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Here is a self-contained argument. First, as Jeremy Rickard observes, $K \otimes K \cong K \otimes_k K$, where $k$ is the prime subfield of $K$ (so $\mathbb{Q}$ if $K$ has characteristic zero and $\mathbb{F}_p$ if $K$ has characteristic $p$). If $K \otimes_k K$ is a field, then as Denis Nardin observes, the multiplication map
$$K \otimes_k K \xrightarrow{m} ...

answered Feb 15 '16 at 4:26

Qiaochu Yuan

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I already wrote this in the comments but I think this might be worth of an answer. I think we can classify all fields $K$ such that $K\otimes K$ is a field.
Claim If $K$ is a field such that $K\otimes_\mathbb{Z}K$ is a field then the multiplication map $K\otimes_\mathbb{Z} K\to K$ is an isomorphism
In fact the multiplication map is always a surjection ...

15

Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product $M \otimes_k N$ by the usual universal property, and it is naturally a left $R \otimes_k S$-module by functoriality. If $M, N$ are both left $R$-modules, then $...

answered Sep 22 '12 at 20:01

Qiaochu Yuan

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No. Let $R = k[x,y]/(y^2-x^3)$. Let $S = k[t]$, with the map $R \to S$ given by $(x,y) \mapsto (t^2, t^3)$. So $S \otimes_R S = k[t,u]/(t^2-u^2, t^3-u^3)$. The element $t-u$ is not zero in the tensor product, because all the relations are in degree $>1$. But $(t-u)^3 = 3 (t-u)(t^2-u^2) - 2 (t^3-u^3)$ is in the ideal.

answered Feb 13 '12 at 18:35

David E Speyer

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If you mean by LF-space a strict inductive limit of Frechet spaces (as it was done by Dieudonne and Schwartz) I think the answer is yes.
Here is what I believe could be made a proof: Since the inductive tensor product
respects inductive limits you have $C^\infty(M) \otimes_i F = \lim C^\infty(M) \otimes_i F_n$ if $F= \lim F_n$ is a strict LF-space. Since $...

13

I think both can be proved without choice, essentially because, in both cases, whenever you're tempted to choose a basis, you can manage with a little care to get by with a basis of a finite dimensional subspace.
For (2), if there's a linear dependence between the $v_i\otimes w_j$ then it involves only finitely many $v_i$ and $w_j$. Also, the linear ...

answered Mar 10 '19 at 10:11

Jeremy Rickard

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Yes. For a self-adjoint element $y$, denote $s(y)=\sup{\rm Sp}(y)$. Then for $\gamma \geq s(y)$, one has
$$\inf \lbrace s(y - \gamma(e\otimes f)) : 0\le e\le 1,\ 0\le f\le 1\rbrace \le 0.$$
Indeed, if $e_n$ and $f_n$ are approximate units, then so is $g_n:=e_n\otimes f_n$ and
$y - \gamma g_n \le y-g_n^{1/2} y g_n^{1/2} \to 0$.
Now let $y_0 := x \le 1$ and ...

11

Not always! However, the closure is a polyhedron.
Not always: Take $P = \{a | 0 \leq a \leq 1\}$. Take $Q= \{ b,c | b\geq 0, c=1\}$. Then under the map $x=ab$, $y=ac$. Since $P$ is the convex hull of $(0)$ and $(1)$, and $Q$ is the ray starting at $(0,1)$ and going in direction $(1,0)$, $P \otimes Q$ is the convex hull of $(0,0)$ and the ray starting at $(...

11

Let $R$ be a Dedekind domain with field of fractions $K$. A module $M$ is symtrivial if and only if its maximal torsion-free quotient $F$ is a submodule of $K$ and its torsion submodule $T$ is a sum over each prime $p$ such that $F$ is $p$-divisible of a $p$-divisible $p$-torsion module and a $p$-power cyclic module $R/p^n$.
If $F$ and $T$ are modules as ...

11

$$ \mathrm{End}_{kG}(U \otimes_k V) = \left(\mathrm{End}_k(U \otimes_k V)^N\right)^{G/N} = \mathrm{End}_k(V)^{G/N} = k. $$
The second equality holds because $\mathrm{End}_{kN}(U) = k$ (thanks to the assumption that $k$ is algebraically closed).
EDIT: Using Chevalley's theorem is probably overkill. Let $k$ be an arbitrary commutative field, assume that $U$ ...

11

Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable subset of $\omega^* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-principal ultrafilters.
To see this, suppose $\mathcal U$ and $\mathcal V$ are two non-principal ultrafilters on $\omega$, and let $\mathcal Z = \mathcal U \times \...

11

Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated result of Kirchberg (see Corollary 13.2.5 in Brown and Ozawa's book) it follows that if $B$ has WEP and $C$ has LLP, then $B\otimes_{\max{}} C = B\otimes_{\min{}} C$...

10

Even at the risk of using slightly too heavy armory, I would like to shortly explain, how I would view the situation in a light, that naturally produces and "explains" the observations collected above:
@Martin: Tthe natural extension on tensor products
@TomGoodwillie: The "skew-Leibniz"-rule one has to use
@JackHuizenga: the even-odd grading associated to ...

10

Why can't you just take the product measure induced by the uniform measure on $\{0,1\}$ (or indeed by any other nontrivial measure on this two-element set)? I suppose my question really is: What exactly do you mean by non-trivial?

answered Nov 21 '13 at 19:57

Andreas Blass

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I think the answer is: It's always continuous if $A$ is subhomogeneous and never continuous otherwise(min or max).
First notice that if the product map is continuous for min, then it's continuous for max because we can factor the product map as $A\otimes_{max}A\rightarrow A\otimes_{min}A\rightarrow A$ where the first map is the canonical quotient and the ...

10

This answer provides a scheme how to construct a constructive proof, though I'm still working to actually explicitly extract the constructive proof, so please don't accept the answer just yet. (Update: See below.) We'll prove the following statement:
Let $R$ be a reduced ring. Let $A$ be a finitely generated $R$-module and let $B$ be an arbitrary $R$-...

9

Let $R^t$ be the torsion submodule and consider the exact sequence
$$0\rightarrow R^t\rightarrow R \rightarrow R/R^t\rightarrow 0$$
Bousfield and Kan show that the ring on the right is a localization of ${\mathbb Z}$, hence flat over ${\mathbb Z}$, so its $Tor$ with $R$ vanishes. Thus if we $Tor$ the above with $R$, we get $Tor(R^t,R)=Tor(R,R)$.
Now ...

answered Apr 25 '12 at 15:39

Steven Landsburg

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9

This is the most complete treatment I could come up with. Let $k \subseteq K^{\operatorname{sep}} \subseteq K^{\operatorname{alg}} \subseteq K$ and $k \subseteq L^{\operatorname{sep}} \subseteq L^{\operatorname{alg}} \subseteq L$ be the decompositions of $k \subseteq K$ and $k \subseteq L$ into a separable algebraic, a purely inseparable algebraic, and a ...

9

I asked David Sherman this question and this was his response:
In the Kuratowski setup, if you omit complementation, you get 7 elements. Adding complementation gives you a disjoint upside-down copy.
In the Grothendieck setup, if you omit duality, you get 8 elements (two five-element chains with common inf and sup, like an O). Adding duality gives you an ...

9

Look at $M=S^7$. This has 28 smooth structures, and their tangent bundles are all trivial - compare Parallelizability of the Milnor's exotic spheres in dimension 7 . Thus their tensor algebras do not carry more information about the smooth structure than $C^\infty(M)$.
On the other hand, for a compact smooth manifold, the algebra of smooth functions is ...

9

Many introductory books on quantum information theory go over the linear algebraic tools necessary to study the topic, including the tensor product (since it indeed models quantum entanglement). Taking the tensor product of two or more quantum states (pieces of quantum information) is analogous to forming a bitstring out of two or more bits (pieces of ...

8

In the analytic category this is indeed true: if $\mathscr{F}$, $\mathscr{G}$ are coherent analytic sheaves on a complex space $X$, then $\mathscr{F} \otimes_{\mathscr{O}_X} \mathscr{G}$ is also coherent.
For a reference, look at [Grauert-Remmert, Coherent Analytic Sheaves], Proposition at the bottom of page 240.
It seems to me that their proof works also ...

answered Jan 19 '12 at 15:59

Francesco Polizzi

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8

Omitting the stars, what you have written down is a two-sided bar construction $B(M,A,N)$,
unreduced since you have not assumed that $A$ is augmented. Note that in degree $n$ it
is the direct sum over $p+q=n$ of the elements of the $p$th term in degree $q$. One
conceptual way of thinking about it is that that we can write
$B(M,A,N) = M\otimes_A B(A,A,N)$...

8

$2^{\mathbb{R}}$, being a product of compact Hausdorff groups, is a compact Hausdorff group, so it has a normalized Haar measure ("flipping uncountably many coins").

answered Nov 22 '13 at 0:34

Qiaochu Yuan

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8

I think most category theorists would answer "yes, obviously", and not bother to write down a proof. But presumably that isn't sufficiently convincing, since you ask the question, so let me try to make it a bit more explicit with some big words. (-:
An $M$-enriched category with set of objects $A$ is equivalently a lax functor $A_{ch}\to B M$, where $A_{...

answered Oct 14 '15 at 4:53

Mike Shulman

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8

Looks like nfdc23 has explanations for (a),(b), and (c).
But: indeed, primes of $F_{k'}$ are not in general minimal if $F/k$ is not algebraic. Let $k'=k(x)$ and $F=k(y)$ be transcendental extensions of $k$.
Then $F_{k'}$ identifies with a subalgebra of the field $k(x,y)$, hence is an integral domain [assertion (c)] - i.e. $\{0\}$ is prime.
There is a ...

8

The first thing that comes to mind is the symmetric algebra functor
$$A \mapsto Sym_{\mathbb Z}(A) = \oplus_{n \in \mathbb N} Sym^n_{\mathbb Z}(A) = \mathbb Z \oplus A \oplus (A\otimes A/ \Sigma_2) \oplus \dots $$
This may be regarded as a functor from abelian groups to commutative rings (it is left adjoint to the forgetful functor); or by composing with ...

7

You have a detailed proof (in a much more general context but easy to read) in Proposition I.4.3.2.1 of Illusie, Complexe Cotangent et Déformations I, Springer LNM 239.

7

In case anyone is still interested in this question:
There is a classification of polynomial exponential functors on the category $\mathcal{V}$ of finite-dimensional inner product spaces in terms of involutive $R$-matrices (i.e. involutive solutions to the Yang-Baxter equation). Basics about polynomial functors can be found in the book "Symmetric Functions ...

7

The proof is okay. You did not specify the norm on the tensor product $H⊗H$
in which you take the closure. I do not know which one you want. Most naturally this would be the Hilbert-space or $\ell^2$ tensor product (which also describes the space of Hilbert-Schmidt operators). But the proof works for all reasonable norms. Another one describes the space of ...

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