Some Sparse Exercises

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

Some Sparse Exercises

#1

Post by Tsakanikas Nickos »

  1. Let \( \displaystyle R \) be a ring. Show that every non zero semisimple left \( \displaystyle R \)-module \( \displaystyle M \) contains a simple submodule.
  2. Let \( \displaystyle R \) be a ring. Show that if \( \displaystyle {}_RM \) is simple, then \( \displaystyle {}_RM \) is semisimple.Is the converse true?
  3. Let \( \displaystyle R \) be a ring. A left ideal \( \displaystyle I \) of \( \displaystyle R \) is minimal if and only if \( \displaystyle I \) is a simple left \( \displaystyle R \)-submodule of \( \displaystyle {}_RR \).
  4. If \( \displaystyle R_{1} , \dots , R_{\kappa} \) are left semisimple rings, then their direct product \( \displaystyle R = R_{1} \times \dots \times R_{\kappa} \) is a left semisimple ring.
  5. Let \( \displaystyle I=\left\{ 1,\dots,\kappa \right\} \) be a set of indices. Let \( \displaystyle D_{i} , \, i \in I \) be division rings, let \( \displaystyle n_{i} , \, i \in I \) be natural numbers and let \( \displaystyle \mathbb{M}_{n_{i}}(D_{i}) , \, i \in I \) be the rings of \( \displaystyle n_{i} \times n_{i} \) matrices over \( \displaystyle D_{i} \). For \( \displaystyle i \in I \) show that every simple \( \displaystyle \mathbb{M}_{n_{i}}(D_{i}) \)-module is simple \( \displaystyle R \)-module, where \[ \displaystyle R = \mathbb{M}_{n_{1}}(D_{1}) \times \dots \times \mathbb{M}_{n_{\kappa}}(D_{\kappa}) \]
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Some Sparse Exercises

#2

Post by Papapetros Vaggelis »

1. Let \(\displaystyle{_{R}\,M}\) be a non-zero semisimple left \(\displaystyle{R}\) - module.

Then, \(\displaystyle{M=\oplus_{i\in I}S_{i}}\), where \(\displaystyle{S_{i}\,,i\in I}\) are simple submodules of \(\displaystyle{_{R}\,M}\).

2. Let \(\displaystyle{_{R}\,M}\) be a simple left \(\displaystyle{R}\) - module. Then, \(\displaystyle{_{R}\,M\neq \left\{0\right\}}\)

and \(\displaystyle{M=\oplus_{i=1}^{1}M}\), which means that \(\displaystyle{_{R}\,M}\) is semisimple. The converse is not true :

Indedd, by setting \(\displaystyle{\left(R,+,\cdot\right)=\left(\mathbb{M}_{n}\,(D),+,\cdot\right)\,\,,\left(M,+\right)=\left(\mathbb{M}_{n}\,(D),+\right)}\)

where \(\displaystyle{\left(D,+,\cdot\right)}\) is a division ring,

we have that \(\displaystyle{_{R}\,M}\) is a semisimple left \(\displaystyle{R}\) - module ( left semisimple ring) cause :

\(\displaystyle{R=\mathbb{M}_{n}\,(D)=E_{11}\,R\oplus E_{22}\,R\oplus...\oplus E_{nn}\,R}\) where

\(\displaystyle{(E_{ii})_{(a,b)}=\begin{cases}
0_{D}\,\,,\left(a,b\right)\neq \left(i,i\right)\\
1_{D}\,\,,\left(a,b\right)=\left(i,i\right)\end{cases}}\)

and \(\displaystyle{E_{ii}\,R}\) are non-zero minimal ( simple submodules) of \(\displaystyle{\left(R,+,\cdot\right)}\) .

3. Suppose that \(\displaystyle{I}\) is a left minimal ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) . The subsets \(\displaystyle{\left\{0\right\}\,,I}\) of

\(\displaystyle{I}\) are left \(\displaystyle{R}\) - submodules of \(\displaystyle{_{R}\,I}\) . Let \(\displaystyle{J}\) be a left \(\displaystyle{R}\) - submodule of \(\displaystyle{_{R}\,I}\) .

Then, \(\displaystyle{J\leq \left(I,+\right)\leq \left(R,+\right)}\) and \(\displaystyle{r\cdot x\in J\,,\forall\,r\in R\,,\forall\,x\in J}\) , that is \(\displaystyle{J}\)

is a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) such that \(\displaystyle{\left\{0\right\}\subseteq J\subseteq I}\). Since the left ideal \(\displaystyle{I}\)

is minimal, we have that: \(\displaystyle{J=\left\{0\right\}}\) or \(\displaystyle{J=I}\).

On the other hand, suppose that \(\displaystyle{_{R}\,I}\) is simple. Consider left ideal \(\displaystyle{J}\) of \(\displaystyle{\left(R,+,\cdot\right)}\) such that

\(\displaystyle{\left\{0\right\}\subseteq J\subseteq I}\) . Then, \(\displaystyle{J}\) is \(\displaystyle{R}\) -submodule of \(\displaystyle{I_{R}}\) and since

\(\displaystyle{I_{R}}\) is simple, we get \(\displaystyle{J=\left\{0\right\}}\) or \(\displaystyle{J=I}\).

4. Let \(\displaystyle{\left(R_{1},+,\cdot\right)\,,\left(R_{2},+,\cdot\right)}\) be left semisimple rings and \(\displaystyle{R=R_{1}\times R_{2}}\) .

Suppose that \(\displaystyle{_{R}\,S}\) is a left simple \(\displaystyle{R_{1}}\) - module. By setting, \(\displaystyle{S^{\star}=S\times \left\{0\right\}}\), we have

that \(\displaystyle{S^{\star}}\) is a left simple \(\displaystyle{R}\) - module cause if \(\displaystyle{M}\) is a submodule of \(\displaystyle{_{R}\,S^{\star}}\), then

\(\displaystyle{M=N\times \left\{0\right\}}\), where \(\displaystyle{N}\) is a submodule of \(\displaystyle{_{R}\,S}\) and thus:
\(\displaystyle{N=\left\{0\right\}}\) or \(\displaystyle{N=S}\), so : \(\displaystyle{M=\left\{\left(0,0\right)\right\}}\) or \(\displaystyle{M=S\times \left\{0\right\}=S^{\star}}\) .

Similarly, every simple \(\displaystyle{R_{2}}\) - module defines a simple \(\displaystyle{R}\) - module \(\displaystyle{T^{\star}=\left\{0\right\}\times T}\) .

Now, since \(\displaystyle{ \left(R_{1},+,\cdot\right)\,,\left(R_{2},+,\cdot\right)}\) are left semisimple rings, then

\(\displaystyle{R_{1}=\oplus_{i=1}^{n}S_{i}\,,R_{2}=\oplus_{j=1}^{m}T_{j}}\) (the rings have unity), where \(\displaystyle{S_{i}\,,T_{j}\,,1\leq i\leq n\,\,,1\leq j\leq m}\)

are simple \(\displaystyle{R_{1}\,,R_{2}}\) - modules, respectively, and then:

\(\displaystyle{R= S_{1}^{\star}\oplus...\oplus S_{n}^{\star}\oplus T_{1}^{\star}\oplus...\oplus T_{m}}\), and \(\displaystyle{S_{i}^{\star}\,,T_{j}^{\star}}\)

are left simple \(\displaystyle{R}\) - modules of \(\displaystyle{_{R}\,R}\), so : \(\displaystyle{_{R}\,R}\) is a semisimple module.

Then, by induction we have the desired result.

5. Suppose that \(\displaystyle{\left(M,+,\cdot\right)}\) is a simple \( \displaystyle \mathbb{M}_{n_{i}} (D_{i}) \) - module, where \( i\in\left\{1,...,k\right\} \) .

The abelian group \(\displaystyle{\left(M,+\right)}\) equipped with the left scalar multiplication

\(\displaystyle{\star:R\times M\longrightarrow M\,,\left(\left(A_{1},...,A_{k}\right),x\right)\mapsto \left(A_{1},...,A_{k}\right)\star x=A_{i}\cdot x}\)

which is well defined, becomes a left \(\displaystyle{R}\) - module because :

\(\displaystyle{\left(A_{1},...,A_{k}\right)\star \left(x+y\right)=A_{i}\cdot (x+y)=A_{i}\cdot x+A_{i}\cdot y=\left(A_{1},...,A_{k}\right)\star x+\left(A_{1},...,A_{k}\right)\star y}\)

\(\displaystyle{\begin{aligned}\left[\left(A_{1},...,A_{k}\right)+\left(B_{1},...,B_{k}\right)\right]\star x&=\left(A_{1}+B_{1},...,A_{k}+B_{k}\right)\star x\\&=\left(A_{i}+B_{i}\right)\cdot x\\&=A_{i}\cdot x+B_{i}\cdot x\\&=\left(A_{1},...,A_{k}\right)\star x+\left(B_{1},...,B_{k}\right)\star x\end{aligned}}\)

\(\displaystyle{\begin{aligned} \left[\left(A_{1},...,A_{k}\right)\cdot \left(B_{1},...,B_{k}\right)\right]\star x&=\left(A_{1}\,B_{1},...,A_{k}\,B_{k}\right)\star x\\&=\left(A_{i}\,B_{i}\right)\cdot x=A_{i}\cdot (B_{i}\cdot x)\\&=\left(A_{1},...,A_{k}\right)\star \left[\left(B_{1},...,B_{k}\right)\star x\right]\end{aligned}}\)

\(\displaystyle{1_{R}\star x=\left(I_{n_{1}},...,I_{n_{k}}\right)\star x=I_{n_{i}}\cdot x=1_{\mathbb{M}_{n_{i}}\,(D_{i})}\cdot x=x}\)

The subsets \(\displaystyle{\left\{0\right\}\,,M}\) are left \(\displaystyle{R}\) - submodules of \(\displaystyle{_{R}\,M}\) .

Let \(\displaystyle{N}\) be a left \(\displaystyle{R}\) submodule of \(\displaystyle{_{R}\,M}\), that is :

\(\displaystyle{N\leq \left(M,+\right)}\) and \(\displaystyle{\forall\,r\in R\,,\forall\,x\in N: r\star x\in N}\) .

Let \(\displaystyle{A\in\mathbb{M}_{n_{i}}\,(D_{i})}\) and \(\displaystyle{x\in N}\). Then,

\(\displaystyle{\left(0,...,A,...,0\right)\in R\,,x\in N}\) and thus:

\(\displaystyle{\left(0,...,A,...,0\right)\star x\in N\implies A\cdot x\in N}\) and we conclude that \(\displaystyle{N}\) is a left \(\displaystyle{R}\) - submodule

of \(\displaystyle{_{\mathbb{M}_{n_{i}}\,(D_{i})}\,M}\) and since this module is simple we have that \(\displaystyle{N=\left\{0\right\}}\) or \(\displaystyle{N=M}\) .

I am not so sure about the last question.
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