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11

In the beginning I just specify what Rob Arthan meant. There are following three theorems. For simplicity considered vector spaces are over filed $\mathbb{R}.$ 1.If $V$ and $W$ are finite dimensional vector spaces then $\dim(V\times W)=\dim(V)+\dim(W).$ Be awere that in category of vector spaces we can use $\times$ and $\oplus$ interchangeably. ...


6

Here's another approach. Consider $A \otimes {\bf 1}_m$, we will show that this matrix can always be brought to the block form $$\left( \begin{array}{cccc} A & 0 & \cdots & 0 \\ 0 & A & \cdots & 0 \\ 0 & \cdots & \cdots & A \\ \end{array}\right) $$ To this end consider the matrix $A$ with components $(a_{ij})$ in some ...


5

The symmetrizer $S: \bigotimes^k V \to \bigotimes^k V$ is idempotent. Hence, $\ker(S) = \mathrm{im}(\mathrm{id}-S)$. This is generated by elements of the form $\alpha-{}^\sigma \alpha$, where $\sigma$ is some permutation.


5

Proof of the second part: We assume that both $U$ and $V$ contain a non-zero vector. Suppose that $u$ and $v$ are non-zero. We may extend each to a basis of $U$ and $V$ respectively. That is, we now have the bases $\{u = u_1,u_2,\dots\}$ and $\{v = v_1,v_2,\dots\}$ of $U$ and $V$. We define the bilinear map $$ f_{u,v}:\left(\sum_{i \in I} a_i ...


5

Since you are considering not necessarily commutative ring and thus is forced to taking hom-set and tensor product of abelian group, it's not reasonable to expect that $\hom_R(M,M')\otimes\hom_R(N,N')$ and $\hom(M\otimes_RN,M'\otimes_RN')$ are comparable. For example, taking $M:=R_R,N:=_RR$, then the two become $R\otimes_{\mathbb Z}\hom_R(M',N')$ and ...


4

This is an error in Atiyah-Macdonald. As Georges Elencwajg says in this MathOverflow thread, This is false since that map is not a ring morphism. The correct structural map $A\to D$ is actually $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.


4

Essentially what you want is stated in Kelly's Basic Concepts Of Enriched Category Theory. The tensor product $\mathcal{C} \otimes \mathcal{D}$ of categories enriched in a symmetric monoidal category $\mathcal{V}$ can be defined in the obvious way. As Kelly discusses (starting in the last paragraph of page 12 and going into page 13), a functor $\mathcal{C} ...


4

That's because $\,m \otimes \Bigl(\sum\limits_{i=0}^n a_i t^i\Bigr) $ can be written as $\,\sum\limits_{i=0}^n (a_im)\otimes t^i$, identified with $\,\sum\limits_{i=0}^n (a_im)t^i$.


3

$\mathbb{Q}(i)\otimes_{\mathbb Q}\mathbb{Q}(\sqrt{2})\simeq\mathbb Q(i)\otimes_{\mathbb Q} \mathbb Q[X]/(X^2-2)\simeq\mathbb Q(i)[X]/(X^2-2)$ which is a field since $X^2-2$ is irreducible over $\mathbb Q(i)$. (This field is in fact $\mathbb Q(i,\sqrt 2)$.) However, if you consider $\mathbb{Q}(i)\otimes_{\mathbb Q}\mathbb{Q}(i)\simeq\mathbb ...


3

Recall that the tensor product is bilinear (over $\mathbb{Q}$ in your example). Therefore, $$(a + b\sqrt{2})\otimes(c + di) = ac(1\otimes 1) + ad(1\otimes i) + bc(\sqrt{2}\otimes 1) + bd(\sqrt{2}\otimes i).$$ Note that $\{1\otimes 1, 1\otimes i, \sqrt{2}\otimes 1, \sqrt{2}\otimes i\}$ is a basis for $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i)$ ...


3

Suppose $M$ and $N$ are both left modules and we define $M\otimes_R N$ as the span of pure tensors $m\otimes n$ satisfying the additivity relations and $rm\otimes n = m\otimes rn$. Then for $r,s\in R$, you get $$(rs)(m\otimes n) = (rs)m\otimes n = r(sm)\otimes n = sm\otimes rn = m\otimes (sr)n = (sr) (m\otimes n).$$ So you get $(rs)(m\otimes n) = ...


3

I believe that the first thing you want to proof (in a "clean" way) is the following theorem Let $U,V,W$ be vector spaces and $f:U\times V\rightarrow W$ be bilinear. For every $u\in U$ and $v\in V$ if $u=\theta_U$ or $v=\theta_W,$ than $f(u,v)=0.$ Proof. Assume that $u=\theta_U,$ than $$f(u,v)=f(\theta_U,v)=f(0\cdot \theta_U,v)=0 \cdot ...


2

I won't even attempt to be the most general with this answer, because I admit, I do not have a damn clue about what perverted algebraic sets admit tensor products, for example, so I will stick with vector spaces, but I am quite sure everything I say about vector spaces works for finitely generated modules over commutative rings as well. And also that the ...


2

The most intrinsic proof is probably showing $\det(V\otimes W) \cong \det(V)^{\otimes\text{rk}(W)} \otimes \det(W)^{\otimes\text{rk}(V)}$, using the universal properties involved, and that under this (natural) isomorphism, $\det(f\otimes g)$ becomes $\det(f)^{\otimes\text{rk}(W)} \otimes \det(g)^{\otimes\text{rk}(V)}$. Here, $f\!: V \to V$ and $g\!: W \to W$ ...


2

HINT: It's enough to show the identity holds for matrices with entries from $\mathbb{C}$. Moreover, it's enough to show it for complex diagonalizable matrices, a dense subset. So show that if $A$ is diagonalizable with eigenvalues $\lambda_i$ and $B$ diagonalizable with eigenvalues $\mu_j$, then $A\otimes B$ is diagonalizable with eigenvalues $\lambda_i ...


2

It cannot be that $f'(s\otimes t) = (f(s),f(t))$ because on one hand $f$ isn't necessarily defined on $T$ (as a polynomial function), while on the other hand $(f(s),f(t))$ isn't an element of $S \otimes T$. Now, note that there is a natural map $S \to S \otimes T$, namely $s \mapsto s \otimes 1$. Thus if $f \in S[X]$ is $$ f(X) = a_n X^n + a_{n-1}X^{n-1} + ...


2

If $\|\cdot\|_B$ is a cross norm, then yes, your argument shows that $D_1 \otimes_a D_2$ is dense in $B_1 \otimes_a B_2$ and therefore also dense in $B$. Without this assumption or something similar, the answer is no, as then there is nothing at all relating the $B$ norm to the $B_1, B_2$ norms. This really has nothing to do with tensor products, so take ...


2

We have $$L(V,W;F) \cong L(V\otimes W;F) = (V\otimes W)^* \cong V^*\otimes W^*$$ by the definition of tensor product by universal property. As $V^*\otimes W^*$ is spanned by the pure tensors $\tau\otimes\theta$, we are done (up to proving that the dual of the tensor is the tensor of the duals). Note: This proof works equally well in infinite dimension, ...


2

Note that every abelian group can be seen as a $\Bbb{Z}$-module and that a $\Bbb{Z}$-linear map between abelian groups is just a group homomorphism. Then this answer shows you how to prove that $A \otimes (B \otimes C) \simeq (A \otimes B) \otimes C$ as modules, hence as groups, where the isomorphism is the obvious one: $$ \varphi \colon \;\, a \otimes (b ...


1

In fact, the kernel of $g$ is $I\otimes_AR+R\otimes_AI$, so the correct isomorphism is $$(R\otimes_AR)/(I\otimes_AR+R\otimes_AI)\simeq R/I\otimes_AR/I.$$


1

First of all, let's be careful: $I\otimes_A I\to R\otimes_A R$ may not be injective, so presumably $(R\otimes_A R) / (I\otimes_A I)$ is supposed to be the quotient by the image of this map. Anyway, this is false even when $f$ is the identity: then the left-hand side is isomorphic to $R/I$, while the right-hand side is isomorphic to $R/I^2$.


1

$0$-tensors are just scalars, so the tensor product in this case is just scalar multiplication.


1

An alternate proof: we can prove that this map is surjective as follows. Take an arbitrary $A \in {\rm End}_K(V)$. With respect to some basis $e_1,\dots,e_n$ of $V$, we can write $$ A = \pmatrix{a_{11} & \cdots & a_{1n}\\ \vdots & \ddots \\ a_{n1} & \cdots & a_{nn}} $$ We then note that $$ A = \Phi \left(\sum_{i=1}^n \sum_{j=1}^n ...


1

I think I got it, please check it for mistakes. For starters, we have that: $$\dim(V^\ast \otimes V) = \dim({\rm End}_K(V)) = n^2.$$ Let $\Phi: V^\ast \otimes V \to {\rm End}_K(V)$ be given by $\Phi(h \otimes a)(x) = h(x)a$. If $h\otimes a \in \ker \Phi$, then $\Phi(h \otimes a) = 0$. Meaning that $h(x)a = 0$ for all $x \in V$. If $a = 0$, done. Otherwise ...


1

I think you want $I$ to be a right ideal and $J$ a left ideal, not the other way round. You are almost done. Consider the canonical homomorphism $\mathbf{m} :A\otimes_{A}A\rightarrow A$ of $\mathbb{Z}$-modules which sends every $a\otimes b\in A\otimes_{A}A$ to $ab\in A$. It is well-known that $\mathbf{m}$ is an isomorphism. Thus, $\left( ...


1

$\require{cancel}$Let me deal with the case of the tensor product of two complexes (I hope after that the case of three complexes should be easy). So let the first complex be $C = (0 \to R_1 \xrightarrow{x_1} R_0 \to 0)$, and the second one $C' = (0 \to R'_1 \xrightarrow{x_2} R'_0 \to 0)$; of course $R_0 = R_1 = R'_0 = R'_1 = R$, but this notation change ...


1

I assume that by transoposition you mean dual mapping $T^*$. That is if $T:V\rightarrow V$ then $T^*:V^*\rightarrow V^*$ in following manner $$\left[T^*\alpha\right](v):=\alpha(T(v)).$$ Let $\alpha_1,\dots,\alpha_n\in V^*.$ Fix arbitrary $v_1,\dots,v_n\in V.$ Then $$\left[\bigwedge^n T^*(\alpha_1\wedge\dots\wedge\alpha_n)\right](v_1\wedge\dots\wedge ...



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