
## 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

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