the tensor product $$ T \otimes S(x_1,x_2,....,x_{k+n}):=T(x_1,...,x_k)S(x_{k+1},...,x_{k+n}) $$ (their product as Real numbers, ...
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22 views

Evaluation of polynomials at tensor products

Let $S,T$ be $R$-Algebras, $f \in S[X]$ a polynomial. in my notes it says you can easily lift $f$ to a ploynomial $f'$ in $(S \otimes T)[X]$. But I have no idea what $f'(s \otimes t)$ is. My guess is ...
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66 views

Example of two field extensions such that their tensor product is not a field

Example of two fields $K$ and $L$, which are extensions over $k$, such that $K\otimes_k L$ is not a field. Here is what I did. But I am a little bit unsure. Can someone suggest anything, or perhaps ...
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Rank of tensor product of morphisms

Let $R$ be a commutative, noetherian, unital ring, $F$ and $G$ two projective $R$ modules, $\phi: F\to G$ a module morphism and $M$ a finitely generated $R$ module such that $$\phi \otimes M := \phi ...
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23 views

Why not take the tensor product of two left modules in this way? [duplicate]

Let $A,B$ be two left $R$-modules. I was wondering if we then can form the tensor product of $A$ and $B$ by the free abelian group on $A \times B$ divided out by the span of the following elements ...
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1answer
30 views

Direct Sum vs. Direct Product vs. Tensor Product

There are a lot of questions like this all over the site, but I cannot find one that resolved my confusion- what are the formal definitions of direct sums, direct products, and tensor products (in the ...
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56 views

Tensor product of bounded analytic functions

Let $H^\infty(\mathbb{D})$ denote the set of functions holomorphic and bounded on $\mathbb{D} = \{z \in \mathbb{C}: |z| < 1\}$. Conseqently, $H^\infty(\mathbb{D}^n)$ denotes the set of bounded ...
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22 views

Dense subsets in tensor products of Banach spaces [duplicate]

Assume $B_1$ and $B_2$ are Banach spaces of univariate functions. Moreover, assume that the sets $D_1 \subset B_1$ and $D_2 \subset B_2$ are dense with respect to the respective norms ...
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52 views

Dense subsets in tensor products of Banach spaces

Assume $B_1$ and $B_2$ are Banach spaces of univariate functions. Moreover, assume that the sets $D_1 \subset B_1$ and $D_2 \subset B_2$ are dense with respect to the respective norms ...
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1answer
119 views

Isomorphism between $\Bbb{R}^2 \times \Bbb{R}^2$ and $\Bbb{R}^2 \otimes \Bbb{R}^2$ [closed]

I hope you can help me with this: Show that $\Bbb{R}^2 \times \Bbb{R}^2$ and $\Bbb{R}^2 \otimes \Bbb{R}^2$ are isomorphic, and specify an isomorphism. Thanks.
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39 views

About the tensor product identity: $A=B \otimes C = B \otimes I + I \otimes C$

I am reading about Chern classes in Nakahara's Geometry, Topology and Physics, and am having trouble understanding the equation $$ A=B \otimes C = B \otimes I + I \otimes C \tag{1}$$ where $A,B,C$ are ...
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49 views

Showing the tensor product spans a vector space, and interpreting things

The question is as follows: Prove that the tensor products $\tau\otimes\theta\in L(V,W;F)$ where $\tau\in V$ (suspect typo for $\tau\in V^*$ - the book even defines this above) and $\theta\in W$ ...
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26 views

Compatibility of operator spaces and tensor product norm

I have a problem with understanding the notion of complete boundedness in tensor notation. One possible way of saying a mapping $\phi\colon \mathcal{A}\to \mathcal{B}$ is completely bounded is to ...
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1answer
30 views

Elements of $M\otimes_{\mathbb K}\mathbb K[t]$?

I'm studying tensor product of modules over algebras and I came across the following statement. Given a $\mathbb K$-module $M$, the elements of $M\otimes_{\mathbb K}\mathbb K[t]$ can be written as ...
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1answer
46 views

If $a\otimes(b\otimes c)=0$ then $(a\otimes b)\otimes c=0$

I'm trying to prove the identity above, while the tensor product is between members of abelian groups $A,B,C$. This seemed trivial to me at first but since the tensor products are quotient groups I ...
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1answer
24 views

Left/Right-module distinction for tensor products

I am working through Cohomology of Groups (Kenneth Brown) and I noticed that in the beginning chapters when tensor products came up, he would emphasize the following: the tensor product $N \otimes_R ...
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1answer
55 views

How to construct a tensor product of two preadditive categories in pure categorical fashion?

Let $\mathsf C$ and $\mathsf D$ be two preadditive categories (by an preadditive category I mean a category together with compatible abelian group structure on every hom-set). I thought of ...
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2answers
206 views

Kernel of a bilinear map and tensor product specificially

I am cementing my understanding of tensors and a book I am reading handwaves and simply says "of course we may denote $0\otimes 0$ as the $0$ of $U\otimes V$" I have proved that the kernel is larger ...
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2answers
41 views

$R/I \otimes_A R/I \cong (R \otimes_A R)/(I \otimes_A I)$?

Let $f: A \to R$ be a homomorphism of commutative rings, and let $I$ be an ideal of $R$. Is it true that $R/I \otimes_A R/I \cong (R \otimes_A R)/(I \otimes_A I)$ ? After obtaining the surjection ...
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227 views

Determinant of the Kronecker Product of Two Matrices

I'd like to know how can be shown that $\det(A \otimes B) = \det(A)^m \det(B)^n$ when $A$ and $B$ are square matrices of size $n$ and $m$ respectively and $\otimes$ represents the Kronecker product of ...
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what is a rank-n tensor

So I am taking this mechanics class over the summer the book introduces the concept of rank 2 tensors for moment of inertia as any physics book it didn't give the concept justice I understand how to ...
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146 views

How to intuitively understand prolongations

This question is concerned with the algebraic side of the theory of prolongations as explained in this paper by V. Guillemin and S. Sternberg. Let me first introduce my notation. We're working with a ...
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37 views

Characterization of tensor products of fields

For which commutative rings $R$ are there field homomorphisms $L \leftarrow K \to L'$ (not assumed to be algebraic or anything) such that $R \cong L \otimes_K L'$? Is there an intrinsic ...
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3answers
55 views

Show that the tensor product is zero.

This time I am asking for help in the following question, regard $M=\mathbb{Z}$ as a $\mathbb{Z}$-module, then $M_i=\mathbb{Z}/m_i\mathbb{Z}$ where $m_i=1,2,...$ Show that $M_1 \otimes M_2=0$ for ...
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1answer
44 views

What is the tensor product with zero factors?

I am trying to understand what tensor products are, but one particular thing confuses me slightly. For $V$ a vector space over a field $\mathbb{k}$, why do we have that $V \otimes \mathbb{k} \cong V$? ...
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1answer
79 views

The realationship of $\hom(M\otimes_RN,M'\otimes_RN')$ and $\hom_R(M,M')\otimes\hom_R(N,N')$.

Let $R$ be a ring with identity, $M$ and $M'$ two right $R$-module, $N$ and $N'$ two left $R$-module. There is a natural way to define a homomorphism ...
4
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2answers
56 views

Linear Maps as Tensors

Let $V$ and $W$ be finite dimensional vector spaces and let $V^{\ast}$ denote the dual $V$. I read that the space $V^{\ast}\otimes W$ may be thought of in four different ways: as the space of linear ...
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2answers
86 views

For what kind of $R$-modules $M$ can we find an element $m\in M$ satisfing that $i:M\to M\otimes_R M, x\mapsto x\otimes m$ is an epimorphism?

Let $R$ be a commutative ring with identity and $M$ a $R$-module. I'm interested in under what condition we can find an element $m\in M$ satisfing that $i:M\to M\otimes_R M, x\mapsto x\otimes m$ is an ...
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1answer
30 views

Is $(A \otimes X) \circ (B \otimes Y)$ equal to $AB \otimes XY$ ? If yes, why?

If $A,B,X,Y$ are linear functions over $V_1$ and $V_2$ vectorial spaces $A : V_1 \to V_1$ $B : V_1 \to V_1$ $X : V_2 \to V_2$ $Y : V_2 \to V_2$
3
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2answers
87 views

Is there a relation between $End(M)$ and $M$ under tensor products?

Let $R$ be a commutative ring and $M$ be an $R$-module (If necessary, you can assume more conditions) Let $\phi$ be an $R$-endomorphism on $M$ and $\overline{\phi}:M^{\otimes n}\rightarrow M^{\otimes ...
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0answers
15 views

Annihilator for a tensor $T\in\wedge V^{\ast}$ [duplicate]

For $T\in\wedge^{k} V^{\ast}$ the annihilator is set $$an(T)= \{\phi\in V^{\ast}\mid \phi\wedge T=0\}$$ Then I need to prove that $dim(an(T))\leq k$ and is equal iff $T$ is decomposable ($i.e.$, ...
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84 views

Tensor Product of Complexes and the definition of the differentials

Suppose we have the following complexes, $$0 \rightarrow R \xrightarrow{x_1} R \rightarrow 0$$ $$0 \rightarrow R \xrightarrow{x_2} R \rightarrow 0$$ $$0 \rightarrow R \xrightarrow{x_3} R \rightarrow ...
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Pullback map distributes over wedge product (proof)

To prove that the pullback map distributes with the wedge product is it first best to prove that it distributes over the tensor product and then use the relation $$dx^{\mu_{1}}\wedge\cdots\wedge ...
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What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$?

What is $\mathbb{Z}[i] \otimes_{\mathbb{Z}[2i]} \mathbb{Z}[i]$? Also, since $\mathbb{Z}[i]$ is a PID, we should be able to write this $\mathbb{Z}[i]$-module as a direct sum of cyclic ...
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1answer
44 views

Perfect pairing induces isomorphism of tensor products

Let $M, N$ be $R$-modules and $(\cdot, \cdot): M \times N \to R$ be a perfect pairing. Wikipedia sais that this means that the map $\varphi: M \to \text{Hom}_R(N, R), m \mapsto (n \mapsto (m, n))$ is ...
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1answer
29 views

Tensor product of homology equivalences

Let $f : C \to C'$ and $g : D \to D'$ be chain maps of non-negative chain complexes of $R$-modules, where $R$ is any commutative ring. Assume that $f$ and $g$ are homology equivalences. Is the same ...
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17 views

Is there a natural link between symmetric polynomials and symmetric algebra?

Let $R$ be a commutative ring and $R[X_1,...,X_n]^{S_n}$ be the ring of symmetric polynomials. I have learned some basic properties of this ring and the results are really similar to those by ...
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Understanding Extension of Scalars in a Vector Space

$\newcommand{\R}{\mathbf R}\newcommand{\C}{\mathbf C}$ Low-Tech Complexification: Let $V$ be a finite dimensional vector space over $\R$. We can forcefully make $W:=V\times V$ into a complex vector ...
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27 views

Tensor independence

Let $(e_{i})$ be a basis in $V$, $( \epsilon_{i} )$ - basis in $V^{*}$ so that $\epsilon_{i} (e_{j})= \delta_{i}^{j}$ (Kronecker delta, $\epsilon_{i} (e_{j}) = 1 \Leftrightarrow i=j$, otherwise it's ...
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2answers
52 views

tensor product of polynomial algebra

Is $R[x] \otimes R[x]$ a free $R \otimes R$-module? Here $R$ is a $k$-algebra and $\otimes = \otimes_k$.
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$\mathbb{Z}G\otimes_{\mathbb{Z}N} \mathcal{N}= \mathcal{N}^G$ is a non-trivial idempotent ideal in $\mathbb{Z}G$

Let $G$ be a group and $I_G$ be the augmentation ideal of the group ring $\mathbb{Z}G$, i.e. $I_G$ consists of formal linear combinations $\sum n_i g_i$ ($n_i\in\mathbb{Z}$, $g_i\in G$) such that ...
2
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1answer
56 views

Is it possible that $1\otimes 1 = 0$?

Let $R$ be a commutative ring. Let $A,B$ be $R$-algebras and consider their product $A\otimes_R B$. Is it possible that $1\otimes 1=0$? What is an example? If $R$ is a field, $1\otimes 1$ is never ...