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?

andas a right $R$-module? – Steven Landsburg Mar 12 '13 at 5:26bimodule...»? – Mariano Suárez-Alvarez♦ Mar 12 '13 at 15:19