# Proving finite presentation [closed]

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.

• View $S$ as a quotient of a polynomial ring and first do the case where it is that polynomial ring. Feb 1 at 9:02
• Since $S$ is finitely presented $R$ algebra then for some $n$ and $m$ I have as an algebra iso $S \cong R[x_1,...,x_n]/(f_1,..f_m)$, so now are you telling me to work with $S$ being $R[x_1,..x_n]$ sir? Feb 1 at 9:53
• @Wilberd van der Kallen: S doesn't need to be commutative.
– tj_
Feb 1 at 12:31

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.