## 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 $k$ 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 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?

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 I just see a related question: mathoverflow.net/questions/33205/… – darij grinberg Feb 7 2011 at 17:11

(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! :) )

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 It' "J. C. McConnell and J. C. Robson". Otherwise, great answer! – darij grinberg Feb 7 2011 at 17:09 projecteuclid.org/… Theorem 1 answers the case of $A$ projective over $k$ as well. – darij grinberg Feb 7 2011 at 17:20