# Exactness of sequence and localization

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 $$a\in A$$?

• All localisations by multiplicatively closed sets are flat.
– Zeek
Aug 22, 2020 at 12:29
• the sequence F is not exact sequence. Aug 22, 2020 at 12:34
• 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 ? Aug 22, 2020 at 12:42
• The tensor is over $A$. It is a generic freeness type question. Aug 22, 2020 at 12:45
• But then $R$ is useless, $M_1,M_2,M_3$ might as well be modules over $A$ Aug 22, 2020 at 12:48

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

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.

• I thikn this should work!. The answer I wrote as a comment above is the similar idea. Though I use the Generic Freeness. Aug 22, 2020 at 14:09

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 ($$\ker/\mathrm{im} = \mathbb Q/\mathbb Z[\frac 1 n]$$)

• 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 Aug 22, 2020 at 13:06