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9

Let $M, N$ be $R-$modules. Then the following holds. If $M$ and $N$ is flat, then so is $M\otimes_{R}N$: see related question here. If $M$ and $N$ are projective, then so is $M\otimes_{R} N$. Indeed, writing $M\oplus M'=F,\ N\oplus N'=F'$, for free $R-$modules $F,\ F'$, one has that $$ F'':=F\otimes_{R}F' $$ is free (tensor product of free modules) and ...


4

"Uniqueness will fail for noncommutative monoids, so here we must use commutativity of our monoids, " This is not correct. You get a morphism in $V$ also for non-commutative monoids, but it won't be a monoid morphism. This is where commutativity is used. Uniqueness holds in general: If $h : M \otimes N \to X$ is a monoid morphism with $h \circ (M \otimes ...


4

This map is actually not even well defined. This is because $f(a \otimes 2) = a$, but $f(2a \otimes 1) = 2a$, even though $a \otimes 2 = 2a \otimes 1$ in $A \otimes_K K$. The correct map is $a \otimes k \mapsto ka$.


4

A basic fact in homological algebra is that the functor $$-\otimes_R P:_R\mathsf{Mod}\to_R\mathsf{Mod}$$ is always right exact. So, applying $-\otimes_RP$ to an exact sequence $$ M\xrightarrow{f}N\to Q\to 0\tag{1} $$ gives an exact sequence $$ M\otimes_R P\xrightarrow{f\otimes_R\DeclareMathOperator{id}{id}\id_P}N\otimes_RP\to Q\otimes_R P\to 0\tag{2} $$ ...


3

There is a short exact sequence $M\xrightarrow{f} N\xrightarrow{g}\operatorname{coker} f\to 0$. Applying the functor $(\mathord-)\otimes_RP$, which is right exact, we get a new exact sequence $M\otimes_RP\xrightarrow{f\otimes1_P} N\otimes_RP\xrightarrow{g\otimes1_P}(\operatorname{coker} f)\otimes_RP\to 0$


3

I think putting tensors in the right context will clear much of this up. I'll stick with $\mathbb R^3$ since that's the example you use. Rank $2$ tensors are elements of the so-called tensor product of $\mathbb R^3$ with itself, which is denoted by $\mathbb R^3 \otimes \mathbb R^3$. This space consists of all linear combinations of expressions of the form $u ...


3

One of your errors is that you think every element of a tensor product $A\otimes_K B$ is an elementary tensor $a\otimes b$. I say this because from the way you are defining $f$ and trying to compute the kernel it seems you think every tensor is elementary. Elements of a tensor product are sums of elementary tensors $a_1\otimes b_1 + \cdots + a_r \otimes ...


3

In general it is very rare for a braided monoidal category to commute "on the nose". For a very concrete example, the usual category of sets and functions is cartesian monoidal, so braided (even symmetric) monoidal, but in general $A \times B \neq B \times A$. As for the braids, there is a PRO (monoidal category with $\mathbb{N}$ as the set of objects and ...


2

Projective modules are always flat. So if $P$ is a projective $R$-module, $M$ is a flat $R$-module and $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact sequence of $R$-modules, then, as $P$ is flat, $0 \rightarrow P\otimes A \rightarrow P\otimes B \rightarrow P\otimes C \rightarrow 0$ is exact, so by flatness of $M$, $0 \rightarrow M ...


2

Your isomorphism should be \begin{align} f: A \otimes_K K& \to A\\ a \otimes_K k& \mapsto k \cdot a \end{align} This is a well-defined isomorphism and is $K$-linear, which your function wasn't. Because the scalar multiplication is only defined for scalars, this isomorphism can only be defined when one of the domain vector spaces is $K$ itself.


2

The external direct sum of $N_1$ and $N_2$ is canonically isomorphic to their internal direct sum by $$N_1\oplus N_2\ \longrightarrow\ N_1+N_2:\ (n_1,n_2)\ \longmapsto\ n_1+n_2,$$ whenever the internal direct sum exists, of course. So nothing changes except for some notation; for the isomorphism you could in stead write $$m\otimes(n_1+n_2)\ \longmapsto\ ...


2

The tensor product $R_p\otimes_R L$ is "just" the localization of the module $L$ with respect to $R\setminus p$. So suppose $ \frac{r}{s}\cdot\frac{m}{t}=0 $ for some $r\in R$, $s,t\in R\setminus p$, $m\in L\setminus 0$. Then by the definition of localization we have $ urm=0 $ for some $u\in R\setminus p$. Hence $r=0$ by assumption about $L$. As one ...


2

$(1\otimes C)^{-1}(A\otimes 1 - 1\otimes A)(1\otimes C) = A\otimes 1 - 1\otimes B$


1

To prove boundedness of $S\otimes T$, it suffices to prove it when $S$ and $T$ are unitaries (since the unitaries span $\mathcal{B}(H)$. Now if $$ z = \sum_{i=1}^n x_i\otimes y_i $$ where $\{y_1, y_2,\ldots, y_n\}$ are orthogonal. So $$ \|(S\otimes T)(z)\|^2 = \|\sum_{i=1}^n S(x_i)\otimes T(y_i)\|^2 $$ Since $\{y_i\}$ are orthogonal $$ = \sum_{i=1}^n ...


1

Only when $M=N=0$. Look at $0 \otimes n = 0 = m \otimes 0$.


1

a) Let $$ v=\alpha^i x_i=0,\quad w=\beta^j y_j=0, $$ and build the product $$ v \otimes w = \alpha^i \beta^j x_i \otimes y_j = 0. $$ Being $(x_i \otimes y_j)_{(i,j)\in I\times J}$ linearly independent, then $$ \alpha^i \beta^j = 0\quad\forall i,j \implies \alpha^i=0,\,\forall i,\quad\beta^j=0\,\forall j $$ so $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ are ...


1

Hint: Tensor products commute with direct sums in each variable.


1

Your interpretation of $(x^\ast \otimes y)(x) = x^\ast(x)y$ is the correct one. Ryan explains several different interpretations of tensors in section 1.3 of his book, among which the one intended here. With this interpretation, the algebraic tensor product $X^\ast \otimes Y$ is simply the space of finite rank operators from $X$ to $Y$. In particular, ...


1

$P$ isn't even well-defined. For the corresponding map $\tilde P \colon X \times Y \to X$, $\tilde P(x,y) = x$ is not bilinear (if $X \ne 0$). If $\tilde P$ is bilinear and hence $P$ well-defined, we must have $$ x = P\bigl(x \otimes (y_1+y_2)\bigr) = P(x \otimes y_1 + x \otimes y_2) = P(x\otimes y_1) +P(x \otimes y_2) = 2x $$ for all $x$, hence $X = 0$. ...


1

Saying that $M$ is finitely generated is the same as saying there is an epimorphism $R^n\to M$. Tensoring preserves epimorphisms, so you get an epimorphism $$ R^n\otimes_R S\to M\otimes_R S $$ Can you go on?


1

We can assume that any torsion element is in the valuation ideal, $PR_P$ which we will just call $P$ by abuse of notation. If not, then we can find an inverse for the torsion number, $n$, in $R_P - P=R_P^\times$, and this would imply $1\cdot v=0$ for some $v\in R_P\otimes_R L$, which implies $v$ was $0$ to begin with. But then it must be that we need only ...



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