TITLE: Definition of Hochschild homology in terms of Tor functor (bar resolutions)
QUESTION [7 upvotes]: I had 2 kind of dumb questions about the definition of Hochschild homology in terms of the Tor functor:
1 - Let $R$ be a $k$-algebra and $M$ an $R$-bimodule, let $H_*(R,M)$ be the Hochschild homology of $R$ with coefficients in $M$, this homology can also be defined in the following way: the "bar resolution" of $R$ (which is itself an $R$-bimodule) as an $R$-bimodule is:
$B(R,R) = \cdots \xrightarrow{b'} R \otimes R \otimes R \otimes R \xrightarrow{b'} R \otimes R \otimes R \xrightarrow{b'} R \otimes R$, $\space \space \space \space$(1)
here we write $\otimes$ for $\otimes_k$, $B(R,R)_n = R^{\otimes n+2}$ and $B(R,R)_0 = R \otimes R$,
$b'_n(a_o \otimes \cdots \otimes a_{n+1}) = \Sigma_{i=0}^n (-1)^i a_0 \otimes \cdots \otimes a_ia_{i + 1} \otimes \cdots \otimes a_{n+1}$,
and $b' \circ b' = 0$.
Let $M$ be an $R$-bimodule (hence a right $R^e$-module), if we apply the functor $(M \otimes_{R^e}-)$ to the complex (1) we're supposed to get back the Hochschild homology:
$\cdots \xrightarrow{b} M \otimes R \otimes R \otimes R \xrightarrow{b} M \otimes R \otimes R \xrightarrow{b} M \otimes R \xrightarrow{b} M$,
which proves that $H_n(R,M) = Tor_n^{R^e}(M,R)$.
The thing here is when I apply the functor $(M \otimes_{R^e}-)$ to the complex (1) I get back:
$\cdots \xrightarrow{b} M \otimes R \otimes R \otimes R \otimes R \xrightarrow{b} M \otimes R \otimes R \otimes R \xrightarrow{b} M \otimes R \otimes R$,
not the complex (1)
What am I doing wrong?

2 - Another thing, "the bar resolution" of a left $R$-module $M$ (as defined on page 283 of Weibel's "Homological algebra") is
$B(R,M) = \cdots \rightarrow R \otimes R \otimes R \otimes M \rightarrow R \otimes R \otimes M \rightarrow R \otimes M \rightarrow M$,
where $B(R,M)_n = R^{\otimes n+1} \otimes M$ and $B(R,M)_0 = R \otimes M$,
I was wondering, if $M$ is an $R$-bimodule
- Is this resolution also a resolution of $M$ as a left $R^e$-module? 
- And is $R \otimes M$ projective as a left $R^e$-module?
:)

REPLY [2 votes]: Regarding your second question: if $M$ is a bimodule, the complex $B(R,M)$ is indeed a complex of $R$-bimodules (the objects are bimodules and the maps are bimodule maps), and it is exact, but in general it is not projective.
Indeed, $R\otimes M$ is not usually  projective.
Notice, in fact, that there are isomorphisms $\hom_{R{-}R}(R\otimes M,N)\cong\hom_{{-}R}(M,N)$ natural in $N$, so that the bimodule $R\otimes M$ is projective iff $M_R$ is projective (for this, we need that $k$ be a field or something along those lines)