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6

Let $A = k\left[x\right]/\left(x^2\right)$, and let us denote the projection of $x \in k\left[x\right]$ onto $A$ by $x$ (by abuse of notation). I assume that your $\left(x\right)$ means the $A$-submodule of $A$ spanned by $x$. And your question is: Is $\left(x\right) \otimes_A \left(x\right) = 0$ ? I claim that the answer is "No". Indeed, there is an ...


5

It is still true because: Any $A$-module is the direct limit of its finitely generated submodules (for any ring $A$). Tensor products commute with direct limits. Direct limit is an exact functor. Btw, a submodule of an $A$-module with this property (the morphism $M'\hookrightarrow M$ is universally exact) is called a pure submodule of $M$, and the ...


4

The image of $* : K[x] \times K[y] \to K[x,y]$ contains $x$ and $y$, but not $x+y$.


3

Choose bases $(E_i)$ and $(F_j)$ for $A$ and $B$ and extend them to bases for $V$ and $W$, respectively. We get a basis $(E_i\otimes F_j)$ for $A\otimes B$, and in the same way the extensions give a basis for $V\otimes W$. Then, we get that $\bar{f}$ maps a basis of $A\otimes B$ to a linearly independent subset of $V\otimes W$ (a subset of a basis). This ...


3

Yes, this is a covariant functor $Set\times\mathcal{C}\to \mathcal{C}$. Given $f:S\to T$ and $g:C\to D$, the induced map $(f\cdot g):S\cdot C\to T\cdot D$ is the unique map such that if $s\in S$ and $i_s:C\to S\cdot C$ is the corresponding inclusion, then $(f\cdot g)i_s=i_{f(s)}g$, where $i_{f(s)}:D\to T\cdot D$ is the inclusion corresponding to $f(s)\in ...


2

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( ...


2

I think this will do: take $U = \Bbb R^2, V = \Bbb R^2$. We have $$ e_1 \otimes e_1 + (e_1 + e_2) \otimes e_2 = e_1 \otimes (e_1 + e_2) + e_2 \otimes e_2 $$ Or, a more interesting example $$ (e_1 + e_2) \otimes e_1 + (e_1 - e_2) \otimes e_2 = e_1 \otimes(e_1 + e_2) + e_2\otimes (e_1 - e_2) $$


2

The following observations are made en passant in the accepted answer. I think they deserve a full proof. Let $R=K[X]/(X^2)$, and $x$ be the residue class of $X$ modulo the ideal $(X^2)$. Then $$\operatorname{Ann}_R(x)=Rx.$$ Since $x^2=0$ we have $Rx\subseteq\operatorname{Ann}_R(x)$. For the converse, let $f(x)\in\operatorname{Ann}_R(x)$. Then ...


2

There is no reason for the two module structures to be the same and, in general, they are not. For an extreme example, consider a ring $R$ with a non-identity endomorphism $\sigma:R\to R$ and let $S=R$, $M=R$ with the obvious right $R$-module structure and the left $R$-module structure given by $r\cdot m=\sigma(r)m$. $N=R$ with the obvious left and right ...


2

$f \otimes g$ is the function $[0,1]^2 \to \mathbb C$ given by $(f \otimes g)(x,y) = f(x) g(y) $.


2

Yes, this is true. In fact, you don't even need $M$ to be torsionfree or $K$ to be the field of fractions of $R$; all you need is for $K$ to be some field equipped with a ring-homomorphism $R\to K$. Given such a homomorphism, $K$ can be considered as an $R$-module, so we can form the tensor product $M\otimes_R K$. Furthermore, this tensor product is a ...


1

Presumably, you mean for $H$ to be a vector space over $\Bbb C$. Note that $\operatorname{Herm}(H^k)$ is a vector space over $\Bbb R$ (as opposed to $\Bbb C$), and that $\dim \operatorname{Herm}(H^k) = k^2$. From there, we can simply apply your earlier reasoning to find that $$ \mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H^k)^{\otimes N})) = \binom{N + k^2 - ...


1

As by the question linked by Michael, it is generally proved in a course about differential forms that $\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$ where $p$ and $q$ are the degrees of $\alpha$, $\beta$. Using this the result is immediate. But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for ...


1

Given a $p$-form $\theta \in \bigwedge^p E^*$, we can define an alternating multilinear map $h \colon E^{n-p} \to \bigwedge^n E$ by $$h(u_1, \ldots, u_{n-p}) = \theta \wedge \tilde{u}_1 \wedge \ldots \wedge \tilde{u}_{n-p}.$$ Let $b \colon \mathbb{R} \to \bigwedge^n E$ be the linear map $$b(t) = t \omega.$$ Because $\bigwedge^n E$ is one-dimensional and ...


1

Since $I$ is a principal ideal, it is isomorphic to $R$ as an $R$-module. Hence $$I \otimes_R I \cong R \otimes_R R \cong R$$ which is obviously torsionfree.


1

1. "Yes" to the second question. More precisely, he is regarding $B$ as a $\left(B,A\right)$-bimodule, where the left $B$-module structure is obvious (i.e., given by multiplication) and where the right $A$-module structure is the one you define. "No" to the first question. He is tensoring a $\left(B,A\right)$-bimodule with a left $A$-module. This yields a ...



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