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Let $A\to R$ be a faithfully flat ring map. If $a,b\in A$ and there exists $r\in R^*$ (invertible element) such that $b=ar$, is it true that $b=ac$ for some $c$ in $A^*$?

I have been stuck on this for a while. Is there some slick to apply the definition of faithfull flatness and get the result instantly? I have managed to prove that $R^*\cap A=A^*$.

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    Use the fact that that faithfully flat implies $A\to R$ is an injection and $A/aA\to R/aR$ is also faithfully flat. – 2017-01-18
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    How do you prove that the $c$ you get that way is invertible? – 2017-01-18
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    $c$ may not be invertible in general. The only thing you can say is $aA=bA$. To write an example, consider $A=k[x,y,z]/z(xy-1)$, $a=z, b=xz$ and $R=A[u, \frac{1}{x+u(xy-1)}]$. – 2017-01-20
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    Thank you very much! Did you find this example geometrically? Could you explain me the geometric reasoning behind it? Also, if you write this as an answer I will gladly accept it. – 2017-01-20
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    Algebra is easier than geometry in this case. This is what one would call the universal example for the given problem. – 2017-01-20

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