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First note that it's not necessary to consider the tensor products $A \otimes K$ for arbitrary fields $K$; $K$ has some prime subfield $k$, and $A \otimes K \cong (A \otimes k) \otimes_k K$, so all th …

Both of these statements are true (at least if $H$ is semisimple). It suffices to prove the first one. By hypothesis there is a nonzero map $V_1 \otimes V_2 \otimes V_3 \to Q$. It dualizes to a nonzer …

Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \part …

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 $\ma …

For a), if $L$ is a finite Galois extension of $k$ with Galois group $G$, then $$L \otimes_k L \cong \prod_{g \in G} L$$ has $|G|$ prime ideals. For b), if $k = \mathbb{F}_p(a)$ and $L = k[x]/(x^p - …

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

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 …

It seems to me that your question is actually about something that is clearer in a more basic context than chain complexes. Fix a field $k$ and let's work instead in the category $C$ of pairs $(V, T)$ …

I'd like to borrow a little from several of the answers given so far to give a "practical" perspective. Let $V$ be an $\mathbb{R}$-vector space. A tensor of type $(m, n)$ is an element of $V^{\otime …

I don't understand the definition you're using, but I'll tell you that the Wikipedia definition is correct. The point of the Kronecker product is that it is a basis-dependent form of the tensor produ …