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4

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 ...


3

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 ...


2

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 ...


2

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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