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5

The general notion of a "bimorphism" and the classifying tensor product has been studied in the paper B. Banaschewski and E. Nelson. Tensor products and bimorphisms. Canad. Math. Bull, 19(4):385-402, 1976. It contains a general theorem which shows that tensor products "almost always" exist. In the case of topological spaces, the construction looks as ...


5

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 ...



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