----- > [!proposition] Proposition. ([[homsets in R-mod are R-modules]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]] and $M,N$ be $R$-[[module|modules]]. The homset $\text{Hom}_{R\text{-}\mathsf{Mod}}(M,N)$ is itself an $R$-[[module]]: since [[linear map|module homomorphisms]] are in particular [[group homomorphism|abelian group homomorphisms]], $\text{Hom}_{R\text{-}\mathsf{Mod}}(M,N) \subset \text{Hom}_{\mathsf{Ab}}(M,N),$ and so [[abelian group|abelian group]] structure comes from [[homsets in ab are abelian groups]]. The [[module|ring action]] is$\begin{align} R\times \text{Hom}_{R\text{-}\mathsf{Mod}}(M,N) \to& \text{Hom}_{R\text{-}\mathsf{Mod}}(M,N)\\ (r, \varphi) \mapsto& r\varphi, \text{ where } r\varphi: M \to N \text{ is defined} \\ (r\varphi)(m) :=& r \ \varphi(m) \text{ for all } m \in M. \end{align}$ > ^proposition > [!specialization] > - In the special case where $R$ is viewed as an $R$-[[module]] over itself [^1] we in fact have have $\text{Hom}_{R\text{-}\mathsf{Mod}}(R,M) \cong M$ as $R$-[[module|modules]]. > - In the special case where $R=k$ is a [[field]] (so that $M,N$ are $k$-[[vector space|vector spaces]]), this is precisely the content of [[vector space of linear maps between two vector spaces]]. ^specialization [^1]: Put $S=R, \alpha=\id$ in [[module induced by a ring homomorphism|here]]. > [!note] Remark. > Compare to [[homsets in ab are abelian groups]]. This is one of many similarities between the [[category|category]] $R$-$\mathsf{Mod}$ and the [[category]] $\mathsf{Ab}$ of [[abelian group|abelian groups]]. > [!proof] Proof of Specialization. > Try the function $m \mapsto \varphi_{m}$, where the latter is given by $\varphi_{m}: R \to M$, $\varphi_{m}(r):=rm$? > > Take any element $\varphi \in \text{Hom}_{R\text{-}\mathsf{Mod}}(R,M)$. Put $m:=\varphi(1_{R})$. Then for any $r \in R$, > $\varphi(r)=\varphi(r 1_{R})=r \varphi(1_{R})=rm,$ > in other words, any [[linear map]] $\varphi: R \to M$ is determined by where it sends $m=\varphi(1_{R})$. We thereby obtain a [[bijection|bijective]] correspondence $m \xleftrightarrow{} \varphi_{m}$, where $\varphi_{m}: R \to M$ is given by $\varphi_{m}(r):=rm$. The claim is that $m \xmapsto{ \Psi}\varphi_{m}$ is a [[module isomorphism]]. All we need to do is show it is a [[linear map]], and indeed it is, for (and this is where we really do enjoy [[commutative ring|commutativity]]!) $\begin{align} > \Psi(rm)(s)= & \varphi_{rm}(s) \\ > = & s(rm) \\ > \stackrel{!}{=} & (rm)(s) \\ > = & r(ms) \\ > = & r\big(\varphi_{m}(s)\big) \\ > = & (r \varphi_{m})(s) \\ > = & \big(r \Psi (m)\big)(s), > \end{align}$ > meaning that $\Psi(rm)=r\Psi(m)$, and also $\begin{align} > \Psi(m+n)(s)= & \varphi_{m+n}(s) \\ > = & s(m+n) \\ > = & sm + sn \\ > = & \varphi_{m}(s) + \varphi_{n}(s) \\ > = & \Psi(m)(s) + \Psi(n)(s) > \end{align}$ > meaning that $\Psi(m+n)=\Psi(m)+\Psi(n)$. ^proof-1 > [!proof]- Proof. ([[homsets in R-mod are R-modules]]) > First it must be shown that $(r\varphi)$ as defined is indeed a [[linear map]] (this is where we need to use that $R$ is [[commutative ring|commutative]]). We have homogeneity: $\begin{align} > (r \varphi)(am) = & r \varphi(am)= (ra) \varphi(m)= (ar)\varphi(m)=a\big( r \varphi(m) \big) > \end{align}$and additivity: $(r \varphi)(m+n)=r \ \varphi(m+n)=r( \varphi(m)+\varphi(n))=r \varphi(m)+r\varphi(n)=(r\varphi)(m)+(r\varphi)(n),$as required. > > Now it must be checked that the map $(r, \varphi) \mapsto r\varphi$ is indeed an action. This is easy. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```