# The Stacks Project

## Tag 0583

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 10.9.7. 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. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 1549–1562 (see updates for more information).

\begin{lemma}
\label{lemma-hom-from-finitely-presented}
Let $R$ be a ring. Let $M$ be a finitely presented $R$-module.
Let $N$ be an $R$-module.
\begin{enumerate}
\item For $f \in R$ we have
$\Hom_R(M, N)_f = \Hom_{R_f}(M_f, N_f) = \Hom_R(M_f, N_f)$,
\item for a multiplicative subset $S$ of $R$ we have
$$S^{-1}\Hom_R(M, N) = \Hom_{S^{-1}R}(S^{-1}M, S^{-1}N) = \Hom_R(S^{-1}M, S^{-1}N).$$
\end{enumerate}
\end{lemma}

\begin{proof}
Part (1) is a special case of part (2).
The second equality in (2) follows from
Lemma \ref{lemma-universal-property-localization-module}.
Choose a presentation
$$\bigoplus\nolimits_{j = 1, \ldots, m} R \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} R \to M \to 0.$$
By
Lemma \ref{lemma-hom-exact}
this gives an exact sequence
$$0 \to \Hom_R(M, N) \to \bigoplus\nolimits_{i = 1, \ldots, n} N \longrightarrow \bigoplus\nolimits_{j = 1, \ldots, m} N.$$
Inverting $S$ and using Proposition \ref{proposition-localization-exact}
we get an exact sequence
$$0 \to S^{-1}\Hom_R(M, N) \to \bigoplus\nolimits_{i = 1, \ldots, n} S^{-1}N \longrightarrow \bigoplus\nolimits_{j = 1, \ldots, m} S^{-1}N$$
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 \ref{lemma-hom-exact})
the exact sequence
$$0 \to \Hom_{S^{-1}R}(S^{-1}M, S^{-1}N) \to \bigoplus\nolimits_{i = 1, \ldots, n} S^{-1}N \longrightarrow \bigoplus\nolimits_{j = 1, \ldots, m} S^{-1}N$$
which is the same as the one above.
\end{proof}

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