Schur's Lemma - Converse

[Papapetros Vaggelis]

1. If \(\displaystyle{_{R}M}\) is a simple left \(\displaystyle{R}\) - module, then show that the

ring \(\displaystyle{\left(\rm{End}_{R}(M),+,\circ\right)}\) is a division ring.

2. Is the converse true ?

[Tsakanikas Nickos]

(1) Let \( f \in End_{R}(M) \). We know that \( Ker(f) \) and \( Im(f) \) are submodules of \( _{R}M \). Therefore, if \( f \neq 0 \), then \( Im(f) \neq 0 \), and since \( M \) is simple, \( Im(f) = M \). Also, \( Ker(f) \neq M \), and since \( M \) is simple, \( Ker(f) = 0 \). This, of course, means that \( f \) is an isomorphism. It follows that \( End_{R}(M) \) is a division ring, since every non-zero element of it is invertible, i.e. an isomorphism.

[Papapetros Vaggelis]

2.

Consider the abelian group \(\displaystyle{\left(\mathbb{Q},+\right)}\) as \(\displaystyle{\mathbb{Z}}\) - module.

This \(\displaystyle{\mathbb{Z}}\) - module is not simple since \(\displaystyle{\mathbb{Z}}\) is a

non trivial and proper \(\displaystyle{\mathbb{Z}}\) - submodule.

Let \(\displaystyle{f\in\rm{End}_{\mathbb{Z}}(Q)}\). Then,

\(\displaystyle{\forall\,x\in\mathbb{Z}: f(x)=f(x\cdot 1)=x\,f(1)\,\,(\star)}\).

If \(\displaystyle{x=\dfrac{a}{b}\in\mathbb{Q}-\mathbb{Z}\,,a\in\mathbb{Z}\,,b\in\mathbb{N}}\), then

\(\displaystyle{b\,f(x)=f(b\,x)=f(a)\stackrel{(\ast)}{=}a\,f(1)\implies f(x)=\dfrac{a}{b}\,f(1)=x\,f(1)}\) .

So, \(\displaystyle{f(x)=x\,f(1)\,,\forall\,x\in\mathbb{Q}}\), where \(\displaystyle{f(1)\in\mathbb{Q}}\) .

On the other hand, if \(\displaystyle{q\in\mathbb{Q}}\), then the map

\(\displaystyle{f_{q}:\mathbb{Q}\to \mathbb{Q}\,,f_{q}(x)=x\,q}\) is \(\displaystyle{\mathbb{Z}}\) -linear.

Indeed,

\(\displaystyle{f(x+m\,y)=(x+m\,y)\,q=x\,q+m\,y\,q=x\,q+m(y\,q)=f(x)+m\,f(y)\,,\forall\,x\,,y\in\mathbb{Q}\,,\forall\,m\in\mathbb{Z}}\) .

So, \(\displaystyle{\rm{End}_{\mathbb{Z}}(\mathbb{Q})=\left\{f_{q}: q\in\mathbb{Q}\right\}}\) .

Consider now the map

\(\displaystyle{\Phi:\mathbb{Q}\to \rm{End}_{\mathbb{Z}}(\mathbb{Q})\,\,,q\mapsto \Phi(q)=f_{q}}\)

which is onto \(\displaystyle{\rm{End}_{\mathbb{Z}}(\mathbb{Q})}\) .

Obviously, \(\displaystyle{\Phi(1_{\mathbb{Q}})=f_{1}=Id_{\mathbb{Q}}=1_{\rm{End}_{\mathbb{Z}}(\mathbb{Q})}}\) .

Also, for every \(\displaystyle{q\,,p\in\mathbb{Q}}\) and each \(\displaystyle{x\in\mathbb{Q}}\) holds :

\(\displaystyle{\begin{aligned} \Phi(p+q)(x)&=x\,(p+q)\\&=x\,p+x\,q=\Phi(p)(x)+\Phi(q)(x)\\&=(Phi(p)+\Phi(q))(x)\end{aligned}}\)

\(\displaystyle{\begin{aligned} \Phi(p\,q)(x)&=x\,(p\,q)\\&=(x\,q)\,p\\&=\Phi(p)(x\,q)\\&=\Phi(p)(\Phi(q)(x))\\&=\left(\Phi(p)\circ \Phi(q)\right)(x)\end{aligned}}\) .

that is, \(\displaystyle{\Phi(p+q)=\Phi(p)+\Phi(q)\,\,\,,\Phi(p\,q)=\Phi(p)\circ \Phi(q)}\) and

\(\displaystyle{\Phi}\) is a ring epimorphism. Finally,

\(\displaystyle{\begin{aligned} \rm{Ker}(\Phi)&=\left\{q\in\mathbb{Q}: \Phi(q)=\mathbb{O}\right\}\\&=\left\{q\in\mathbb{Q}: x\,q=0\,,\forall\,x\in\mathbb{Q}\right\}\\&=\left\{0\right\}\end{aligned}}\) .

So, \(\displaystyle{\left(\rm{End}_{\mathbb{Z}}(\mathbb{Q}),+,\circ\right)\simeq \left(\mathbb{Q},+,\cdot\right)}\)

and \(\displaystyle{\left(\mathbb{Q},+,\cdot\right)}\) is a division ring.

To be continued...

[Papapetros Vaggelis]

Here is another example :

Let \(\displaystyle{\mathbb{K}}\) be a field. Consider the ring \(\displaystyle{S=T_{n}(\mathbb{K})\,,n\geq 2}\)

of all the matrices \(\displaystyle{A\in\mathbb{M}_{n}(\mathbb{K})}\) with te property \(\displaystyle{a_{i\,j}=0\,,i>j}\) .

As you can easily check, the abelian group \(\displaystyle{\left(V=\mathbb{M}_{n\times 1}(\mathbb{K}),+\right)}\)

is a left \(\displaystyle{S}\) - module with the usual multiplication.

Then, \(\displaystyle{V}\) is not a simple \(\displaystyle{S}\) - module, but

\(\displaystyle{\rm{End}_{S}(V)\simeq \mathbb{K}}\) .