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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 ...


6

At some level, the tensor product of maps is, like most things in math, a convenient choice of definition. However, it arises naturally, in a precise sense. The standard definition of a tensor product of two spaces, $V \otimes W$, actually provides more than a vector space constructed from $V$ and $W$. It is a universal construction, meaning that it ...


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 $$ M\otimes_R P\xrightarrow{f\otimes_R\DeclareMathOperator{id}{id}\id_P}N\otimes_RP\to Q\otimes_R P\to 0\tag{2} $$ ...


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 ...


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 $M\otimes_RP\xrightarrow{f\otimes1_P} N\otimes_RP\xrightarrow{g\otimes1_P}(\operatorname{coker} f)\otimes_RP\to 0$


2

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


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

If you want polynomial expressions whose variables are vectors of $V$, then the full symmetric algebra ${\rm Sym}(V)$ is the way to go: this allows you to form and then multiply arbitrary monomials of vectors, linearly combine them, and distribute addition of vectors and more complicated expressions over multiplication, all exactly as desired. If you want ...


2

try to show in general that: let $A$ a commutative ring and $I$ , $J$ two ideals, then $A/I\otimes A/J\simeq A/(I+J)$. And remark that $12\mathbb{Z}+10\mathbb{Z}=2\mathbb{Z}$.


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 ...


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

Let's recall the constructive definition of the tensor product: Suppose $M$ and $N$ are $A$-modules. Let $F$ be the free $\mathbb{Z}$-module on the set of symbols $\{e_{m,n} \ | \ (m,n)\in M\times N\}.$ Let $I$ be generated by $e_{m+m',n} - e_{m,n} - e_{m',n}$ etc (I'm sure you know the other generators). Then $F/I$ is a tensor product. You want to define ...


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 ...


1

More generally, if $V$ is any vector space over $K$, then $\dim_K(V) = \dim_L(V \otimes_K L)$. In fact, $V \cong \bigoplus_B K$, where $B$ is some basis of $V$, so that $V \otimes_K L \cong \bigoplus_B L$ (using that tensor products commute with direct sums and that $K \otimes_K L = L$).


1

Since $y = W\cdot h$, you can consider $h$ to be the independent variable and hold $W$ constant. This gives you $$dy = W\cdot dh$$ and as you observed, we can immediately identify the derivative as $$\frac{\partial{y}}{\partial{h}} = W$$ Now let's hold $h$ constant and vary $W$: $$\eqalign{ dy &= dW\cdot h \cr &= I\cdot dW\cdot h \cr &= ...


1

There is a canonical linear map $V^* \otimes W^* \to (V \otimes W)^*$. It is an isomorphism when $W=K$ (both sides identify with $V^*$ and the linear map becomes the identity). The class of $W$s for which it is an isomorphism is closed under finite direct sums - this is because both sides are additive (contravariant) functors in $W$. It follows that it is an ...


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 $0 \otimes n = 0 = m \otimes 0$.



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