Let $k$ be a commutative ring (with unity). Let $R$ be a $k$-algebra (with unity, but not of necessity commutative).


Let $M$ be an $\left(R,R\right)$-bimodule where $k$ acts in the same way from the left and from the right (I'd call this an $\left(R,R\right)\_k$-bimodule, but I haven't seen this notation anywhere). There are two ways to define the Hochschild homology and cohomology of $R$ with coefficients in $M$: either as the homology of the standard complex tensored with $M$ rsp. the cohomology of Hom of the standard complex and $M$, or as $\mathrm{Tor}$ and $\mathrm{Ext}$. As far as I understand, these two definitions are equivalent only if $R$ is a projective $k$-module, which I **don't** want to require here.


**Question 1:** So let me define Hochschild cohomology and homology through the standard complex. Then, [Löfwall's text](http://www2.math.su.se/~clas/hochschild/lesson1.pdf) seems to silently hint at the fact that if a $k$-algebra $R$ satisfies $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$, then it also satisfies $\mathrm{H}\_1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$. While this is clear from homological algebra in the case when $R$ is a projective $k$-module, is this true otherwise? And how is it proven?


(*Remark:* A $k$-algebra $R$ such that $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$ is said to be *zero-dimensional* (in Löfwall's text) or *separable* (in the modern sense of this word).)


**Question 2:** The same text gives a counterexample for the opposite direction (if $\mathrm{H}\_1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$, then $\mathrm{H}^1\left(R,M\right)=0$ for all $\left(R,R\right)$-bimodules $M$). In this counterexample, $k$ is a field and $R$ is commutative, but infinite-dimensional. I assume that counterexamples fade when we impose some more restrictive conditions on $k$ and $R$. What about finite-dimensional $R$? What about finitely-generated-as-algebras $R$? If $k$ is not a field anymore? If $R$ is not commutative anymore?



Hochschild H^1 (R,M) = 0 vs. H_1 (R,M) = 0 where R is a ring and M is an (R,R)-bimodule
(Let me write $A$ for your $R$, because I will mix letters up if not...)


The standard complex for $A$ over $k$ is exact, independently of the projectiveness of $A$ over $k$; call $d$ its differential. Then we have a short exact sequence 
$$
0 
\to 
\frac{A\otimes\_kA\otimes\_kA}{d(A\otimes\_kA\otimes\_kA\otimes\_kA)}
\to A\otimes\_k A
\to
A
\to
0
$$
Call $K$ the quotient appearing here, and call $\iota$ the map $K\to A\otimes\_kA$


Now suppose $H^1(A,K)=0$ (with cohomology defined using the standard complex, as you wanted; this hypothesis is implied by the hypothesis that $H^1(A,M)=0$ for all $M$ that you wanted to consider, of course!) This means that every bimodule map $f:K\to K$ can be factorized through $\iota$, so there exists a $\bar f:A\otimes\_kA\to K$ such that $\bar f\circ\iota=f$. In particular, if we take $f=\mathrm{id}\_K$ we see that the injective map $\iota$ splits. It follows that the short exact sequence of $A$-bimodules above itself splits, and that the map $A\otimes\_kA\to A$ given by multiplication is split by a map $\phi:A\to A\otimes\_kA$. Let $e=\phi(1)=A\otimes\_kA$.


Consider now the standard complex, 
$$
\cdots
\to
A\otimes\_kA\otimes\_kA\otimes\_kA
\to
A\otimes\_kA\otimes\_kA
\to
A\otimes\_kA
\to
A
$$
Using $\phi$ (or $e$), you can construct a retraction of this complex, so that when you tensor it with an arbitrary bimodule $M$ over $A^e$ you get an acyclic complex. In particular, $H\_1(A,M)=0$, defined again as you wanted.


The first two components of that retraction are
$$
a\in A\mapsto ae\in A\otimes\_kA,
$$
$$
a\otimes b\in A\otimes\_kA\mapsto ae\otimes b\in A\otimes\_kA\otimes\_kA,
$$
and it keeps going like that.




---


As for your second question (and I consider now the situation in which $A$ is $k$-projective):


When $H\_1(A,M)=0$ for all $M$, one says that the the *weak dimension* of $A$ as an $A^e$-module is zero and writes $\operatorname{wdim}A=0$. When $H^1(A,M)=0$ for all $M$, then one says that the *projective dimension* of $A$ as a bimodule is zero, and writes $\operatorname{pdim}A=0$.


As you know, $$\operatorname{pdim}A=0\implies\operatorname{wdim}A=0$$ and the converse is false. If $A^e$ is noetherian, then the converse holds (see, for example, McConnell and Robson's bible on noetherian rings); this includes the case in which $A$ is finitely dimensional over $k$ or, more generally, if $A$ is *commutative* and finitely generated over $k$ as an algebra. If $A$ is not projective over $k$, I have no idea what happens (and you should probably ignore that situation if you are reading up on Hochschild cohomology for the first few times! :) )


