Let $k$ be a commutative ring, let $R$ be a $k$-algebra, a $R$-Bimodule $M$ over $R$ is a $k$-module with two actions of $R$ on $M$, on the left and on the right, the classical example of this being $R$ itself with left and right multiplications as the actions. A $R$-Bimodule $M$ can also be considered a left $R^e = R \otimes R^{op}$-module.
Now consider the $R$-Bimodule $R$, according to this text I'm reading on Hochschild (co)homology the $k$-module $R \otimes R$ is an $R$-Bimodule by multiplication on the left and the right, there's a surjective map $R \otimes R \rightarrow R$ of bimodules and a resolution of $R$ as an $R$-Bimodule:
$\cdots \rightarrow R \otimes R \otimes R \otimes R \rightarrow R \otimes R \otimes R \rightarrow R \otimes R \rightarrow R$.
Just two things the author mentioned and I was curious about:
1 - The bimodule $R \otimes R$ converted to a left $R^e = R \otimes R^{op}$-module is $R \otimes R^{op}$ itself and it is free.
2 - If $R$ is free (or projective) as a $k$-module then $R \otimes R^{\otimes n} \otimes R$ is free (or projective).
I wanted to know:
Why is $R \otimes R$ (and the subsequent modules $R \otimes R^{\otimes n} \otimes R$) free (or projective)? Is that the case too even if $R$ is not free (or projective)?
Suppose $R$ is a $k$-algebra, with $k$ a commutative ring.
If $M$ is a $k$-module, we can construct the $k$-module $R\otimes_kM\otimes_kR$, which is automatically an $R$-bimodule or, equivalently, an $R^e$-module.
If $N$ is an $R$-bimodule, there is a canonical isomorphism $$\hom_{R^e}(R\otimes_kM\otimes_kR, N)\cong\hom_k(M,\bar N)$$ with $\bar N$ the underlying $k$-module of $N$. Since going from $N$ to $\bar N$ is an exact functor, we thus see that the functor $\hom_{R^e}(R\otimes_kM\otimes_kR,\mathord-)$ is exact if $\hom_k(M,\mathord-)$ is exact, that is, $R\otimes_kM\otimes_kR$ is a projective $R$-bimodule if $M$ is a projective $k$-module.
If $M$ and $N$ are projective $k$-modules, then $M\otimes_kN$ is projective. Indeed, there exist $k$-modules $M'$ and $N'$ such that $M\oplus M'$ and $N\oplus N'$ are free. One easily checks that the tensor product of free modules is free, so that $(M\oplus M')\otimes_k(N\oplus N')$ is projective, and $M\otimes_kN$ is a direct summand of this (because $\otimes$ is an additive functor).
Using this last fact, we see that if $R$ is $k$-projective, then $R^{\otimes n}$ is $k$-projective for all $n\geq1$, and then the first bullet point above implies that $R\otimes R^{\otimes n}\otimes R$ is a projective $R$-module.
Suppose now $R$ is a $k$-algebra, with $k$-commutative, that we have a morphism $\epsilon:R\to k$ of $k$-algebras, and let us suppose that $R$ is not $k$-projective. There exists then a short exact sequence $$0\to M'\to M\to M''\to0\tag{$\star$}$$ of $r$-modules such that the associated sequence $$0\to\hom_k(R,M')\to\hom_k(R,M)\to\hom_k(R,M'')\to0\tag{$\clubsuit$}$$ is not exact.
Now every $k$-module $N$ can be viewed as a $R$-bimodule, defining the left and right actions of $R$ on $N$ by $$r\cdot n\cdot s=\epsilon(rs)n, \qquad\forall r,s\in R,\quad\forall n\in N.$$ This operation of turning $k$-modules into $R$-bimodules is a functor which is obviously exact (because it does not change the underlying abelian groups, say) It allows us to view the short exact seequence $(\star)$ as a sequence of $R$-bimodules.
Now consider the $R$-bimodule $R\otimes R\otimes R$. For every $R$-bimodule $Q$ we have $\hom_{R^e}(R\otimes R\otimes R,Q)\cong\hom_k(R,Q)$, as we observed before. It follows that if we apply the functor $\hom_{R^e}(R\otimes R\otimes R,\mathord-)$ to the short exact sequence $(\star)$ we get exactly the same exact sequence $(\clubsuit)$, which is not exact. It follows that $R\otimes R\otimes R$ is not a projective bimodule
For a concrete example, take $k=\mathbb Z$, $R=\mathbb Z\oplus \mathbb Z/2\mathbb Z$ with multiplication given by $(a,b)(c,d)=(ac,ad+bc)$ and $\epsilon:(a,n)\in R\mapsto a\in k$.