A Functor fom Grp to Pos and its application to Schur's Lemma in module theory.

Definition (Simple module). A non-zero $R$$R$-module $L$$L$ is simple if $0$$0$ and $L$$L$ are its only submodule.

From the lattice point of view, For any $R-$$R-$module $L$$L$, you could view $L$$L$ as lattice respect to join $L_1+L_2$$L_1+L_2$ and meet $L_1\cap L_2$$L_1\cap L_2$.

Which is $sup(L_1,L_2)$$\sup(L_1,L_2)$ and $inf(L_1,L_2)$$\inf(L_1,L_2)$ is the poset respect to $\subseteq$$\subseteq$.

$$\begin{array}{}\text{(1)}& f(r_1{l}_1+r_2{l}_2)=r_{1}f(l_1)+r_{2}f(l_2)⟹f(L_1+L_2)=f(L_1)+f(L_2)\end{array}$$

Usually, for a function $f:U⟶S$$f:U\longrightarrow S$, we do not have $f(U_1\cap U_2)=f(U_1)\cap f(U_2)$$f(U_1\cap U_2)=f(U_1)\cap f(U_2)$.

It is straightforward to see that $f(U_1\cap U_2)\subseteq f(U_1)\cap f(U_2)$$f(U_1\cap U_2)\subseteq f(U_1)\cap f(U_2)$.

Two explain why $f(U_1)\cap f(U_2)$$f(U_1)\cap f(U_2)$ is larger, consider two elements $u_1\in U_1,u_2\in U_2,u_1,u_2\in U_1\mathrm{\Delta }{U}_2$$u_1\in U_1,u_2\in U_2,u_1,u_2\in U_1\Delta U_2$.

Then maybe $f(u_1)=f(u_2)\ne f(u),\mathrm{\forall }u\in U_1\cap U_2$$f(u_1)=f(u_2)\ne f(u),\forall u\in U_1\cap U_2$.

Hence every $R-$$R-$ module homomorphism induces a join-lattice homomorphism, also a poset homomorphism.

Remark This also works for groups, and each poset homomorphism has the property that fixes the element $0$$0$.

A Functor from $\mathrm{Grp}$$\mathrm{Grp}$ to $\mathrm{P}\mathrm{o}{\mathrm{s}}_{\ast }$$\rm Pos_*$.

Indeed, we already have a functor $F:\mathrm{G}\mathrm{r}\mathrm{p}\to \mathrm{P}\mathrm{o}{\mathrm{s}}_{\ast }$$F:\rm Grp\to Pos_*$. Here $\mathrm{Grp}$$\mathrm{Grp}$ is the category of groups and ${\mathrm{Pos}}_{\ast }$$\mathrm{Pos}_*$ means the category of posets with the base point $0$$0$.

For the object part, $F:G⟼(\mathrm{Sub}(G),\subseteq )$$F:G\longmapsto (\mathrm{Sub}(G),\subseteq)$. For the morphism part, $F:f⟼F(f)$$F:f\longmapsto F(f)$ convert a group homomorphism to a poset homomorphism.

We will use this functor to prove the next proposition and Schur's Lemma.

Proposition. The following are equivalent.

$(\mathrm{i})$$\rm (i)$ $L$$L$ is a simple $R-$$R-$module

$(\mathrm{i}\mathrm{i})$$(\rm ii)$ $L=Rx$$L=Rx$ for any nonzero $x\in L$$x\in L$.

$(\mathrm{i}\mathrm{i}\mathrm{i})$$(\rm iii)$ $L\cong R/I$$L\cong R/I$ for a maximal ideal $I\subset R$$I\subset R$.

Proof.

Firstly, let $L$$L$ be a simple module, then $L$$L$ is simple if and only if $F(L)\cong P(\{\ast \})$$F(L)\cong P(\{*\})$.

$(\mathrm{i})⟹(\mathrm{i}\mathrm{i})$$\rm (i)\implies (ii)$. Let $x\in L$$x\in L$ be a non-zero element. Define a $R-$$R-$module homomorphism $\varphi :r⟼rx$$\phi:r\longmapsto rx$.

$\varphi (r_1+r_2)=(r_1+r_2)x=r_{1}x+r_{2}x=\varphi (r_1)+\varphi (r_2)$$\phi(r_1+r_2)=(r_1+r_2)x=r_1x+r_2x=\phi(r_1)+\phi(r_2)$, $\varphi (srx)=srx=s(rx)=s\varphi (rx)$$\phi(srx)=srx=s(rx)=s\phi(rx)$.

It is not hard to see that $\mathrm{Im}\varphi =Rx\subseteq L$$\mathrm{Im}\phi=Rx\subseteq L$.

We only need to prove that $F(\varphi )L=Rx\ne 0$$F(\phi)L=Rx\ne 0$, hence $F(\varphi )L=Rx$$F(\phi)L=Rx$ have to be $L$$L$. Since there are only two elements in $F(L)\cong P(\{\ast \})$$F(L)\cong P(\{*\})$ . But it is obvious, since $\varphi (1)=x\ne 0$$\phi(1)=x\ne 0$.

$(\mathrm{i}\mathrm{i})⟹(\mathrm{i})$$\rm (ii)\implies(i)$. $L=Rx$$L=Rx$ for any nonzero $x\in L$$x\in L$ means $F(L)\cong P(\{\ast \})$$F(L)\cong P(\{*\})$

$(\mathrm{i}\mathrm{i})⟹(\mathrm{i}\mathrm{i}\mathrm{i})$$\rm(ii)\implies(iii)$. Consider the annihilator of $L$$L$, we claim that $\mathrm{ann}(L)$$\mathrm{ann}(L)$ is a maximal ideal.

Since assume $\mathrm{ann}(L)\subset J\subset R$$\mathrm{ann}(L)\subset J\sub R$, then $\mathrm{\exists }r\in J,r\notin \mathrm{ann}(L)$$\exists r\in J,r\notin\mathrm{ann}(L)$. Then $R(rx)$$R(rx)$ will be a non-trivial proper submodule. Contradiction.

The homomorphism $\varphi (r)=rx$$\phi(r)=rx$ induce isomorphism $R/\mathrm{ann}(L)\cong Rx$${R}/\mathrm{ann}(L)\cong Rx$.

$(\mathrm{i}\mathrm{i}\mathrm{i})⟹(\mathrm{i})$$\rm(iii)\implies(i)$ Obviously. Since the definition of ideal is just sub $R-$$R-$module of $R$$R$.

Schur's Lemma. Let $L$$L$ be a simple module, then ${\mathrm{End}}_{\mathrm{R}-\mathrm{m}\mathrm{o}\mathrm{d}}(L)$$\mathrm{End}_{\rm R-mod}(L)$ is a division ring.

Proof. Use the functor $F$$F$, we see that $F(f):P(\{\ast \})\to P(\{\ast \})$$F(f):P(\{*\})\to P(\{*\})$. As a poset $P(\{\ast \})\cong F_2$$P(\{*\})\cong F_2$, i.e. $0\le 1$$0\le 1$.

And $F(f)$$F(f)$ have to fix the point $0$$0$. Hence there are only two choices. One is a zero map, another is identity.

Hence $f$$f$ have to be zero maps or isomorphism.