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Suppose $A$ is a finitely generated $\mathbb{Z}$-algebra and $R$ is a finitely generated $A$-algebra. We have a sequence of finitely generated $R$-modules \begin{align*} \mathbb{F}:M_1\rightarrow M_2\rightarrow M_3 \end{align*} such that the composite of the maps in the sequence is zero (not exact) and we know that $\mathbb{F}\otimes \mathrm{Frac}(A)$, where $\mathrm{Frac}(A)$ is the fraction field of $A$, is exact at $M_2\otimes\mathrm{Frac}(A)$. Then does it follow that $\mathbb{F}\otimes A_a$ is exact at $M_2\otimes A_a$ for some nonzero $a\in A$?

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  • $\begingroup$ All localisations by multiplicatively closed sets are flat. $\endgroup$
    – Zeek
    Aug 22, 2020 at 12:29
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    $\begingroup$ the sequence F is not exact sequence. $\endgroup$
    – dongrugose
    Aug 22, 2020 at 12:34
  • $\begingroup$ Is the tensor over $A$ ? If so, what's the point of $R$ ? Are $M_1,M_2,M_3$ assumed to be finitely generated, or not ? $\endgroup$ Aug 22, 2020 at 12:42
  • $\begingroup$ The tensor is over $A$. It is a generic freeness type question. $\endgroup$
    – dongrugose
    Aug 22, 2020 at 12:45
  • $\begingroup$ But then $R$ is useless, $M_1,M_2,M_3$ might as well be modules over $A$ $\endgroup$ Aug 22, 2020 at 12:48

2 Answers 2

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$\newcommand{\im}{\mathrm{im}}$

With the added conditions, this becomes true (I'm assuming the notation $\mathrm{Frac}(A)$ assumes $A$ is an integral domain).

Consider the inclusion $\im\subset \ker$. $A_a\otimes \im \subset A_a\otimes \ker$ is still an inclusion, as $A_a$ is flat, so we just need to prove that it becomes an equality for some $a$.

But note that this inclusion is still $R$-linear (even though we're tensoring over $A$). So if the LHS contains generators of the RHS, the inclusion is an equality.

$\ker$ is finitely generated ($R$ is noetherian, as it's finitely generated over $\mathbb Z$, and $M_2$ is finitely generated by hypothesis, therefore so is any submodule); so let $x_1,...,x_n$ denote a set of generators.

$\mathrm{Frac}(A) \otimes \im \to \mathrm{Frac}(A)\otimes \ker$ is the directed colimit of the $A_a\otimes \im\to A_a\otimes \ker$.

So let $y_1,...,y_n\in A_a\otimes \im$ be elements that become antecedents of $x_1,...,x_n$ under $A_a\otimes \im \to \mathrm{Frac}(A)\otimes \im$.

It follows that the images of $y_1,...,y_n$ in $A_a\otimes \ker$ become identified with $x_1,...,x_n$ in $\mathrm{Frac}(A)\otimes \ker$. Since there are only finitely many of them, they become identified with $x_1,...,x_n$ in some $A_b\otimes\ker$ for some $b$ divisible by $a$, and so $A_b\otimes \im\to A_b\otimes \ker$ is $R$-linear and its image contains $x_1,...,x_n$, so we are done.

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  • $\begingroup$ I thikn this should work!. The answer I wrote as a comment above is the similar idea. Though I use the Generic Freeness. $\endgroup$
    – dongrugose
    Aug 22, 2020 at 14:09
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The answer is no without further hypotheses.

Indeed, take $A=R=\mathbb Z$, $M_1 = M_2 = \mathbb Q, M_3 = \mathbb{Q/Z}$, the sequence $\mathbb F$ is $id_\mathbb Q$ followed by the canonical projection.

It's clearly not exact in $M_2$ ($id_\mathbb Q$ is surjective, but the canonical projection is not $0$), similarly if you tensor with $\mathbb Z[\frac 1 n]$ for any $n$.

However, if you tensor it with $\mathbb Q$, you get $\mathbb{Q\to Q}\to 0$ which is indeed exact.

If you want a sequence where the composite is $0$, you can do that too:

$\mathbb Z \overset{(1,0)}\to \mathbb{Q\oplus Q}\overset{(0,1)}\to \mathbb Q$.

The composite is of course $0$, if you tensor it with $\mathbb Q$, you get a split short exact sequence; however if you tensor it with $\mathbb Z[\frac 1 n]$ it will still not be exact ($\ker/\mathrm{im} = \mathbb Q/\mathbb Z[\frac 1 n]$)

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  • $\begingroup$ Of ocurse i forgot to add the composite of the maps is zero and $M$'s should be finitely generated $R$ modules. Forgot to add that. Adding it now $\endgroup$
    – dongrugose
    Aug 22, 2020 at 13:06

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