## Tag `08YQ`

Chapter 10: Commutative Algebra > Section 10.13: Base change

Lemma 10.13.5. Let $R \to S$ be a ring map. Given $S$-modules $M, N$ and an $R$-module $P$ we have $$ \mathop{\rm Hom}\nolimits_R(M \otimes_S N, P) = \mathop{\rm Hom}\nolimits_S(M, \mathop{\rm Hom}\nolimits_R(N, P)) $$

Proof.This can be proved directly, but it is also a consequence of Lemmas 10.13.4 and 10.11.8. Namely, we have \begin{align*} \mathop{\rm Hom}\nolimits_R(M \otimes_S N, P) & = \mathop{\rm Hom}\nolimits_S(M \otimes_S N, \mathop{\rm Hom}\nolimits_R(S, P)) \\ & = \mathop{\rm Hom}\nolimits_S(M, \mathop{\rm Hom}\nolimits_S(N, \mathop{\rm Hom}\nolimits_R(S, P))) \\ & = \mathop{\rm Hom}\nolimits_S(M, \mathop{\rm Hom}\nolimits_R(N, P)) \end{align*} as desired. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 2412–2419 (see updates for more information).

```
\begin{lemma}
\label{lemma-hom-from-tensor-product-variant}
Let $R \to S$ be a ring map. Given $S$-modules $M, N$ and an $R$-module $P$
we have
$$
\Hom_R(M \otimes_S N, P) = \Hom_S(M, \Hom_R(N, P))
$$
\end{lemma}
\begin{proof}
This can be proved directly, but it is also a consequence of
Lemmas \ref{lemma-adjoint-hom-restrict} and \ref{lemma-hom-from-tensor-product}.
Namely, we have
\begin{align*}
\Hom_R(M \otimes_S N, P)
& =
\Hom_S(M \otimes_S N, \Hom_R(S, P)) \\
& =
\Hom_S(M, \Hom_S(N, \Hom_R(S, P))) \\
& =
\Hom_S(M, \Hom_R(N, P))
\end{align*}
as desired.
\end{proof}
```

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