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You're right-- this sort of question is studied a lot. As you have defined things, you're looking at a semiring, instead of a ring because there are no additive inverses to the direct sum operation. Of course distributivity goes through, since $L \otimes (K \oplus J) \cong L \otimes K \oplus L \otimes J$ via $l \otimes(j \oplus k) \mapsto l \otimes j ... 4 If you consider any$\mathbb{Z}$-module$M$and its torsion part$t(M)$, we have the exact sequence $$0\to t(M)\to M\to M/t(M)\to0$$ that, tensored with$\mathbb{Q}$, says $$\mathbb{Q}\otimes_{\mathbb{Z}}M\cong\mathbb{Q}\otimes_{\mathbb{Z}}M/t(M)$$ So, if$1\otimes x=0$in$\mathbb{Q}\otimes_{\mathbb{Z}}M$(for some$x\in M$), then also ... 3 What you want is a self-enriched category, that is a monoidal category$\mathcal V$which is the underlying category of a$\mathcal V$-category. Any (symmetric) monoidal closed category is naturally self-enriched. Martin Brandenburg's answer is a special case (when the monoidal product is a cartesian one). You might want to take a look at Kelly's reprint ... 3 There are two trivial answers and one more profound answer: 1)$m \otimes n = m' \otimes n'$means that$(m,n) - (m',n')$lies in the mentioned submodule of bilinear relations 2)$m \otimes n = m' \otimes n'$means that$\beta(m,n)=\beta(m',n')$for all$R$-bilinear maps$\beta : M \times N \to T$, where$T$is any abelian group. 3) We have the following ... 3$Ax\simeq A/\operatorname{ann}(x)$and$A/I\otimes_A M\simeq M/IM$, so$Ax\otimes_AM=0$iff$M=\operatorname{ann}(x)M$.$(U/V)\otimes M\simeq (U\otimes M)/(V\otimes M)$and$(\ker g)\otimes M=\ker g_M$. 3 Hints in steps: The map$\;\varphi: I\times I\to A\;,\;\;\varphi(r,s):=rs\;$is bilinear, so we get a unique homomorphism$\;\phi: I\otimes I\to A\;$. Assume$\; 2\otimes 2+ x\otimes x= r\otimes s\;$. Prove that then$\;4 + x^2=rs\;$. Reach a contradiction by analyzing the (two) differentpossibilites of$\;r,s\;$: or$\;\deg r=0\;$or ... 3 You actually have a short exact sequence$0\to S\stackrel{f}\to R^2\stackrel{g}\to S\to 0$, where$g(a,b)=X^2a+X^3b$, and by tensoring this with$S$want to prove that it is not exact, that is,$S$is not$R$-flat. Since$R^2\otimes_RS\simeq S^2$by$(a,b)\otimes c\mapsto(ac,bc)$, we can see$f\otimes 1: S \otimes_R S \to S^2$sending$p\otimes_R q$to ... 3 As I mentioned in the comments, the method to find the weights of an irreducible representation with highest weight$\lambda$is: Take the orbit of$\lambda$under the Weyl group. Since the set of weights of any representation is preserved by the Weyl group, all these weights are in the weight space of$V_{\lambda}$. For an irreducible representation, the ... 2 There is an isomorphism of rings$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$(Hint: Use$\mathbb{C}=\mathbb{R}[x]/(x^2+1)$and then CRT), but$\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C} = \mathbb{C}$. So these are not isomorphic rings, since$\mathbb{C} \times \mathbb{C}$has zero divisors for example. But they are ... 2 The symmetric power$S^p(V)$of a vector space$V$is defined as the quotient$V^{\otimes p} / (v_1 \otimes \dotsc \otimes vp = v_{\sigma(1)} \otimes \dotsc \otimes v_{\sigma(p)} : v_i \in V, \sigma \in \Sigma_p)$. The natural isomorphism$V^{\otimes p} \otimes V^{\otimes q} \to V^{\otimes p+q}$extends to a linear map$S^p(V) \otimes S^q(V) \to S^{p+q}(V)$. ... 2 Note that$\mathbb{Z} \otimes \mathbb{Z}/n$is canonically isomorphic to$\mathbb{Z}/n$, and that the first arrow$\mathbb{Z} \otimes \mathbb{Z}/n \to \mathbb{Z} \otimes \mathbb{Z}/n$identifies with the multiplication by$m$map$\mathbb{Z}/n \to \mathbb{Z}/n$in the sense that the composition $$\mathbb{Z}/n \cong \mathbb{Z} \otimes \mathbb{Z}/n \to ... 2 Not completely sure about complex numbers, but... If A^\top A=B then, B is symmetric positive semi-definite. So you can take the eigenvalue/vector decomposition of B, such that B=U\Lambda U^\top. Then A=(U\Lambda^{1/2})^\top. 2 From matrix analysis, if B is complex-symmetric, there is a unitary matrix U such that B=UDU^{T} where the columns of U are eigenvectors of BB^{*} = B \bar{B} and D is diagonal and the entries are the positive square roots of the corresponding eigenvalues. Now, if the columns of U are real, then U is orthogonal and so B is orthogonally ... 2 Vector spaces with \oplus and \otimes do not form a semiring, since associativity etc. do not hold - the laws only hold up to isomorphism. These isomorphisms fit together in a certain way, and what we get is called a 2-semiring or 2-rig. Just like a semiring is a "fusion" of two monoids (one being commutative), a 2-semiring is a "fusion" of two ... 2 Since R/J\otimes_R A=0, the second exact sequence implies that the inclusion$$J\otimes_R A\to R\otimes_R A\cong A$$is surjective. J\otimes_R A is generated by elements of the form j\otimes a for some a\in A and j\in J. The homomorphism J\otimes_R A\to A\otimes_R R\cong A sends j\otimes a\mapsto j\otimes a=ja. The image is thus JA (as JA ... 2 For your second question, I really like Exercise 1.6.H in Vakil's notes, which deals with the interaction between left/right exact functors and homology. I think the hint there is pretty generous and you'll learn three useful facts. The upshot is that tensoring by M "preserves" kernels and cokernels, and one has \operatorname{im} f = ... 2 The definition in Cartan-Eilenberg is wrong. For, t_i = T(A_1,\ldots, s_i,\ldots,A_r) defines a map T(A_1, \ldots, A_i,\ldots A_r) \to T(A_1, \ldots, A'_i,\ldots A_r). But for a homotopy you need a map T(A_1, \ldots, A_r) \to T(A'_1, \ldots, A'_r). The correct definition of the homotopy in your case is u=:(s \otimes g_1, f_2 \otimes t): A \otimes ... 2 "I have seen direct proofs of the general statement [...] but I am not looking that.": Since you haven't specified what a direct proof or a category theoretic proof is for you, I am not sure if the following proof will satisfy your requirements. If not, please tell me why and we will try to find something else. Proof 1. Recall that in general M \otimes_R ... 2 End(O) is a set (or a monoid), it belongs to the category of sets and nothing else. Perhaps you are interested in the notion of a cartesian closed category. There one has an internal hom object \underline{\hom}(x,y) for all objects x,y, in particular \underline{\mathrm{End}}(x):=\underline{\hom}(x,x). 1 That element you want to form is just an elementary tensor x\otimes y in the algebraic tensor product \mathcal H\otimes\mathcal H. Then you want to have sums of those guys, and then limits of them. 1 Ok, in 3-dimensional real vector space, V, each tensor of rank two can be considered as a bilinear for in each of the following 4 cases: If B\in V^*\otimes V^* then B is a pairing V\times V\to\Bbb{R} via (v,w)\to B(v,w)=v^{\top}Bw or in components B(v,w)=v^sw^tB_{st}; If B\in V\otimes V then B is a pairing V^*\times V^*\to\Bbb{R} via ... 1 Here is an alternative approach which however only works for tensor products that can be considered as localizations: If R is a commutative ring and S\subset R is a multiplicative subset of R, then given any R-module M the R_S-module R_S\otimes_R M together with the map M\to R_S\otimes_R M, m\mapsto 1\otimes m, is a localization of M at ... 1 If I understood you correctly, we have (as \;\Bbb Z- modules = abelian groups) for any basic tensor \;a\otimes b\in \Bbb Z_2\otimes\Bbb Z_2\;:$$f\otimes1(a\otimes b):=f(a)\otimes b:=(2a)\times b=2(a\otimes b)=a\otimes (2b)=\ldots$$1 B \subset C doesn't imply A/B \subset A/C. It implies A/B surjects to A/C. The argument in Dummit & Foote must not be finished. To show what you want (which presumably comes later in Dummit & Foote): The key is to use (x+y) \otimes (x+y) = x \otimes y + y \otimes x + x\otimes x + y\otimes x. This shows that x\otimes y + y \otimes x is ... 1 For simplicity assume that V and W both have dimension 2 and have bases \{e_1, e_2\} and \{f_1, f_2\} respectively. I assume that \tau is defined by$$ \tau (a_1e_1 + a_2e_2, b_1f_1 + b_2f_2) = a_1b_1(e_1\otimes f_1) + a_2b_1(e_2\otimes f_1) + a_1b_2(e_1\otimes f_2) + a_2b_2(e_2\otimes f_2).$$Now say that$\tau$surjective, then there you ... 1 As Tobias said, when you see something like$M \otimes N$with unadorned tensor product where$M$and$N$are modules, what's usually happening is that there is some standard fixed base ring that is being suppressed in the notation. For instance, rather than rings I generally work with algebras over some field$k$- for me$\otimes$is always$\otimes_k$, ... 1 Hint: Take a linear combination$L=k_1a+k_2b+k_3c+k_4d$and evaluate at a vector$X$for which$a(X)=1$,$b(X)=0$,$c(X)=0$and$d(X)=0$. 1 The reason why they appeared in the wrong order is because you applied$\tau(g)$and$\tau(h)$to$L$in the wrong order. Call$\tau(h)(L) = L'$(since it is also a bilinear form). Then$(\tau(g)\tau(h))(L)$is defined as$\tau(g) (\tau(h)(L)) = \tau(g)(L')$. The reason why it is defined like this is because the operation in$GL(V)$is defined as composition ... 1 An$R$submodule of$\Bbb H$would have to also be an$\Bbb H$submodule by restriction of$R$'s action to the subring$\Bbb H\otimes 1\cong \Bbb H$. Thus a nontrivial$R$submodule would yield a nontrivial$\Bbb H$submodule, but of course$_\Bbb H\Bbb H$is simple, so there is no nontrivial submodule. Alternatively, you can just show that$R\$ acts ...