## 10.10 Internal Hom

If $R$ is a ring, and $M$, $N$ are $R$-modules, then

$\mathop{\mathrm{Hom}}\nolimits _ R(M, N) = \{ \varphi : M \to N\}$

is the set of $R$-linear maps from $M$ to $N$. This set comes with the structure of an abelian group by setting $(\varphi + \psi )(m) = \varphi (m) + \psi (m)$, as usual. In fact, $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ is also an $R$-module via the rule $(x \varphi )(m) = x \varphi (m) = \varphi (xm)$.

Given maps $a : M \to M'$ and $b : N \to N'$ of $R$-modules, we can pre-compose and post-compose homomorphisms by $a$ and $b$. This leads to the following commutative diagram

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ R(M', N) \ar[d]_{- \circ a} \ar[r]_{b \circ -} & \mathop{\mathrm{Hom}}\nolimits _ R(M', N') \ar[d]^{- \circ a} \\ \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \ar[r]^{b \circ -} & \mathop{\mathrm{Hom}}\nolimits _ R(M, N') }$

In fact, the maps in this diagram are $R$-module maps. Thus $\mathop{\mathrm{Hom}}\nolimits _ R$ defines an additive functor

$\text{Mod}_ R^{opp} \times \text{Mod}_ R \longrightarrow \text{Mod}_ R, \quad (M, N) \longmapsto \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$

Lemma 10.10.1. Exactness and $\mathop{\mathrm{Hom}}\nolimits _ R$. Let $R$ be a ring.

1. Let $M_1 \to M_2 \to M_3 \to 0$ be a complex of $R$-modules. Then $M_1 \to M_2 \to M_3 \to 0$ is exact if and only if $0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M_3, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_2, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_1, N)$ is exact for all $R$-modules $N$.

2. Let $0 \to M_1 \to M_2 \to M_3$ be a complex of $R$-modules. Then $0 \to M_1 \to M_2 \to M_3$ is exact if and only if $0 \to \mathop{\mathrm{Hom}}\nolimits _ R(N, M_1) \to \mathop{\mathrm{Hom}}\nolimits _ R(N, M_2) \to \mathop{\mathrm{Hom}}\nolimits _ R(N, M_3)$ is exact for all $R$-modules $N$.

Proof. Omitted. $\square$

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. $\square$

Comment #624 by Wei Xu on

Typo: should be

Comment #8365 by Hao Zhang on

In lemma 0583, the second equality in (2) should be the result of the lemma 07JY.

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).