Let $R_1$ and $R_2$ be two subrings of the ring $R$ which commute in $R$ so that we have a ring homomorphism $R_1\otimes_\mathbb{Z} R_2\rightarrow R$. Assume that $R$ is flat over $R_1$ and $R_2$. Is then $R$ also flat over $R_1\otimes_\mathbb{Z} R_2$? Is there an easy counterexample?
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Take $R_1 = R_2 = R = {\mathbb Z}[x]$. Then $R_1\otimes_{\mathbb Z} R_2 = {\mathbb Z}[x_1,x_2]$ and $R = {\mathbb Z}[x]$ is not flat over it.

$\begingroup$ Thx for your answer. Do you have a reference or easy argument why Z[x] is not flat over Z[x1,x2]? $\endgroup$ – Werner Thumann Apr 21 '13 at 9:17

$\begingroup$ Use free resolution $$ 0 \to {\mathbb Z}[x_1,x_2] \xrightarrow{x_1x_2} {\mathbb Z}[x_1,x_2] \to {\mathbb Z}[x] \to 0 $$ to compute $Tor_1^{{\mathbb Z}[x_1,x_2]}({\mathbb Z}[x],{\mathbb Z}[x]) = {\mathbb Z}[x] \ne 0$. $\endgroup$ – Sasha Apr 21 '13 at 10:41
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Ref, H. Matsumura, Commutative Ring Theory P1620 probably... The counter example is just @Sasha mentioned.