About standard complex (or bar construction): $\otimes$ means $\otimes_A$? or $\otimes_k$? I'm sorry I'm not good at homological algebra. I have an elementary question about standard complex (or bar construction).
Let $k$ be a field, $A$ be an associative unital $k$-algebra and $M$ be a left $A$-module.
In many literatures, standard complex is defined as
$$
A\otimes A^{\otimes n} \otimes M
$$
or
$$
A\otimes (A/k)^{\otimes n} \otimes M
$$
and they are free resolutions of $M$.
Now, I cannot judge whether $\otimes$ means $\otimes_A$ or $\otimes_k$. And I cannot prove that $A\otimes A^{\otimes n} \otimes M$ or $A\otimes (A/k)^{\otimes n} \otimes M$ is a free $A$-module.
Please tell me the detail.
 A: The tensor is over $k$, since tensoring over $A$ would give a rather uninteresting complex! The acyclic bar complex $B(A,A)$ has in degree $n$ a free $A$-module with basis $A^{\otimes n}$ (or sometimes the reduced version, $\bar{A}^{\otimes n}$), that is, $A\otimes \bar{A}^{\otimes n}$. Usually, in the setting of Hochschild (co)homology and other homological matters orbiting $k$-algebras (where $k$ is fixed), unadorned functors such as $\hom$ and $\otimes$ are meant to be take over $k$. In fact, it is quite traditional that authors clarify this at the beginning of chapters or sections. 
One can in fact take a two sided bar construction, $B(A,A,A)$, which has a free $A$-bimodule with basis $\bar{A}^{\otimes n}$ in degree $n$. This is $k$-contractible, so for a left $A$-module $N$, the complex defined by $B(A,A,N) := B(A,A,A)\otimes_A N$ is a left $A$-free resolution of $N$. In this way, if $M$ is a right $A$-module, the complex $B(M,A,N) := M\otimes_A B(A,A,A)\otimes_A N$ computes $\text{Tor}^A_n(M,N)$. In particular, for commutative rings, one deduces immediately that $\text{Tor}$ is symmetric, which takes a bit more work to pull off with arbitrary resolutions.
