Timeline for Torsion in tensor products over noncommutative rings

5 events

when toggle format	what		by	license	comment
Mar 20, 2011 at 15:42	vote	accept	TonyS
Mar 20, 2011 at 15:41	comment	added	TonyS		Ahh, very good. Thanks for your help. This are two conditions for $\omega_R$ i can work with.
Mar 20, 2011 at 15:20	comment	added	user91132		Yes. If $\omega_R$ is free as a right $R$-module, then it's flat as such. Now $M$ is torsion-free so $M \hookrightarrow R^k$ as a left $R$-module for some integer $k$. Therefore by flatness, $\omega_R \otimes_R M \hookrightarrow \omega_R \otimes_R R^k \cong \omega_R^k$ as a left $R$-module. So if you also know that the left $R$-module $\omega_R$ is torsion-free, then so is $\omega_R \otimes_R M$. For this argument to work it is necessary and sufficient for $\omega_R$ to be projective as a right $R$-module and torsion-free as a left $R$-module.
Mar 20, 2011 at 15:09	comment	added	TonyS		Thanks! If $dim(A)=1$ and $rk(R)=4$, one can even compute $\omega_R$: $\omega_R=p^{-1}R$, where $p$ generates the radical ideal, i.e. $rad(R)=pR$. I also found the notion of Gorenstein orders, see for example arxiv.org/abs/math/0401425, these are orders such that $\omega_R$ and $R$ are isomorphic as left and right $R$-modules, but not necessarily as $R$-bimodules. But would this be enough for $\omega_R\otimes_R M$ to be torsion free over $A$, if we consider $\omega_R$ just as a left $R$-module, i.e. forget the left module structure?
Mar 20, 2011 at 13:14	history	answered	user91132	CC BY-SA 2.5