0
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ View $S$ as a quotient of a polynomial ring and first do the case where it is that polynomial ring. $\endgroup$
    – Wilberd van der Kallen
    Feb 1 at 9:02
  • $\begingroup$ 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? $\endgroup$
    – user443060
    Feb 1 at 9:53
  • 1
    $\begingroup$ @Wilberd van der Kallen: S doesn't need to be commutative. $\endgroup$
    – tj_
    Feb 1 at 12:31

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes I got it thank you and yes there is no need of flatness here . $\endgroup$
    – user443060
    Feb 1 at 11:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.