3

I am trying to solve an exercise given in Vakil's Algebraic Geometry notes. Suppose $M$ is a finitely presented $A$-module. The $M$ fits inside an exact sequence $A^q\rightarrow A^p\rightarrow M\rightarrow 0$. I'd like to understand why in this case we get an isomorphism $S^{-1}\text{Hom}_A(M,N)\cong \text{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}M)$. This problem is towards the beginning of the book, so in particular there should be a way to solve it without heavy duty commutative algebra.

So far, I've only come up with the following: We can use the universal property of localization of modules so that for any map from $\text{Hom}_A(M,N)$ to $\text{Hom}_{S^{-1}A} (S^{-1}M,S^{-1}M)$ (in which the elements of $S$ are invertible), there exists a unique map from $S^{-1}\text{Hom}_A(M,N)$ to $\text{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}M)$.

However, what should this map explicitly be? Is this the way to go about showing these two are isomorphic?

EDIT: There is a question about the same problem, but I am specifically asking about how to construct a map between the two sets. The solution in the related question uses facts about flat modules which I am trying to eschew.

ponchan
  • 2,656

2 Answers2

3

As $S^{-1}$ is a functor we have a map $$\mathrm{Hom}_A(M,N)\rightarrow \mathrm{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}N)$$ and as multiplication by an element $s\in S$ gives an isomorphism in the module $\mathrm{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}N)$ this map extends to a map $$\tag{$\star$} S^{-1}\mathrm{Hom}_A(M,N)\rightarrow \mathrm{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}N)$$

Now you have that

  • The map ($\star$) is an isomorphism for $M=A$. More generally, it is an isomorphism for $M=A^n$.
  • The map ($\star$) is natural in $M$, in particular if we take an exact sequence $$A^q\rightarrow A^p\rightarrow M\rightarrow 0.$$ Then we get a diagram $$\begin{array}{c} 0 &\rightarrow & S^{-1}\mathrm{Hom}_A(M,N) & \rightarrow & S^{-1}\mathrm{Hom}_A(A^p,N) & \rightarrow & S^{-1}\mathrm{Hom}_A(A^q,N)\\ &&\downarrow && \downarrow && \downarrow\\ 0 &\rightarrow & \mathrm{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}N) & \rightarrow & \mathrm{Hom}_{S^{-1}A}(S^{-1}A^p,S^{-1}N) & \rightarrow & \mathrm{Hom}_{S^{-1}A}(S^{-1}A^q,S^{-1}N) \end{array}$$

By the previous point the maps on the 3rd and 4th columns are isomorphism. Hence by Five Lemma the map in the second column is an isomorphism and we finish.

nowhere dense
  • 2,816
0

Regarding the problem, solve it for $M=A^p$ and then use the exact sequence. Regarding the map, the universal property of localization tells you what it is: if $f\in \text{Hom}_A(M,N)$, then the image of $f/s$ is the map taking $m/t$ to $f(m)/st$.

A.G
  • 1,423
  • Where does the factor of $st$ in the denominator come from? – ponchan Jun 28 '20 at 13:41
  • $s$ beacuse it must be a homomorphism of $S^{-1}A$-modules and $t$ so that it extends the map $\text{Hom}A(M,N) \to \text{Hom}{S^{-1}A} (S^{-1}M,S^{-1}M)$. All you have to worry about is to check that this map is a well defined homomorphism of $S^{-1}A$-modules extending $\text{Hom}A(M,N) \to \text{Hom}{S^{-1}A} (S^{-1}M,S^{-1}M)$. Since by the universal property there exists only one such homomorphism, you have found it. – A.G Jun 28 '20 at 15:19