Let $R_1$ and $R_2$ be two subrings of a ring $R$ (not necessarily commutative) which commute in $R$ so that we have a ring homomorphism $R_1\otimes_\mathbb{Z} R_2\rightarrow R$ and $R$ is a module over $R_1\otimes_\mathbb{Z}R_2$. Assume also that $R$ is flat both as $R_1$- and $R_2$-module. Is then $R$ flat over $R_1\otimes_\mathbb{Z}R_2$? Is there an easy counterexample?
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1Counterexample: $R_1=R_2=R=\mathbb{Z}[x]$ – Riccardo de Rossa Apr 22 '13 at 10:02
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1Please consider expanding your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 24 '13 at 17:58
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Do you mean flat as an $R_1$-$R_2$ bimodule? Or flat over both rings individually? – rschwieb Jun 25 '13 at 01:46