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6
votes
Accepted
Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to ...
Let $k$ be a field, $I$ and $J$ infinite sets, and $A$ the $k$-subalgebra of $$k(t)[x_i,y_j: i\in I,j\in J]$$
generated by
$$\{x_i,y_j,tx_i,t^{-1}y_j: i\in I, j\in J\}.$$
Then
$$(tx_i)_{i\in I}\ot …
5
votes
Free Module with a Projective Sub- Module, and Tensor Products
No. Take $N=M=A$, where $A$ is any non-trivial algebra over a field $k$, and $B=k$.
6
votes
Accepted
linear independent families in a tensor product
This answer to a related question gives a way of constructing counterexamples.
For a similar but more concrete example, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$, so $R$ is a $3$-dimensional alge …
13
votes
Accepted
Axiom of choice and algebraic tensor product
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 dimension …