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8

Sometimes it is better to look at a more general situation. In fact, this makes it easier to see what is really going on. 1) Let $V$ be a $K$-vector space and let $L/K$ be a field extension. Then $L \otimes_K V$ carries the structure of a vector space over $L$ via (linear extension of) $\alpha(\beta \otimes x) = \alpha\beta \otimes x$. 2) If ...


7

This is not true in general. For example, $\mathbb{Q}\otimes_{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z}) \cong 0 \cong 0\otimes_{\mathbb{Z}}(\mathbb{Z}/p\mathbb{Z})$ but $\mathbb{Q}$ and $0$ are not isomorphic as $\mathbb{Z}$-modules.


7

Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $A\otimes 0=0$ is finitely generated for any $A$). Choose a finite set of generators of $A\otimes_\mathbb{C} B$; each of these is a finite sum of tensors $a\otimes b$. Let $A_0\subseteq A$ be the subalgebra generated by all the $a$'s appearing in these tensors. Then $A_0$ is ...


5

No: We have $M \otimes 0 \cong 0 \cong M' \otimes 0$ for all modules $M,M'$. Even if $N$ is a very nice $R$-module, say free of finite rank, and non-zero, then $M \otimes N \cong M' \otimes N$ does not imply $M \cong M'$: This is because there are non-isomorphic modules $M,M'$ with $M^2 \cong M'^2$, i.e. $M \otimes R^2 \cong M' \otimes R^2$ (see here, take ...


5

You are asking about vector spaces over fields and modules over rings. Since the latter are more general, let's start with those. We know that tensor products are distributive over direct sums. In other words, for every triple $A, B, C$ of $R$-modules, we have a canonical isomorphism between $A\otimes (B \oplus C)$ and $(A\otimes B) \oplus (A \otimes C)$. ...


5

Let $\mathbb{F}$ be a field and let $V$ be a module over $\mathbb{F}$ (i.e. a vector space over $\mathbb{F}$), then $V^*$ is also a module over $\mathbb{F}$. The tensor product $V^*\otimes_{\mathbb{F}}V$ is canonically isomorphic to $\operatorname{End}_{\mathbb{F}}V$ via the map induced by the bilinear map $V^*\times V \to ...


4

Take $R$ to be a field $k$ and take $A = B = k^2$, with basis $e_1, e_2$. The tensor product $A \otimes_k B$ is $k^4$ with basis $e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2$, and most elements of it are indecomposable. For example, $e_1 \otimes e_1 + e_2 \otimes e_2$ is indecomposable. There's no more reason to expect all tensors to ...


4

Sorry, my previous answer was incorrect: I attempted to be too fancy. It looks like a straightforward approach is probably best. Take a simple tensor in $R \otimes_R N$ which will be of the form $r \otimes n = r(1 \otimes n) = r \iota (n)$. Arbitrary elements are finite sums of simple tensors so since $\iota$ is a homomorphism, we are done.


4

Assuming that you have seen the fact that $\Bbb{Q}$ is a flat $\Bbb{Z}$-module you can argue as follows. Because $A$ is assumed to be torsion free, the submodule generated by the element $a$ is isomorphic to $\Bbb{Z}$. So $f:\Bbb{Z}\to A, n\mapsto na$ is an injective homomorphism. Flatness means that $$ f\otimes id: \Bbb{Z}\otimes\Bbb{Q}\to A\otimes\Bbb{Q}, ...


3

As I commented, your approach is doomed because there are no homomorphisms from $\mathbb{Q}$ to $\mathbb{Z}$. Here are a couple different possible approaches you could take. Approach 1: Consider the localization $A_{(0)}$ of the $\mathbb{Z}$-module $A$ obtained by inverting every nonzero element of $\mathbb{Z}$. Explicitly, $A_{(0)}$ consists of ...


3

This is a natural isomorphism of abelian groups, so it will respect any extra module structures which may be present; naturality in $M$ implies that if $M$ is a $(C, A)$-bimodule for some other ring $C$ then this is an isomorphism of $C$-modules, and naturality in $P$ implies that if $P$ is a $(B, D)$-bimodule for some other ring $D$ then this is an ...


3

Clearly $2\otimes 1$ generates $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$ as a module over $\mathbb{Z}$. So it is enough to prove that $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$ is nonzero. But what would it mean if it was zero? This would mean: any $\mathbb{Z}$-bilinear map from $2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ to ...


3

Let $\mathcal B=(e_1,\ldots,e_n)$ be a basis of $V$ and $\mathcal B^*=(e_1^*,\ldots, e_n^*)$ the associated dual basis of $V^*$. The map $$V\otimes V^* \rightarrow End(V)$$ sends $e_i\otimes e_j^*$ to the linear operator : $$v=(v_1,\ldots,v_n)\mapsto e_j^*(v)e_i=\sum_{k=1}^nv_je_i.$$ Hence, the matrix representation of $e_i\otimes e_j^*$ in the basis ...


2

It's the standard adjunction with the Hom functor, namely $$ M_B\cong\operatorname{Hom}_A(A,M) $$ regarding $A$ as a $B$-module by means of the given homomorphism. Another way to see this is by building the natural bijection by hand. If $f\colon N\otimes_BA\to M_B$ is a homomorphism of $B$-modules, define $\tilde{f}\colon N\to M$ by $\tilde{f}(x)=f(x\otimes ...


2

If $I\subseteq R$ is any $G$-invariant ideal, then $I\otimes V$ is a $G$-invariant submodule of $R\otimes V$. So if $R$ has any nontrivial $G$-invariant ideals (e.g., if $G$ acts trivially on $R$ and $R$ is not a field), then $R\otimes V$ will have nontrivial $G$-invariant submodules.


2

Maybe you mean that $\mathrm{sl}_n(k) \otimes_k V \approx \mathrm{sl}_n(V)$. The isomorphism is induced by the bilinear map $\mathrm{sl}_n(k) \times V \rightarrow \mathrm{sl}_n(V)$ sending a pair $(A,v)$ where $A = (a_{ij})$ is an $n \times n$ traceless scalar matrix and $v \in V$ is a vector to the vector-valued matrix $$ \begin{pmatrix} a_{11} v & ...


2

The $\delta$s stand for Kronecker's delta: $$\delta_{ij} = \delta_i^j = \begin{cases} 1, & \text{if } i=j \\ 0, &\text{if } i\neq j \end{cases} $$ I will prove that the $\{f_\gamma^{kl}\}$ form a basis for $L(E\otimes F,G)$, and you do the other one. Let $\lambda_\gamma^{kl}$ be scalars such that the finite combination is zero: $$\sum ...


2

If you dont have any further knowledge on $M \otimes N$, this is the very first step and an instant consequence of the universal property of the tensor-product: $m \otimes n \in M \otimes N$ is non-zero iff there exists some $R$-Module $T$ and a bilinear map $M \times N \to T$, which maps $(m,n)$ to non-zero. Proof: If $m \otimes n \neq 0$, let $T = M ...


2

There are a couple of things "generate" could mean here. The weakest reasonable one, which suffices in this case, is "generates under taking tensor products, finite direct sums, and direct summands." In general, depending on the situation, you might want to take other kinds of colimits. For references try looking up anything that deals with Tannaka ...


2

The second sentence of the proof of Corollary 9 is kind of poorly phrased. Here's a clearer phrasing: Suppose now that $L$ is an $S$-module, $K\subseteq N$ is a submodule, and $\psi:N/K\to L$ is an injective $R$-module homomorphism. Write $\varphi:N\to L$ for the composition of $\psi$ with the quotient homomorphism $N\to N/K$, so $\ker(\varphi)=K$. ...


1

Here's a counterexample. Let $k$ be a field and take $A=B=k[x]/(x^2)$ and $M=C=k$ (considered as an $A$-algebra via $x\mapsto 0$). Then $\mathrm{Hom}_B(M,B)\cong k$ (spanned by $1\mapsto x$), so $\mathrm{Hom}_B(M,B) \otimes_A C\cong k$ as well. But $1\mapsto x$ maps to $0$ in $\mathrm{Hom}_B(M,B \otimes_A C)$, since $x=0$ in $B\otimes_A C=C$. On the ...


1

In the end for a couple of tensors $A_{ijk}$ and $v^s$ (described by components) their juxtaposition $$A_{ijk}v^s,$$ give you the components of a new tensor, whose rank is the sum of their ranks, this juxtaposition is only a plain multiplication of numbers. One can get the components of a rank two if we "contract" some indexes as in $A_{ijs}v^s$.


1

Not necessarily! The tensor product $\mathbb Q\otimes_{\mathbb Z}\mathbb Z/n\mathbb Z$ is zero.


1

There's a homomorphism from $2\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$ to $\mathbb Z\otimes_{\mathbb Z} \mathbb Z/2\mathbb Z$ defined by $2s \otimes q \mapsto s \otimes q$. So now: can you prove that $$ 1 \otimes 1 $$ is nonzero in the latter group?


1

The relation $(s_1+s_2,n)-(s_1,n)-(s_2,n)$ in the case $s_2=-s_1$ implies that $s_1\otimes n+(-s_1)\otimes n=0\otimes n$. Using the same relation with $s_1=s_2=0$ gives $0\otimes n+0\otimes n=0\otimes n$, so $0\otimes n=0$. Thus $-(s_1\otimes n)=(-s_1)\otimes n$. So in your expression for $y$, any term of the form $-(s_i\otimes n_i)$ can be replaced by ...


1

For simplicity, I will discuss categories with colimits of $\kappa$-small diagrams, where $\kappa$ is a regular cardinal. Specifically, consider the following category $\mathbf{K}$: The objects are small categories equipped with chosen $\kappa$-ary coproducts and coequalisers of parallel pairs. The morphisms are functors that strictly preserve the chosen ...


1

No, this is not true. For instance, take $n=k=2$, so $M=F[x]^2$, $\Lambda^k_{F[x]}M=F[x]$, and $\Pi:M^2\to F[x]$ is the map that sends $((a,b),(c,d))$ to $ad-bc$ (for $a,b,c,d\in F[x]$). Write $a_n$ for the degree $n$ coefficient of a polynomial $a\in F[x]$. Let $N=F$ and define $f:M^2\to N$ by $f((a,b),(c,d))=a_0d_1-b_1c_0$. Then $f$ is $F$-bilinear and ...


1

This is isomorphic to $Hom(TM,TM)$. Namely it is the endomorphism bundle of the tangent bundle. In general given two vector spaces(or bundles) we have an isomorphism between $W \otimes V^*$ and $Hom(V,W)$.


1

I think it might be false. Take for instance $R=\mathbb{F}[t]$ with trivial $G$-action, and $N=tM$.


1

I think that the cleanest way is to define this as an equivalence relation. Show that this is indeed an equivalence relation. As notation, your notation or $\equiv$ are good. In my opinion it would be bad to just use equal, there would most likely be some places where confusing the two different meanings of the $=$ sign would lead to wrong results.



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