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For a field $k$, a $k$-algebra $A$ and an $A^e$-module $N$, where $A^e=A \otimes_k A^{op}$, I have proved that the $A^e$-module $Hom_k(A,N)$, with action $(a\otimes b) f (x) := a f(x) b$ is an injective $A^e$-module.

Now I am trying to see if it is flat or not, let $f:X \to Y$ be a monomorphism of $A^e$-modules, we have to proof that the morphism: $$ f \otimes 1_{Hom_k(A,N)} : X \otimes_{A^e} Hom_k(A,N) \to Y \otimes_{A^e} Hom_k(A,N) $$

is a monomorphism.

I also know that every flat module is a direct limit of finitely generated free modules. But I do not see how $Hom_k(A,N)$ is a direct limit of finitely generated free $A^e$-modules, I think that $Hom_k(A,N)$ might be "too small" to be flat, depending on $N$, but I want $N$ to be arbitrary.

Any help proving that $Hom_k(A,N)$ is a flat $A^e$-module?

Marco Armenta
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1 Answers1

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This is not the case in general. Consider the $k$-algebra $A = k[t,t^{-1}]$ of Laurent polynomials. Then $A^e = k[t,t^{-1},s,s^{-1}]$ is the algebra of Laurent polynomials in two commuting variables. Let $N=k$ with the trivial action of $t$, $s$ and their inverses. In terms of group rings $ A = k[\mathbb{Z}]$ and $A^e = k[\mathbb{Z} \times \mathbb{Z}]$. On the other hand, since your action on $\text{Hom}_k(A,N)$ is only on $N$, we have

$$ \text{Hom}_k(A,N) \cong \prod_{\mathbb{Z}} k $$

as a $k$-module and the corresponding action on the right is again the trivial action of $t$, $s$ and their inverses. Of course $k$ acts by multiplication on each component. This $A^e$-module is not flat because

$$ \text{Tor}_1^{A^e}\left(k,\prod_{\mathbb{Z}} k\right) = \text{Tor}_1^{k[\mathbb{Z} \times \mathbb{Z}]}\left(k,\prod_{\mathbb{Z}} k\right) = H_1 \left(\mathbb{Z} \times \mathbb{Z};\prod_{\mathbb{Z}} k\right) $$

where the first $k$ in the Tor terms again has the trivial action of $t$, $s$ and their inverses, the last term is group homology, and therefore

$$ H_1 \left(\mathbb{Z} \times \mathbb{Z};\prod_{\mathbb{Z}} k\right) = H_1 \left( S^1 \times S^1;\prod_{\mathbb{Z}} k \right) \cong \prod_{\mathbb{Z}} k \times \prod_{\mathbb{Z}} k \neq 0 $$

If you are not familiar with group homology, this may seem a bit magical, but if you want a concrete computation to see that tensoring with this module does not preserve monomorphisms consider the short exact sequence of $A^e$-modules

$$ 0 \to \text{Ker} \to A^e \to k \to 0 $$

where the map $A^e \to k$ is the augmentation map, that is, it sends all the powers of $t$ and $s$ to $1$. Tensoring this with $\prod_{\mathbb{Z}} k$ over $A^e$ will not give you an exact sequence.

Goa'uld
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