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11

Think of the tensor product of two modules as an object that is built out of the original modules, but with linearity encoded in each variable separately. This means that, as opposed to taking a sum in both variables simultaneously as in a direct product: $(m,n) + (m',n') = (m+m',n+n')$, the operations are naturally "one variable at a time", as in: $m \... 8 What you're missing is that the ring you're tensoring over affects the scalars you can slide from left to right in the tensor product.$\def\Q{\mathbb{Q}}$In fact, the first of your tensor products is$\mathbb{Q}$and the second is zero. We'll show the second one first (your proof also works and is the better proof because it generalizes, but I want to ... 8 arctic tern's answer is very nice and complete. But less formally, the adjunction really says something quite simple: By the universal property of the tensor product, a linear map out of$U \otimes V$is the same thing as a bilinear map from$U \times V$. (I.e. the tensor product is built precisely to encode bilinearity.) And a bilinear map from$U \times V$... 7 Please see the following paper by Bhargava and Satriano here, which contains the details of this computation and other related results. M. Bhargava and M. Satriano, On a notion of “Galois closure” for extensions of rings, Journal of the European Mathematical Society 16, 1881-1913 (2014). The aim of the authors is to define for commutative ring ... 6 (1) If$V\cong\bigoplus_I k$is isomorphic to a direct sum of$I$-indexed copies of$k$, then$V^*\cong\prod_I k$is isomorphic to a direct product of$I$-indexed copies of$k$. The duality map$V\otimes V^*\to k$can be reinterpreted as$\big(\bigoplus_I k\big)\otimes\big(\prod_I k\big)\to k$via the obvious dot product. In particular, since$k[x]$has the ... 5 There is a de-linearized version of tensor-hom adjunction, called currying in computer science, which states there is a bijection$\hom(U\times V,W)\cong \hom(U,\hom(V,W))$in the category$\mathsf{Set}$. Consider an arbitrary$\phi\in \hom(U\times V,W)$, i.e. a function$\phi:U\times V\to W$. By fixing$u\in U$, we can consider$\phi(u,-)$to be a function ... 4 Note that$\mathbb{Q}[X]/(X^{2}+1) \cong \mathbb{Q}(i)$, so $$\mathbb{Q}[X]/(X^{2}-2) \otimes_{\mathbb{Q}} \mathbb{Q}[X]/(X^{2}+1) \cong \mathbb{Q}[X]/(X^{2}-2) \otimes_{\mathbb{Q}} \mathbb{Q}(i) \cong \mathbb{Q}(i)[X]/(X^{2}-2)$$ Since$X^{2}-2$is irreducible over$\mathbb{Q}(i)$, the last expression is a field, namely$\mathbb{Q}(i, \sqrt{2})$. 4 This is somewhat similar to one of your previous questions. Note that$\mathbb{F}_{p}(\sqrt{X}) \cong \mathbb{F}_{p}(X)[T]/\langle T^{2}- X\rangle$. This polynomial is irreducible by Eisenstein's criterion, and is separable if$p \neq 2$. In this case, we have $$\mathbb{F}_{p}(\sqrt{X}) \otimes_{\mathbb{F}_{p}(X)} \mathbb{F}_{p}(\sqrt{X}) \cong \mathbb{F}_{p}... 4 Let's start with bilinear maps. In particular, choose a map f:M\times N\to P. Now consider the pairs (u,\alpha v) and (\alpha u,v) where u\in M, v\in N and \alpha is a scalar. We find that f(u,\alpha v)=\alpha f(u,v) = f(\alpha u, v), and this does not depend on what bilinear map we have chosen. In other words, as far as bilinear maps are ... 3 Other that simplifying into \bigwedge R=R\oplus R, you are completely right. You can also go for a different approach, by defining the degree-n terms as follows. Namely by \bigwedge^nM=M^{\otimes n}/N, where N is the submodule generated by the dublicate terms. Then we have \bigwedge M=\bigoplus_n \bigwedge^n M. From this definition, it is ... 3 Hint: Use formula (1) with c=0. 3 The problem is that the "obvious" inclusion mapping I \otimes B \to R \otimes B, given by tensoring the inclusion mapping with the identity, need not be injective. Recall that, by definition, a module is flat if tensoring with that module preserves the head of exact sequences; but that is the same as saying that tensoring an injective homomorphism with ... 3 Since 2 \cdot 2 = 1 \mod \mathbb{Z}/3\mathbb{Z} we obtain$$\frac{n}{2^k} \otimes x = \frac{1}{2^{k'}} \otimes nx$$with k' = 0,1. If k' = 1 we obtain$$\frac{n}{2^k} \otimes x = 1 \otimes 2nx$$and otherwise we have$$\frac{n}{2^k} \otimes x = 1 \otimes nx.$$Thus every element is of the form 1 \otimes x. Therefore \mathbb{Z}[1/2] \otimes \mathbb{Z}... 3 According to this MO question, and more precisely this answer by Jacob Lurie, the answer is in general no. To sum up the answer, if you have a monoidal functor F : \mathcal{A} \to \mathsf{Mod}_R, then F(I) \cong R, and thus when F if fully faithful you can recover R as \operatorname{End}(I). This then implies that F is given by F(A) = \... 3 Let D be a divisible abelian group, and let D' be a divisible torsion abelian group. The inclusion j:D\to D\oplus D' satisfies your requierements, because Since D is divisible we know \mathbb F_p\otimes D=D/pD=0 for every prime p, and the same holds for D'. Since D' is torsion, j\otimes \mathbb Q is the map 1_D\otimes \mathbb Q, which ... 2 If you have already proved that the tensor product M\otimes N is unique up to isomorphism, the heavy lifting is already done. What you need to prove now is that M\otimes N satisfies the condition for being N\otimes M. That is, you need to show that a \phi':N\times M\to M\otimes N such that (etc etc etc). This ought to be easy enough because you ... 2 If you want to go with the universal property route, assuming you've defined N \otimes M as a quotient of the free module with basis N \times M by the bilinearity relations: Consider the A-bilinear map \phi:M \times N \to N \otimes M given by \phi(m,n) = n \otimes m. Given P and a bilinear map f:M\times N \to P, the only way to define a ... 2 Fix a basis \{e_1, \ldots, e_n\} of V, and consider the dual basis \{f_1, \ldots, f_n \} of V^\ast. Then we have a basis$$\{e_1\otimes f_1,\ldots, e_i \otimes f_j, \ldots, e_n \otimes f_n\}$$for V \otimes V^\ast, and the matrix$$A = (a_{ij})$$is just a way of representing the element$$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \; e_i \otimes f_j \in V \... 2 Can linear duals (i.e. linear functionals) be represented using the geometric algebra formalism? Yes and no. In geometric algebra, dual vectors can be computed through Hodge duality. Let$\{u_1, u_2, \ldots, u_n\}$be an orthogonal basis set for an$n$-dimensional vector space. Let$I$be their geometric product, which is grade-$n$due to orthogonality. ... 2 How about$P=\mathbb Z/a\mathbb Z$and$f(x,y)=xy$? (More explicitly:$f(x,y)=x\psi(y)$where$\psi$is the natural homomorphism$\mathbb Z/ab\mathbb Z\to \mathbb Z/a\mathbb Z$). 2 A standard basis for$S^2(\mathbb{Z}^3)$is given by$(e_1e_1,e_1e_2,e_1e_3,e_2e_2,e_2e_3,e_3e_3)$, where juxtaposition denotes the symmetric product and$(e_i)_{i=1}^3$is the standard basis of$\mathbb{Z}^3$. A symmetric integral$3 \times 3$matrix is uniquely determined by$6$integer entries which lie on the coordinates$(1,1),(1,2)(1,3),(2,2),(2,3),(3,...

2

Your guess is correct, except $2$ is unit modulo $3$, so that $\;\mathbf Z[1/2]\otimes\mathbf Z/3\mathbf Z\simeq \mathbf Z/3\mathbf Z$. To see it in a rigourous way, note that $$\mathbf Z[1/2]\otimes\mathbf Z/3\mathbf Z\simeq\mathbf Z[X]/(2X-1)\otimes\mathbf Z/3\mathbf \simeq\mathbf Z/3\mathbf Z[X]/(-X-1)\simeq\mathbf Z/3\mathbf Z.$$

2

The space $L^1(\mu) \tilde\otimes_\pi L^1(\nu)$ is isometrically isomorph (under the canonical isomorphism) to $L^1(\mu \times \nu)$, see, e.g., chapter 7 in Defant & Floret: "Tensor norms and operator ideals". I do not know whether the space $L^1(\mu) \tilde\otimes_\varepsilon L^1(\nu)$ has an easy description. Maybe you could try first with finite ...

2

The entanglement of the qubits does not matter. All you need to do is tensor $U$ with $\begin{pmatrix}1&0 \\ 0 & 1\end{pmatrix}$, which is the effect of $U$ on the third qubit. You'll get $U' = \begin{pmatrix}U&0 \\ 0 & U\end{pmatrix}$ or a permutation thereof. As a linear transformation, $U' = U \otimes \text{id}$, so you'd have $$U' \left|... 2 Every object in an additive category is a group and a cogroup, canonically, with respect to the monoidal structure of direct sum. This reflects the fact that vector spaces, and more generally, modules, are actually groups. So there's nothing there. Hopf algebras weren't invented to cause pain, much less because of the desire to decorate the monoidal ... 2 You ask: wouldn't it make the most sense to just use the monoidal product from the "mother category"? but this question does not make sense, really. To judge if something makes more sense or less sense than something else, you have to spell out what you are trying to achive. Sometimes, the direct sum is the correct operation to turn modules into a ... 2 By the universal property you have a set theoretic map l : V \times V^* \to V \otimes V^*, l(v,f) = v \otimes f, and a linear map \phi : V \otimes V^* \to \mathcal{L}(V,V) such that \phi \circ l = g. Hence \phi takes v\otimes f \mapsto vf(\cdot). Show \phi is an isomorphism. 2 You have to because it shows \ker\phi=0. Indeed, by linearity, for a general tensor$$\phi\bigl(\sum_i v_i\otimes f_i\bigl)=\sum_i\phi(v_i\otimes f_i)\stackrel{\text{def}}{=}\sum_i g(v_i,f_i),$$so that$$\sum_\limits ig(v_i,f_i)=0\iff \sum_i(v_i\otimes f_i)\in\ker\phi.

1

By the universal property of $A \otimes_R B$, there is a natural homomorphism $A \otimes_R B \to (A \otimes B)/H$, $a \otimes_R b \mapsto a \otimes b + H$. To go the other way around, note that the universal property of $A \otimes B$ gives us a homomorphism $A \otimes_R B \leftarrow A \otimes B$; this map descends to a homomorphism $A \otimes_R B \leftarrow (... 1 One example is the case$R = \Bbb{Z}_4$,$B = I = \{0, 2\}$that Osbourne is talking about at that point in the book. In this example, you find that$I \otimes B \simeq R \otimes B \simeq \Bbb{Z}_2$, but the natural mapping from$I\otimes B$to$R \times B$is not an injection. In fact,$2 \otimes 2$generates$I \otimes B$but maps to zero in$R \otimes B\$, ...

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