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The fact that $\mathbb{Z}[i]\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{C}$ only as $\mathbb{Z}$-modules follows as $\mathbb{Z}[i]\cong\mathbb{Z}^{\oplus 2}$ and $\mathbb{Z}^{\oplus 2}\otimes_{\mathbb{Z}}\mathbb{R}\cong\mathbb{R}^{\oplus 2}\cong\mathbb{C}$. If you need to use the universal property of tensor products somewhere, it can be used in showing that ...
The map you wrote is not an isomorphism. It is instead a map that arguably parametrizes (though not uniquely) all multiples of the diagonal map $M\rightarrow M^{\oplus n}$ (where the multiple, depending on an element $(r_1,\ldots,r_n)$ of $R^{\oplus n}$, is $\prod_ir_i$). This is why, as you observe, you could make a similar argument for $R^{\oplus n}\times ... 3 The set of$R$-bilinear maps$f: R\times N\to P$is in natural bijection with the set of$R$-linear maps$g:N\to P$. The correspondence is obtained by setting$f(r,n) = g(rn)$. 3 Suppose$\alpha_{m+1}f(v_{m+1})+\dots+\alpha_nf(v_n)=0$. This can be rewritten as $$f(\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n)=0$$ that is, $$\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n\in\ker(f)$$ In particular, $$\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n= \beta_{1}v_1+\dots+\beta_mv_m$$ for some scalars$\beta_1,\dots,\beta_m$. Therefore ... 3 The definition of the tensor product is$(f \otimes g)(u,v) = f(u)g(v)$. You should be able to check that whenever$f,g \in T^*$(i.e. are linear), their product$f \otimes g$is bilinear; and the product$f \otimes f$is symmetric and weakly positive definite. Thus the action of$dx \otimes dx$on a pair$u,v$of tangent vectors is $$(dx \otimes dx)(u,v) = ... 3 One reason why one needs the bilinear map (or multilinear in general), instead of just going ahead to define a map on the tensor product, is that one needs to show that the latter is well-defined. So it may be easy to define some A \otimes B \rightarrow C, but it actually may be quite hard to prove that it is well-defined. On the other hand, once a ... 2 The proof constructs two subalgebras C and D of A. You want to show that C\otimes D\simeq A, so you need to give an isomorphism C\otimes D\to A. The tensor product of two algebras X and Y has the following universal property : for any algebra Z, any two morphisms f: X\to Z and g: Y\to Z such that f(x)g(y)=g(y)f(x) for all x\in X, ... 2 Consider G=\Bbb Z[x], H= \Bbb R treated as \Bbb Z-modules. Then for everything to be an elementary tensor would mean that every polynomial in \Bbb R[x]\cong\Bbb Z[x]\otimes_{\Bbb Z}\Bbb R is of the form r\cdot p(x) for some r\in\Bbb R and p(x)\in\Bbb Z[x]. 2 In general this is not true: for instance that \mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_q \cong \{ 0\} if \gcd(p,q)=1. 2 No. To show it's injective, you have to show that, if \;\sum r_i\otimes m_i\mapsto\sum r_im_i=0, then \;\sum r_i\otimes m_i=0. But that is because \;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0. 2 An element of the domain of T is of the form (f_1,\dots, f_k, v_1, \dots, v_l),\, f_1,\dots,f_k\in V^*, v_1,\dots,v_l \in V, so the first k elements of T should be able to take elements of V^* and the next l terms should be able to take elements of V, hence v_{\mu_1} \otimes \cdots \otimes v_{\mu_k}(f_1,\dots, f_k) first, and v^{{\nu_1}^*} ... 2 The key is that, for any injective representations (i.e., *-homormophisms, so in particular f and g),$$ \|\sum a_j\otimes b_j\|_\min=\|(f\otimes g)\left(\sum a_j\otimes b_j\right)\| $$(technically, this might require using a further set of faithful representations to get embeddings A\hookrightarrow B(H), B\hookrightarrow B(K), but it doesn't ... 2 HINT: show that every element of \mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z} is equal to one of the form [k]\otimes 1 for some k\in\{0, . . . , m-1\}. It will be enough to show this for elements of the form [x]\otimes y. 2 Here's a counterexample. Take A=k[x]/(x^2) and M=N=A/(x). Then \operatorname{Hom}(M,A)\cong M, generated by the map f:1\mapsto x, and \operatorname{Hom}(A,M)\cong M, generated by the map g:1\mapsto 1. The tensor product \operatorname{Hom}(M,A)\otimes \operatorname{Hom}(A,M) is then also isomorphic to M, generated by f\otimes g. But ... 2 Let U and V be free modules over a nontrivial commutative ring R and let \alpha:U\to R and \beta:V\to R be two R-linear maps which have 1 in their image; such things are easily seen to exist using freeness. Then using the properties of tensor products you can show that there is a morphism of abeelian groups f:U\otimes_RV\to R such that for ... 1 The notation in the C^* literature for tensor products is far from uniform. But yes, as K(H) is nuclear, it doesn't matter which C^*-norm you put on K(H)\otimes A, they are all the same. 1 It's much easier to show that if A, B and C are R-modules (commutative R, for simplicity), then$$ (A\oplus B)\otimes_R C\cong (A\otimes_R C)\oplus (B\otimes_R C) $$The bilinear map (A\oplus B)\times C is$$ ((a,b),c)\mapsto (a\otimes c,b\otimes c) $$and it's quite easy to show that this satisfies the universal property. By induction, we get ... 1 Intuitively, any R bilinear map f:R \times N \to M is really a linear map from N to M in disguise, for if g is a linear map from N to M, then we may define f by letting f(1,n) = g(n), extended to a unique bilinear extension. The details are then trivial. Let f: R \times N \to M be a bilinear map. Given your projection map (r,n) \mapsto ... 1 how exactly to represent specific linear transformations in tensor notation. Indeed every linear map A:V\rightarrow W has a representation as a tensor T\in V^{*}\otimes W such that \forall x\in V:Ax=T(x)\in W. In chosen bases,$$ T=\sum a_{ij}\;v_j^*\otimes w_{i} $$where a_{ij} is the matrix of the linear map and \{v_j^*\} is the basis in ... 1 We have that \mathbb Z[X]\otimes_\mathbb Z\mathbb R\cong \mathbb R[X] because both are free commutative \mathbb R-algebras over one element. It immediately follows that \mathbb Z[i]\otimes_\mathbb Z\mathbb R\cong \mathbb R[i]\cong\mathbb C 1 Here is another way to do it: you can just write down a module homomorphism \mathbb{Z}[i]\otimes \mathbb{R}\to \mathbb{C} by a+bi\otimes r \mapsto ra+rbi, and show that it is an isomorphism. To see injectivity for example, you can say that if ra + rbi = 0 then ra = rb = 0 and hence r = 0 or a=b=0 in which case the tensor a+bi\otimes r = 0, so ... 1 Ok, lets formalise all that have been said in the comments: Let (v_1,\dots,v_n) and (v^1,\dots,v^n) be basis of V_p and V^*_p, respectively. Take a tensor \tau \in \mathcal{T}^r_s(V_p), and pick indexes k \leq r, l \leq s. Then we define the contraction C^k_l\tau \in \mathcal{T}^{r-1}_{s-1}(V_p) as ... 1 Tensor products of c.p. faithful maps with respect to the minimal norm are faithful: D. Avitzour, Free products of C*-algebras, Trans. Amer. Math. Soc. 271 (1982), 423–435. Certainly, *-homomorphisms are c.p. 1 For finite dimensional vector spaces, we have an isomorphism V \cong V^{**}, so you are really using a basis of V^{**} 1 So with the help of @Winther, the solution is as follows:$$ U = \frac{1}{r^3}\left(\vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2}\right) \equiv \frac{1}{r^3}P_iT_j^iQ^j\\ \vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2} = P_iT_j^iQ^j\\  P^ke_kQ_le^l- ... 1 Not necessarily. A counter-example would be the sub-algebra of the algebra of$3\times 3$matrices described by $$A = \begin{pmatrix} k & k & k \\ 0 & 0 & k \\ 0 & 0 & k \end{pmatrix}.$$ For this algebra,$\mu:A\otimes A\to A$is surjective, but there cannot be an identity element, since there are no$X\in A$such that $$X\cdot ... 1 Here is a very low dimensional example: consider a two dimensional vector space V with basis \left\{v_1,v_2\right\}. Then \left\{v_1\otimes v_1, v_2\otimes v_1,v_1\otimes v_2, v_2\otimes v_2\right\} is a basis of V\otimes V. You can easily show that$$v_1\otimes v_2+v_2\otimes v_1\neq u\otimes w$$for all$u,w\in V$. Edit: Be sure to work out the ... 1 Tensor product of vector spaces and algebras are both special cases of tensor product of modules. Indeed a vector space is just a module over some field and an algebra is just a module that has a ring structure. So you really need to understand the tensor product of modules to get the big picture. To make things easy it is better to focus on commutative ... 1 Tensor product is an "external" operation. It takes two completely unrelated algebras$A$and$B$and it spits out a new algebra$A \otimes B$that only depends on the data of$A$and$B$taken separately.$A$and$B$could be the same algebra, subalgebras of a bigger algebra, it won't matter from the point of view of$\otimes$. The product you define is an ... 1 (This works only for commutative$k$-algebras.) Hint. One can define a ring homomorphism$A\otimes_kA\to A$by$\sum a_i\otimes b_i\mapsto\sum a_ib_i\$.