This is a quite general question, I know. Given a commutative ring $K$ and an algebra $K\to A$, there's an endofunctor $A\otimes_K-\colon K\text{-}\mathrm{Mod}\to K\text{-}\mathrm{Mod}$. I was wondering if the condition on this functor to be faithful was known to be related in any sort of way with some other more "easy to check" properties of the algebra $A$.

Thanks in advance for any suggestion ðŸ™‚

## Best Answer

EditI just realized this is not a complete answer, since OP assumed $K$ only to be a commutative ring. My pitfall was the German convention that $K$ denotes KÃ¶rper (fields), while general rings are denoted $R$. My answer hence only remains valid in the case that $A$ is a flat algebra. I donâ€™t think that $A\otimes -$ being faithful implies $A$ being flat, but I couldnâ€™t find a counterexample. Sorry for this.Algebras over fields are flat (they are free modules). Thus the assertion that $A\otimes -$ is faithful is equivalent to $A$ being faithfully flat.

Indeed, since $A\otimes-$ is additive, the condition that for any two vector spaces $U,V$ the map $\mathsf{Vect}_K (U,V) \rightarrow \mathsf{Vect}_K(A\otimes U, A\otimes V)$ is injective is equivalent to the condition that $A\otimes f=0$ iff $f=0$. Since $A$ is flat, this condition is equivalent to the assertion that $A$ is faithfully flat.

This last equivalence may need further justification. If $A$ is not faithfully flat, then there is some vector space $W \neq 0$ with $A\otimes W =0$. Then $\operatorname{id}_W\neq 0$ but $A\otimes \operatorname{id}_W =0$. Conversely, if $A$ is faithfully flat and $A\otimes f=0$, then $$0=\operatorname{img}(A\otimes f) \cong A \otimes \operatorname{img}(f),$$ where the isomorphism is due to $A$ being flat, hence preserving images. $A$ faithfully flat then implies $\operatorname{img}(f)=0$, thus $f=0$.