Theorem 9.1.7 in Weibel's homological algebra reads as follows (I will change the notation slightly):

Let $f:k\to \ell$ be a morphism of commutative rings. Denote $\otimes = \otimes_k$. Let $A$ be a $k$-algebra. Consider $\ell \otimes A$: it is an $\ell$-algebra where $\ell$ acts on the left.

Let $Q$ be an $(\ell \otimes A)$-bimodule. In particular, it is an $A$-bimodule by restriction of scalars. Then:

$\operatorname{HH}^\ell_*(\ell \otimes A,Q) \cong \operatorname{HH}^k_*(A,Q)$.

He proves it as follows: consider the canonical $k$-isomorphism

$$ Q \otimes_\ell (\ell \otimes A)^{\otimes_\ell n} \cong Q \otimes A^{\otimes n}$$

He claims this yields an isomorphism of the Hochschild complexes, but I fail to see how it commutes with the last face map. So let's look at it in detail. The isomorphism is

$$q\otimes((\lambda_1\otimes a_1)\otimes \dots \otimes (\lambda_n \otimes a_n)) \mapsto (q\cdot (\lambda_1\dots \lambda_n \otimes 1)) \otimes a_1\otimes \dots \otimes a_n.$$

We need to check that the following square commutes. $$ \require{AMScd} \begin{CD} Q \otimes_\ell (\ell \otimes A)^{\otimes_\ell n} @>>> Q \otimes A^{\otimes n} \\ @VVV {} @VVV \\ Q \otimes_\ell (\ell \otimes A)^{\otimes_\ell n-1} @>>> Q \otimes A^{\otimes n-1} \end{CD} $$

But I believe that going first horizontally and then vertically makes $q\otimes((\lambda_1\otimes a_1)\otimes \dots \otimes (\lambda_n \otimes a_n))$ land in

$$ (1\otimes a_n) \cdot q \cdot (\lambda_1 \cdots \lambda_n \otimes 1) \otimes a_2\otimes \dots \otimes a_n$$

and going in the other direction makes it land in

$$ (\lambda_n \otimes a_n) \cdot q \cdot (\lambda_1 \cdots \lambda_{n-1} \otimes 1) \otimes a_2 \otimes \dots \otimes a_n$$

and I don't see why these two should be equal. If the bimodule $Q$ were symmetric, this would be true, but why would it be true in general?

So where is my mistake?


I think the question is what Weibel means by "$R_\ell$-$R_\ell$-bimodule" in his Theorem 9.1.7. When algebraists talk of an "$\left(A,B\right)$-bimodule" (Weibel writes "$A$-$B$-bimodule"), they rarely mean literally an additive group endowed with a left $A$-module structure and a right $B$-module structure satisfying the "associativity law"

(1) $\left(ac\right)b=a\left(cb\right)$ for all $a \in A$, $b \in B$ and $c \in C$.

Usually, what they mean is a relative version of this notion. Namely, they usually work over a commutative "base ring" $k$, for which both $A$ and $B$ are $k$-algebras. Then, they define an $\left(A,B\right)$-bimodule to mean a $k$-module endowed with a $k$-bilinear left $A$-module structure and a $k$-bilinear right $B$-module structure satisfying the "associativity law" (1). Strictly speaking, it is a bad idea to suppress the $k$ from the notation, as it leads to confusion (see, e.g., math.stackexchange #889130), but in practice it works most of the time when the $k$ is really clear from the context. (I personally prefer to keep the $k$ explicit and speak of "$\left(A,B\right)_k$-bimodules".)

Now, I think that when Weibel speaks of "$R_\ell$-$R_\ell$-bimodule", he wants $\ell$ (not $k$) to be the common base ring. If that is so, then your diagram commutes. This is less clear in your restatement of the theorem, because the tensor product $\ell \otimes A$ does not suggest a full base change as strongly as the notation $R_\ell$. But of course, the right solution is not to rely on suggestion, and clarify the notation.

  • $\begingroup$ Thanks a lot! I didn't know this convention: as you suspected, I was only aware of the "ring"-version of a bimodule... It's nice to learn it by making a small tedious routine verification and realizing that something was missing. $\endgroup$ Jan 2 '16 at 1:59

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