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Here's the quick way to describe what's going on: In linear algebra, you learned primarily about linear transformations. In particular, $T:V \to W$ is a function that takes a single vector, and produces another vector in a linear way. In particular $T$ satisfies $T(ax + by) = aT(x) + bT(y)$. It turns out that linear transformations can naturally be ...
$A\otimes_R B$ is not always reduced. In fact, this is good, because it may be used to see intersection multiplicities. Here is an example. Let $R=k[X,Y]$ be the coordinate ring of the plane. Let $A=k[X,Y]/(Y)$ be the coordinate ring of the line $Y=0$ in this plane, and $B=k[X,Y]/(Y-X^2)$ be the coordinate ring of the parabola $Y=X^2$. Then $A\otimes_R ... 2 In my opinion this is in fact mostly a question of taste, but not completely. Different approaches have different advantages. Coordinate system based approaches (those with all the indices, this is what you are referring to when you are referring to the transformation rules. These are just the statement of coordinate invariance of the geometrical content of ... 2 No. To show it's injective, you have to show that, if$\;\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$. 2 As$\Bbb R$-vector space,$\Bbb R$has basis$\{1\}$, thus$\Bbb R\otimes\Bbb R$has basis$1\otimes1$with associated bilinear map $$\phi:\Bbb R\times\Bbb R\longrightarrow\Bbb R\otimes\Bbb R, \qquad \phi(x,y)=xy(1\otimes1)$$ If$f:\Bbb R^2\rightarrow\Bbb R$is the bilinear map$f(x,y)=xy$, then$\bar f:\Bbb R\otimes\Bbb R\rightarrow\Bbb R$, defined as ... 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 The easiest way I can think of is to create a non-trivial bilinear map from$\mathbb Z_{10}\times \mathbb Z_{12}\to \mathbb Z_{2}$. Which one? The one we learn at the mother's teat: Multiplication itself! Let $$\phi([a]_{10},[b]_{12})=[ab]_2$$ Easy exercise: Check that this is well-defined and bilinear and non-zero. Remember that every bilinear map from ... 2 It suffices to show that if$K$and$L$are two fields extending a field$k$and$L$is algebraic over$k$, then$\dim K\otimes_k L=0$(let$K=A_p/m_pA_p$and$L=\overline{k}$). Let$R=K\otimes_k L$; to show that$R$has dimension$0$, it suffices to show that for any homomorphism$R\to F$where$F$is a field, the image of$R$in$F$is a field (if ... 2 Since$k\to \bar k$is integral so is$k(p)\to k(p)\otimes_k\bar k$(Atiyah-Macdonald: Chap.5, Ex.3). But then$\dim k(p)=\dim k(p)\otimes_k \bar k$(Matsumura: Exer. 9.2), so that indeed$\dim k(p)\otimes_k \bar k=0$. Edit: the fibers are actually finite Since the ring$k(p)\otimes_k \bar k$is noetherian of dimension zero, it is artinian and has thus ... 2 If we partially apply your second definition to a section$s \in \Gamma(E)$, we get a$C^\infty(M)$-linear map$\nabla s : \Gamma(TM) \to \Gamma(E)$. You should know the fact that$C^\infty(M)$-linear maps between sheaves of sections are always derived from bundle homomorphisms ("$C^\infty-linear maps are tensors"); i.e. ... 2 Expanding by linearity and removing all terms we know are zero, and using symmetry to gather the rest of the terms, we get \begin{align} 0 = &\,\frac13T(\vec v_1 + \vec v_2 + \vec v_3,\vec v_1 + \vec v_2 + \vec v_3, \vec v_1 + \vec v_2 + \vec v_3)\\ =&\, T(\vec v_1, \vec v_1, \vec v_2)+ T(\vec v_1, \vec v_2, \vec v_2) \\ &+ T(\vec v_1, \vec ... 1 It is just different conventions. In general one can choose to view any tensor product W_1\otimes\cdots\otimes W_p\otimes V^* \otimes U_1\otimes\cdots\otimes U_q $$as the vector space of linear transformations$$ V \to W_1\otimes\cdots\otimes W_p\otimes U_1\otimes\cdots\otimes U_q $$and the two conventions you quote are just particular instances of that ... 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 First a general remark, that really helps when trying to build isomorphisms between quite abstract objects : some objects are made for constructing morphisms from them, and some are made for contructing morphisms to them. Often this is can be seen in terms of a left/right adjunction. Tensor algebras are made for constructing morphisms from them. What is the ... 1 Consider the map \phi : T_A(M) \longrightarrow A\ltimes M defined by \phi(\sum_{i=0}^{l}m_i)=m_0+m_1, where m_i\in M^{\otimes i}. Clearly this map is onto with ker(\phi)= I. Therefore we have an F- linear isomorphism between T_A(M)/I and A\ltimes M. That this isomorphism preserves the multiplication can be easily checked. 1 Hint. One can define a ring homomorphism A\otimes_kA\to A by \sum a_i\otimes b_i\mapsto\sum a_ib_i. (This works for commutative k-algebras.) 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 Try starting from the other direction and use (F^*\omega) \otimes (F^*\eta)(v_{\sigma(1)},\ldots, v_{\sigma(p+q)}) = (F^*\omega)(v_{\sigma(1)},\ldots,v_{\sigma(p)})\cdot (F^*\eta)(v_{\sigma(p+1)},\ldots, v_{\sigma(p+q)}) = \omega(F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p)}))\cdot\eta(F(v_{\sigma(p+1)}),\ldots,F(v_{\sigma(p+q)})) = \omega \otimes \eta ... 1 If \;r\otimes m\to rm=0\; then$$r\times m=r(1\otimes m)=1\otimes rm=1\otimes0=0$$and we've finished. I am assuming, of course, the tensor product is over \;R\; without any further conditions. 1 We can add something to the already excellent answer given by Omnomnomnom. The wonderful Gravitation by Misner,Thorpe and Wheeler (affectionately called MTW) on page 75 describes the most general (m,n) tensor as:" a linear machine with n input slots for n 1-forms and m input slots for m vectors, given the requested input, it puts out a real number..." ... 1 It is reather easy to see that F is surjective. Choose a basis w_1,\dots,w_m for w and consider a linear map \Phi:V\to W. Expanding \Phi(v)=\sum_i\Phi_i(v)w_i defines \Phi_1,\dots,\Phi_m\in V^* and one immediately verifies that \Phi=F(\sum_i\Phi_i\otimes w_i). For injectivity, I think a bit of additional work has to be done. The main point is ... 1 Consider bases of V and W, \{v_1,\dots,v_m\} and \{w_1,\dots,w_n\}. For each i and j, consider the linear map$$ f_{ij}\colon V\to W, \qquad f_{ij}(v_k)=\begin{cases} w_j & \text{ifk=i$}\\[4px] 0 & \text{if$k\ne i$} \end{cases} $$It's easy to prove that \{f_{ij}:1\le i\le m,1\le j\le n\} is a basis for \operatorname{Hom}(V,W): ... 1 Tensor product for Abelian groups is right-exact; so if you take the multiplication map: u: \mathbb Z \to \mathbb Z, u(x) = mx, on the one hand the cokernel is \mathbb Z_m, and on the other hand if you take the tensor product with G, u \otimes G is multiplication by m on G so the cokernel is G/mG. By right-exactness this is isomorphic to G ... 1 Consider the map$$ t\colon G\times\mathbb{Z}_m\to G/mG, \qquad t(g,k+m\mathbb{Z})=kg+mG $$(show t is well defined). Let \alpha\colon G\times\mathbb{Z}_m\to H be a bilinear map, where H is any abelian group. Define$$ \beta\colon G/mG\to H$$by$\beta(x+mG)=\alpha(x,1+m\mathbb{Z})$and show:$\beta$is a group homomorphism$\beta\circ t=\alpha$... 1 There is a tensor-hom adjunction for the tensor product of algebras, but it exists at the level of the Morita 2-category, rather than the 1-category of algebras. Namely, the Morita 2-category has objects$k$-algebras, and the category of morphisms$A \to B$is the category$\text{Mod}(A^{op} \otimes B)$of$(A, B)$-bimodules, where composition is given by ... 1 It's basically the same, it's mainly a matter of implicit identifications. Namely, in a Hilbert space$H$there is a natural isomorphism$H\to H^*$(which is the bra-ket duality), so$H\otimes H\simeq H\otimes H^*$. Now there is a map$H\otimes H^*\to L(H)$which corresponds to what you call the outer product. So for two kets$|\psi\rangle\$ and ...