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It is a common misconception that every element of $S \otimes T$ has the form $s \otimes t$ and that you can define multiplication on $S \otimes T$ via such a formula $(s \otimes t) \cdot (s' \otimes t') = ss' \otimes tt'$ and verify the ring axioms with elements. Rather, one has to use the universal property of the tensor product in order to construct a ...

3

The definition of the tensor product is its universal property, which is quite simple. What you are struggling with is the construction of the tensor product - this is something different. If you want to see a construction of the tensor product which avoids free modules at all, see here. $K$ is by definition a submodule, since it is defined as the submodule ...

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The map $$g_1^1 \left(v_1 \otimes \alpha_1), (v_2 \otimes \alpha_2) \right) = g(v_1,v_2) + g^*(\alpha_1, \alpha_2)$$ isn't even well-defined: For all $\lambda \in \mathbb{R} - \{0\}$ we have $$v_1 \otimes \alpha_1 = (\lambda^{-1} v_1) \otimes (\lambda \alpha_1),$$ but $$g(\lambda^{-1} v_1,v_2) + g^*(\lambda \alpha_1, \alpha_2) = \lambda^{-1} g(v_1,v_2) + ... 3 If R is any ring and M is a left R-module, then there is always an isomorphism of abelian groups R \otimes_R M \cong M given by r \otimes m \mapsto rm and m \mapsto m \otimes 1 in the other direction. In fact, (r,m) \mapsto rm is R-balanced, hence induces a homomorphism R \otimes_R M \to M of abelian groups. Clearly, m \mapsto m \otimes ... 2 The tensor product over the field k is the coproduct in the category of commutative k-algebras. This means that you are given maps k[x] \to k[x] \otimes_k k[y] and k[y] \to k[x] \otimes_k k[y], both given by f(x) \mapsto f(x) \otimes 1 and g(y) \mapsto 1 \otimes g(y). Since you have inclusion maps k[x] \to k[x,y] and k[y] \to k[x,y], the ... 2 If you multiply v_1 by \lambda and \alpha_1 by 1/\lambda, you don't change their tensor product v_1 \otimes \alpha_1. Hence any formula that one proposes for the scalar product must also be left unchanged by such an operation, and this rules out your formula with "plus" instead of "times". 2 Consider the multiplicative set S= \mathbb Z\setminus \{0\}. Then$$\mathbb Q\simeq S^{-1}\mathbb Z.$$Now, in general you have that, if M is an A-module and S is a multiplicative set of A, then$$S^{-1}\otimes_AM \simeq S^{-1}M, $$that is naturally a S^{-1}A-module. This shows that this definition of rank of an A-module holds whenever A is ... 2 Maybe an example makes this more clear. Let A=\mathbb{Z}, and B=\mathbb{C}, with the map A\rightarrow B being the inclusion \mathbb{Z} \hookrightarrow \mathbb{C}. Now let M be the polynomial ring \mathbb{Z}[X,Y], which is clearly a \mathbb{Z}-module. Then$$ B\otimes_A M = \mathbb{C} \otimes_\mathbb{Z} \mathbb{Z}[X,Y] \cong \mathbb{C}[X,Y], ...

2

This is true if $M$ is finitely generated. By the structure theorem, there exists a presentation $M = \mathbb{Z} \langle e_i \rangle$ where the only relations are of the form $n_ie_i = 0$, for some $n_i \in \mathbb{Z}$. Then $M \otimes M = \mathbb{Z}\langle e_i \otimes e_j \mid (n_i, n_j) \ne 1 \rangle$, and to define a group homomorphism on $M \otimes M$ it ...

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In general, if $A$ is a commutative ring, $I$ an ideal, and $M$ an $A$-module, then $A/I\otimes_A M\simeq M/IM$. To see this, consider the bilinear mapping $A/I\times M\to M/IM$ given by $(\bar{x},m)\mapsto \overline{xm}$. By the universal property of the tensor product, this induces a well-defined map $A/I\otimes_A M\to M/IM$ given by $\bar{x}\otimes ... 2 The answer for your first question is yes. The answer for your second is no. The next proposition is well known. It was proved here. Proposition: Let$V,W$be vector spaces over$k$. Let$w=\sum_{i=1}^ra_i\otimes b_i=\sum_{i=1}^sv_i\otimes w_i\in V\otimes W$. If$\{a_1,\ldots, a_r\}$is linear independent then$\text{span ...

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I claim that $I=J=0$ already yields a family of counterexamples to $R\otimes_R R \cong R$. We must have $R\cdot R \neq R$ for this to work (I think this condition might also be sufficient for the natural map to not be an isomorphism, but I haven't worked out all the details). $R=n\mathbb{Z}$ is such a rng: $n\mathbb{Z} \otimes_{n\mathbb{Z}} n\mathbb{Z} ... 1 A natural measure for the similarity of two vectors is their inner product, and it has a geometric interpretation. Matrices can also be regarded as vectors, and the corresponding inner product is $$A:B = \sum_{i,j}a_{ij}b_{ij} = \operatorname{Tr}(A^TB).$$ If the matrices$A$and$B$are symmetric, they can be written in terms of their eigenvectors and ... 1 This question has been answered in comments: The implication$2\implies 1$does not hold. Take for instance$\mathbb{Z}$-modules$A=\mathbb{Q}$,$B=\mathbb{Z}/2\mathbb{Z}$,$C=\mathbb{Z}/3\mathbb{Z}$. Then$A\otimes B=A\otimes C=0$, but$B$and$C$are not isomorphic$\mathbb{Z}$-modules. – adrido Jan 25 at 8:04 and Even easier: Take$A=0$to see ... 1 Yes, there is an abstract nonsense proof. You can show this by only using the universal properties of the tensor product and quotient algebras. Let more generally$A,B$be some commutative algebras over a commutative ring$R$with ideals$I \trianglelefteq A$and$J \trianglelefteq B$. Then we can show$A/I \otimes_R B/J \cong (A \otimes_R B)/(I'+J')$as an ... 1 With respect to a real field of scalars,$\mathbb{R} \otimes \mathbb{R}$is just one dimensional (scalar multiples can pass from one side to the other), while$\mathbb{R} \oplus \mathbb{R}$is two dimensional. More generally a basis for tensor product$V \otimes W$of two vectors spaces over field$\mathbb{F}$can be taken to be tensor products of basis ... 1 This is a misconception in the definition of the module structure. Given$b \in B$, one defines a homomorphism of abelian groups$S \otimes T \to B$via the universal property of the tensor product, mapping$s \otimes t \mapsto sbt$. Then, one defines the result of$r$to be$br$. We have$b(r+r')=br+br'$by construction, namely$r \mapsto br$really is ... 1 This is always the case. The obvious isomorphism of$\mathbb{Z}$-modules $$i^{*}(M \otimes_{\mathbb{Z}} N) \rightarrow i^{*}(M) \otimes_{\mathbb{Z}} i^{*}(N)$$ defined by$i^{*}(m \otimes n) \mapsto i^{*}(m) \otimes i^{*}(n)$(really the identity map as$\mathbb{Z}$-modules) is always a map of$\mathbb{Z}[H]$-modules. This is true for modules over Hopf ... 1 If$R$has a unit, this is still true.$R$is an$(R,R)$-bimodule,$M$is a left$R$-module, therefore$R \otimes_R M$is an$R$-left module with the action given by$r \cdot (x \otimes m) = rx \otimes m$. Define$f : M \to R \otimes_R M$by$f(m) = 1 \otimes m$. Then$f$is a morphism of$R$-left modules (it's obviously additive):$\$f(r \cdot m) = 1 \otimes ...

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