Lemma 10.10.2. Let $R$ be a ring. Let $M$ be a finitely presented $R$-module. Let $N$ be an $R$-module.

1. For $f \in R$ we have $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)_ f = \mathop{\mathrm{Hom}}\nolimits _{R_ f}(M_ f, N_ f) = \mathop{\mathrm{Hom}}\nolimits _ R(M_ f, N_ f)$,

2. for a multiplicative subset $S$ of $R$ we have

$S^{-1}\mathop{\mathrm{Hom}}\nolimits _ R(M, N) = \mathop{\mathrm{Hom}}\nolimits _{S^{-1}R}(S^{-1}M, S^{-1}N) = \mathop{\mathrm{Hom}}\nolimits _ R(S^{-1}M, S^{-1}N).$

Proof. Part (1) is a special case of part (2). The second equality in (2) follows from Lemma 10.9.5. Choose a presentation

$\bigoplus \nolimits _{j = 1, \ldots , m} R \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} R \to M \to 0.$

By Lemma 10.10.1 this gives an exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \bigoplus \nolimits _{i = 1, \ldots , n} N \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} N.$

Inverting $S$ and using Proposition 10.9.12 we get an exact sequence

$0 \to S^{-1}\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \bigoplus \nolimits _{i = 1, \ldots , n} S^{-1}N \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} S^{-1}N$

and the result follows since $S^{-1}M$ sits in an exact sequence

$\bigoplus \nolimits _{j = 1, \ldots , m} S^{-1}R \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} S^{-1}R \to S^{-1}M \to 0$

which induces (by Lemma 10.10.1) the exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _{S^{-1}R}(S^{-1}M, S^{-1}N) \to \bigoplus \nolimits _{i = 1, \ldots , n} S^{-1}N \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} S^{-1}N$

which is the same as the one above. $\square$

There are also:

• 3 comment(s) on Section 10.10: Internal Hom

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).