On Simple Modules

Groups, Rings, Domains, Modules, etc, Galois theory
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Tsakanikas Nickos
Community Team
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

On Simple Modules

#1

Post by Tsakanikas Nickos »

  • Let \( \displaystyle R \) be a ring and let \( \displaystyle M \) be a left \( \displaystyle R \)-module. Show that if \( \displaystyle M \) is simple, then \( \displaystyle M \) is cyclic. Is the converse true?
  • Let \( \displaystyle R \) be a ring and let \( \displaystyle M \) be a simple left \( \displaystyle R \)-module. Show that the ring of endomorphisms of \( \displaystyle M , \, End_{R}(M) , \, \) is division ring. Is the converse true?
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: On Simple Modules

#2

Post by Papapetros Vaggelis »

1st part

Suppose that \(\displaystyle{\left(M,+\right)}\) is a simple left \(\displaystyle{R}\) - module.
Then, \(\displaystyle{M\neq \left\{0\right\}}\) and the only submodules of \(\displaystyle{M}\) are \(\displaystyle{\left\{0\right\}\,\,,M}\) .
There is \(\displaystyle{x\in M}\) such that \(\displaystyle{x\neq 0}\) . Consider the submodule \(\displaystyle{N=R\,x}\) of \(\displaystyle{_{R}M}\), produced by \(\displaystyle{x}\) . We have that \(\displaystyle{N\neq \left\{0\right\}}\) because \(\displaystyle{x\in N-\left\{0\right\}}\) and
according to the hypothesis we get \(\displaystyle{R\,x=M}\) , which means that (\displaystyle{M}\) is cyclic.

The converse is not true : Indeed, setting \(\displaystyle{\left(R,+,\cdot\right)=\left(\mathbb{Z},+,\cdot\right)\,\,,\left(M,+\right)=\left(\mathbb{Z},+\right)}\) we have that \(\displaystyle{M=\mathbb{Z}=(1)}\) is cyclic but not simple cause \(\displaystyle{2\,\mathbb{Z}}\) is a proper submodule of \(\displaystyle{\mathbb{Z}}\) .

2nd part

The set

\(\displaystyle{\rm{End_{R}\,(M)}=\left\{f:M\longrightarrow M : f(x+y)=f(x)+f(y)\,,\forall\,x\,,y\in M\,\,,f(r\cdot x)=r\cdot f(x)\,,\forall\,r\in R\,,\forall\,x\in M\right\}}\)

is eqquiped with the usual operation of addition

\(\displaystyle{+:\rm{End_{R}\,(M)}\times \rm{End_{R}\,(M)}\longrightarrow \rm{End_{R}\,(M)}\,\,,(f,g)\mapsto f+g: M\longrightarrow M\,,(f+g)(x)=f(x)+g(x)}\)

and the composition as its multiplication. The zero element is the zero-function \(\displaystyle{\mathbb{O}:M\longrightarrow M\,,x\mapsto 0_{M}}\) and \(\displaystyle{I_{M}:M\longrightarrow M\,,x\mapsto x}\) is the unity of this ring.

Let \(\displaystyle{f\in \rm{End_{R}\,(M)}-\left\{\mathbb{O}\right\}}\) . The subset \(\displaystyle{\rm{Ker}(f)=\left\{x\in M: f(x)=0_{M}\right\}}\)
of \(\displaystyle{M}\) is a proper submodule of \(\displaystyle{_{R}M}\) and since \(\displaystyle{_{R}M}\) is simple we have that :

\(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\) or \(\displaystyle{\rm{Ker}(f)=M}\) .

If \(\displaystyle{M=\rm{Ker}(f)}\) , then \(\displaystyle{f(x)=0_{M}\,,\forall\,x\in M\implies f=\mathbb{O}}\), a contradiction.

So, \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\) and thus the function \(\displaystyle{f}\) is \(\displaystyle{1-1}\) .

Following the same strategy we also get \(\displaystyle{f\,(M)=M}\). Then, there exists as a function \(\displaystyle{f^{-1}:M\longrightarrow M}\) and \(\displaystyle{f\circ f^{-1}=I_{M}=f^{-1}\circ f}\), which means that \(\displaystyle{f}\) is invertible.
Therefore, the ring \(\displaystyle{\left(\rm{End_{R}\,(M)},+,\circ\right)}\) is a divsion ring.

The converse is not true :

\(\displaystyle{\left(R,+,\cdot\right)=\left(\mathbb{Z},+,\cdot\right)\,\,,M=\mathbb{Q}}\).

The left \(\displaystyle{\mathbb{Z}}\) - module \(\displaystyle{\left(\mathbb{Q},+\right)}\) is not simple because it contains the proper submodule \(\displaystyle{\mathbb{Z}}\) .

Let \(\displaystyle{f\in \rm{End_{\mathbb{Z}}\,(\mathbb{Q})}}\) . It's known that

\(\displaystyle{f(q_1+q_2)=f(q_1)+f(q_2)\,\,,f(n\,q)=n\,f(q)\,,q_1\,,q_2\,,q\in\mathbb{Q}\,\,,n\in\mathbb{Z}}\) .

For each \(\displaystyle{n\in\mathbb{Z}}\) holds: \(\displaystyle{f(n)=f(n\cdot 1)=n\,f(1)\,,(I)}\) .

Let \(\displaystyle{q=\dfrac{a}{b}\in\mathbb{Q}-\left\{0\right\}}\) (obviously, \(\displaystyle{f(0)=0}\)) . Then

\(\displaystyle{a\in\mathbb{Z}-\left\{0\right\}\,,b\in\mathbb{N}}\) and

\(\displaystyle{b\,q=a\implies f\,(b\,q)=f(a)\stackrel{(I)}{\implies} b\,f(q)=a\,f(1)\implies f(q)=q\,f(1)}\)

and thus : \(\displaystyle{f(q)=q\,f(1)\,,q\in\mathbb{Q}}\) . On the other hand, if

\(\displaystyle{f_{c}:\mathbb{Q}\longrightarrow \mathbb{Q}\,,f(q)=c\,q}\)

for some \(\displaystyle{c\in\mathbb{Q}}\), then \(\displaystyle{f_{c}\in \rm{End_{\mathbb{Z}}\,(\mathbb{Q})}}\), so :

\(\displaystyle{\rm{End_{\mathbb{Z}}\,(\mathbb{Q})}=\left\{f_{c}:\mathbb{Q}\longrightarrow \mathbb{Q}: c\in\mathbb{Q}\right\}}\) .

We define \(\displaystyle{g:\mathbb{Q}\longrightarrow \rm{End_{\mathbb{Z}}\,(\mathbb{Q})}}\) by \(\displaystyle{g(c)=f_{c}}\) .

If \(\displaystyle{c_1\,,c_2\in\mathbb{Q}}\), then for every \(\displaystyle{q\in\mathbb{Q}}\) we have :

\(\displaystyle{\begin{aligned}(g(c_1+c_2))(q)&=f_{c_1+c_2}(q)\\&=\left(c_1+c_2\right)\,q\\&=c_1\,q+c_2\,q\\&=f_{c_1}(q)+f_{c_2}(q)\\&=\left(f_{c_1}+f_{c_2}\right)(q)\\&=\left(g(c_1)+g(c_2)\right)(q)\end{aligned}}\)

and

\(\displaystyle{\begin{aligned}(g(c_1\cdot c_2))(q)&=f_{c_1\cdot c_2}(q)\\&=\left(c_1\cdot c_2\right)\,q\\&=c_1\,\left(c_2\,q\right)\\&=c_1\,f_{c_2}(q)\\&=f_{c_1}\,(f_{c_2}(q))\\&=\left(f_{c_1}\circ f_{c_2}\right)(q)\\&=\left(g(c_1)\circ g(c_2)\right)(q)\end{aligned}}\)

Also, \(\displaystyle{g(1)=f_{1}=Id_{\mathbb{Q}}=1_{\rm{End_{\mathbb{Z}}\,(\mathbb{Q})}}}\)

and we deduce that

\(\displaystyle{g(c_1+c_2)=g(c_1)+g(c_2)\,\,,g(c_1\cdot c_2)=g(c_1)\circ g(c_2)}\) .

So, the function \(\displaystyle{g}\) is homomorphism. Now, if \(\displaystyle{f\in\rm{End_{\mathbb{Z}}\,(\mathbb{Q})}}\) , then

\(\displaystyle{f=f_{c}}\) for some \(\displaystyle{c\in\mathbb{Q}}\) and \(\displaystyle{g(c)=f_{c}=f}\), so the function \(\displaystyle{g}\) is onto.

Finally, if \(\displaystyle{c_1\,,c_2\in\mathbb{Q}}\) we have that :

\(\displaystyle{\begin{aligned} g(c_1)=g(c_2)&\implies f_{c_1}=f_{c_2}\\&\implies \forall\,q\in\mathbb{Q}: c_1\,q=c_2\,q\\&\implies c_1\cdot 1=c_2\cdot 1\\&\implies c_1=c_2\end{aligned}}\)

and \(\displaystyle{g}\) is 1-1 . Therefore,

\(\displaystyle{\left(\mathbb{Q},+,\cdot\right)\simeq \left(\rm{End_{\mathbb{Z}}\,(\mathbb{Q})},+,\circ\right)}\) and since

\(\displaystyle{\left(\mathbb{Q},+,\cdot\right)}\) is division ring, so is \(\displaystyle{\left(\rm{End_{\mathbb{Z}}\,(\mathbb{Q})},+,\circ\right)}\) .
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