---- > [!definition] Definition. ([[module]]) > > Let $R$ be a [[ring]]. [[endsets in ab are rings|Recall that]] if $M$ is an [[abelian group]], then $\text{End}_{\mathsf{Ab}}(M)$ is a [[ring]] in a natural way. A **left-action of $R$ on $M$** is a [[ring homomorphism]] $\sigma: R \to \text{End}_{\mathsf{Ab}}(M).$ We say **$\sigma$ makes $M$ into a left-$R$-module**. Explicitly: > > > A **left-$R$-module** structure on an [[abelian group]] $M$ consists of a map $R\times M \to M$, $(r,m)\mapsto r m$, such that > - $r(m+n)=rm+rn$; > - $(r+s)m=rm + sm$; > - $(rs)m=r(sm)$; > - $1m=m$. > > **Right-$R$-modules** are defined analogously. If $R$ is [[commutative ring|commutative]] then the left-right distinction is immaterial. > > (Left-)$R$-modules are objects of [[category]] $R$-$\mathsf{Mod}$. The homset $\text{Hom}(M,N)$ is the set of [[linear map|module homomorphisms]] from $M$ to $N$. [[homsets in R-mod are R-modules|It is itself]] an $R$-module. ^definition > [!basicproperties] The Category $R$-$\mathsf{Mod}$. > - The [[group|trivial group]] $(0)$ has a unique [[module]] structure over any [[ring]] $R$ — as in $\mathsf{Grp}$, it is a [[terminal object|zero-object]] in $R$-$\mathsf{Mod}$. > - ^properties > [!justification]- > Just as we needed to relate [[group action]] to [[homomorphism induced by group action]], we need to relate what appear to be two different definitions provided above. > For now, call by $\rho$ the function $R \times M \to M$ given by $(r,m) \mapsto rm$. > The aforementioned relation, as usual, comes from [[currying]]: $\sigma(r)$ corresponds to the curried function $\rho_{r} \in \text{End}_{\mathsf{Ab}}(M)$ given by $\rho_{r}(m)=\sigma(r)(m)=rm$. Now we show that $\sigma$ satisfies these properties: $r(m+n)=\rho_{r}(m+n)=\sigma(r)(m+n)=\sigma(r)(m) + \sigma(r)(n)=rm+rn$ > where we used that $\sigma(r)$ is a [[group homomorphism]]. Similarly, $\begin{align} > (r+s)m = & \rho_{r+s}(m) \\ > = & \sigma(r+s)(m) \\ > = & \sigma(r)(m)+\sigma(s)(m) \\ > = & rm + sm, > \end{align}$ > where this time we used that *$\sigma$* is a [[ring homomorphism]]. Finally, $\begin{align} > (rs)m = & \rho_{rs}(m) \\ > = & \sigma(rs)(m) \\ > = & \sigma (r) \circ \sigma(s) (m) \\ > = & \sigma(r)(sm) \\ > = & \rho_{r}(sm) \\ > = & r(sm) \\ > \end{align}$ > where multiplicativity of $\sigma$ was used, and $\begin{align} > 1m = & \rho_{1}(m) \\ > = & \sigma(1)(m) \\ > = & \id(m) \\ > = & m > \end{align}$ > where it was that $\sigma$ preserves identity. Next, as a sort of converse, we show that a map $R \times M \to M$ satisfying those four properties induces a [[ring homomorphism]] $\sigma: R \to \text{End}_{\mathsf{Ab}}(M)$. We of course [[currying|curry]] said map into maps $\rho_{r}: M \to M$, $\rho_{r}(m)=rm$, and define $\sigma$ as $\sigma(r)(m)=\rho_{r}(m)$. It just needs to be shown $\sigma$ is a [[ring homomorphism]]. $\begin{align} > \sigma(rs)(m)= & \rho_{rs}(m) \\ > = & (rs)m \\ > = & r(sm) \\ > = & \rho_r \circ \rho_{s}(m) \\ > = & \sigma(r) \circ \sigma(s) (m) > \end{align}$ > and $\begin{align} > \sigma(r + s)(m) = & \rho_{r+s}(m) \\ > = & \rho_{r}(m) + \rho_{s}(m) \\ > = & \sigma(r)(m) + \sigma(s)(m), > \end{align}$ > and $\sigma(1)(m)=\rho_{1}(m)=1m = m,$ > > hence $\sigma$ respects multiplication, addition, and identity— as required. ^justification > [!basicproperties] > - For all $m \in M$, $0 \cdot m=0$; > - For all $m \in M$, $(-1) \cdot m=-m$; ^properties > [!basicexample] > - [[every abelian group is a Z-module, in exactly one way]]. Thus, "[[abelian group]]" and "$\mathbb{Z}$-module" are one and the same notion. > - Every [[ring]] $R$ can be viewed as a (left-,right-) module over itself, as the special case of [[module induced by a ring homomorphism]] wherein $R=S$ and $\alpha=\id$ so that the 'ring action' of $R$ on its underlying [[abelian group]] is just ring multiplication. Note that $R$ is [[submodule generated by a subset|generated by]] $\{ 1 \}$ as a [[module]] over itself. ^basic-example > [!proof]- Proof of Basic Properties > **1.** If $m \in M$, then $0 \cdot m=(0+0)\cdot m=0 \cdot m + 0 \cdot m$ and the [[cancellation law for groups]] concludes $0=m \cdot 0$. > **2.** If $m \in M$, then $(-1) \cdot m + m=(-1) \cdot m + (1) \cdot m$, action definition says this equals $(-1 + 1) \cdot m=0 \cdot m = 0$. Hence $(-1) \cdot m$ is the inverse of $m$ in $M$. ^proof > [!intuition] > The motivation for defining (this kind of) ring action in this way should be clear in analog to the definition of [[group action]]; the difference being that because ring elements need not have inverse (in this case, such is saying that [[group homomorphism|group homomorphisms]] need not have inverses) the best we can do is consider endomorphisms rather than [[automorphism|automorphisms]]. ---- #### ---- #### 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 > ```