Let $R$ be an integral domain, $S$ be a finitely presented $R$ algebra. Then for a flat $R$ module $M$ which is also a finitely generated $S$ module I need to show that $M \otimes\_{R}T$ is a fintely presented $S\otimes\_{R}T$ module where $T$ is the quotient field of $R$.


My attempt :
Since $M$ is finitely generated $S$ module then we have an $S$ module surjection $$S^{n\_1} \rightarrow M.$$ Since tensor product is right exact we would have a surjection $$S^{n\_1}\otimes\_{R}T \rightarrow M \otimes\_{R}T$$, now if I can show that the kernel of this map is finitely generated I will be done and here is what I feel like finitely presented $R$ algebra property of $S$ might be used , but I am unable to progress further from here.


Any hint or suggestion is much appreciated, thank you.



Proving finite presentation
I am following the notation in the comments. I do not have privilege to make a comment, do I should write and answer. Also from the comments I understand that you are working just with commutative rings.


Notice that $R[x\_1,\dots, x\_n]/(f\_1,\dots,f\_m)\otimes T$ is isomorphic to a quotient of $T[x\_1,\dots,x\_n]$ so it is a noetherian ring (because T is a field). Then any finitely generated module is finitely presented.


What I do not understand is why you neet M\_R to be flat.


