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Let $k$ be a commutative ring, a $R$-Bimodule $M$ over a $k$-algebra $R$ 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 $R \otimes R^{op}$-module.

Now what I want to know:

1 - Let this $R$-Bimodule $M = R$, let $R$ (considered as a left $k$-module) be projective (actually, let $k$ be a field, so $R$ is a vector space), is the corresponding $R \otimes R^{op}$-module a projective module too?

2 - 1 - Let this $R$-Bimodule $M = R$, let $R$ (considered as a left $k$-module) be free (actually, let $k$ be a field, so $R$ is a vector space), is the corresponding $R \otimes R^{op}$-module a free module too?

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    $\begingroup$ I don't understand the hypothesis. Is $M$ to be assumed free as a left (or right) $R$-module? Or is it assumed free both as a left $R$-module and as a right $R$-module? $\endgroup$ – Steven Landsburg Mar 12 '13 at 5:26
  • $\begingroup$ $M$ is a $R$-$R$-Bimodule, that is, a left and right $R$ module. I'm interested in $M$ as a left $R$-module. I want to know if $M$, seen as a left $R$-module, is projective, I want to know if $M$, seen as the corresponding left $R \otimes R^{op}$-module, is also projective. $\endgroup$ – Richard Jennings Mar 12 '13 at 6:17
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    $\begingroup$ R will be projective if and ony if the Hochschild cohomology vanished above dimension 0 which almost never happens. $\endgroup$ – Benjamin Steinberg Mar 12 '13 at 13:55
  • $\begingroup$ I don't understand the question, really: in your last sentence do you mean «if $R$ is a (free) projective bi module...»? $\endgroup$ – Mariano Suárez-Álvarez Mar 12 '13 at 15:19
  • $\begingroup$ There's been some confusion (I was tired), I actually rewrote the question. I mean if $M = R$ happens to be projective (as a left $k$-module), is the corresponding $R \otimes R^{op}$-module a projective module too? $\endgroup$ – Richard Jennings Mar 13 '13 at 11:44
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No. A ring $R$ for which $R$ is projective as a left $R\otimes_{\mathbb Z}R^{op}$-module is sometimes called separable over ${\mathbb Z}$.

This is equivalent to the splitting, as a surjection of $R\otimes_{\mathbb Z}R^{op}$-modules, of the multiplication map $R\otimes_{\mathbb Z}R\to R$. So applying $-\otimes_R M$, the $R$-action map $R\otimes_{\mathbb Z}M\to M$ has to split as a surjection of $R$-modules for every left $R$-module $M$. In particular, if $M$ is ${\mathbb Z}$-free, so that $R\otimes_{\mathbb Z}M$ is free as an $R$-module, $M$ must be projective as an $R$-module.

But it's not true that every ${\mathbb Z}$-free $R$-module is projective if (as in Steven's answer) $R=\mathbb{Z}[X]$ or $R=\mathbb{Z}G$ for a non-trivial group $G$, or a host of other examples.

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  • $\begingroup$ Thank you, thank you, thank you so much. There's been some confusion (I was tired), I actually rewrote the question. I mean if a bimodule $M = R$ happens to be projective as a left $k$-module (actually it is a vector space), is the corresponding $R \otimes R^{op}$-module a projective module too? $\endgroup$ – Richard Jennings Mar 13 '13 at 11:53
  • $\begingroup$ You can just replace ${\mathbb Z}$ by $k$ in what I wrote (apart, sometimes, for the bit about ${\mathbb Z}G$). If $R$ is free as an $R\otimes_k R^{op}$-module, then in particular every ($k$-free, which is automatic if $k$ is a field) $R$-module is projective, which restricts $R$ severely. $\endgroup$ – Jeremy Rickard Mar 13 '13 at 15:42
  • $\begingroup$ Thanks, the wording in this paper on Hochschild (co)homology that I was reading confused me and led me to believe that. But I'm curious, you say that if $R$ is free as a left $R \otimes R^{op}$-module then every $k$-free $R$-module is projective, you have a reference for that? or where I can find more info on enveloping algebras and modules over them? I'd like to know what kind of $R$-Bimodules make the corresponding $R \otimes R^{op}$-module projective, or just know more about this topic. thank you. $\endgroup$ – Richard Jennings Mar 15 '13 at 2:25
  • $\begingroup$ You could look up "separable algebra" in Cohn's Algebra 3 (or many other algebra or ring theory books: that's just the first I tried). $\endgroup$ – Jeremy Rickard Mar 15 '13 at 11:25
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I'm not sure what hypothesis you intend, but I don't think there's any reasonable interpretation under which it implies your conclusion.

Let $R=M={\mathbb Z}[X]$, acting on itself by multiplication from both the left and the right. Then $R\otimes R^{op}\approx {\mathbb Z}[X,Y]$ is a domain, and $(X-Y)$ annihilates $M$, whence $M$ cannot be free.

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  • $\begingroup$ First of all: thank you so much. There's been some confusion (I was tired), I actually rewrote the question. I mean if a bimodule $M = R$ happens to be projective as a left $k$-module (actually it is a vector space), is the corresponding $R \otimes R^{op}$-module a projective module too? $\endgroup$ – Richard Jennings Mar 13 '13 at 11:53
  • $\begingroup$ The same counterexample still applies; if you want $k$ to be a field, use $k[x]$ instead of ${\mathbb Z}[x]$. $\endgroup$ – Steven Landsburg Mar 13 '13 at 12:16
  • $\begingroup$ Thank you again. I just was reading this paper on Hochschild (co)homology and the author wrote: "The $R$-bimodule $R$ mentioned above is not free, not even projective in general, in the sense that the corresponding $R \otimes R^{op}$ is not free (or projective)" whick kind of led me to believe that $R$ projective $\Rightarrow$ $R$ (as a left $R \otimes R^{op}$-module) projective. Thank you. $\endgroup$ – Richard Jennings Mar 15 '13 at 2:07

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