Semisimple Ring - Equivalent Conditions

[Tsakanikas Nickos]

Let \( \displaystyle R \) be an associative ring with unity \( \displaystyle 1_{R} \). Show that the following are equivalent:

1. \( \displaystyle R \) is a semisimple ring.
2. Every left or right \( \displaystyle R \)-module is semisimple.
3. Every left or right ideal of \( \displaystyle R \) is of the form \( \displaystyle Re \) or \( \displaystyle eR \), respectively, where \( \displaystyle e \) is an idempotent element of \( \displaystyle R \), that is \( \displaystyle e^{2}=e \).

Conclude that a simisimple ring does not contain any nonzero nilpotent left or right ideals.

[Papapetros Vaggelis]

Hi Nickos.

The implication \(\displaystyle{2)\implies 1)}\) is obviously true.

\(\displaystyle{\bullet\,\,:1)\implies 3)}\) Suppose that the ring \(\displaystyle{\left(R,+,\cdot\right)}\) is left semisimple. If \(\displaystyle{I}\)

is a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\), then \(\displaystyle{I}\) is a submodule of \(\displaystyle{_{R}\,R}\) and since \(\displaystyle{_{R}\,R}\)

is semisimple, there exists a submodule \(\displaystyle{J}\), equivalently, there exists a left ideal \(\displaystyle{J}\) of \(\displaystyle{\left(R,+,\cdot\right)}\)

such that \(\displaystyle{R=I\oplus J}\) . Then, \(\displaystyle{1_{R}=e+f}\) where \(\displaystyle{e\in I}\) and \(\displaystyle{f\in J}\) .

We''ll prove that \(\displaystyle{I=R\,e}\) and \(\displaystyle{e^2=e}\) .

The relation \(\displaystyle{R\,e\subseteq I}\) is direct because \(\displaystyle{I}\) is a left ideal of \(\displaystyle{\left(R,+,\cdot\right)}\)

and \(\displaystyle{e\in I}\). Let now \(\displaystyle{x\in I}\) . Then,

\(\displaystyle{x=x\cdot 1_{R}=x\cdot \left(e+f\right)=x\cdot e+x\cdot f\implies \left(x-x\cdot e\right)+(-x)\cdot f=0}\),

where \(\displaystyle{x-x\cdot e\in I}\) and \(\displaystyle{(-x)\cdot f\in J}\) since \(\displaystyle{I\,,J}\) are left ideals of \(\displaystyle{\left(R,+,\cdot\right)}\) .

Due to the fact that the sum \(\displaystyle{I+J}\) is direct, we have that

\(\displaystyle{x-x\cdot e=0\,,(-x)\cdot f=0}\) and thus: \(\displaystyle{x=x\cdot e\in R\,e}\), so: \(\displaystyle{I\subseteq R\,e}\)

which means that \(\displaystyle{I=R\,e}\). For the same reason,

\(\displaystyle{e=e\cdot 1_{R}=e^2+e\,f\implies \left(e-e^2\right)-e\,f=0\implies e-e^2=0\implies e=e^2}\) .

\(\displaystyle{\bullet\,\,: 3)\implies 1)}\) Let \(\displaystyle{I}\) be a left ideal ( submodule) of \(\displaystyle{\left(R,+,\cdot\right)}\) .

If \(\displaystyle{I=\left\{0\right\}=R\cdot 0}\), then \(\displaystyle{R=R\oplus I}\) because \(\displaystyle{R\cap I=\left\{0\right\}}\) .

If \(\displaystyle{I=R=R\cdot 1_{R}}\), then: \(\displaystyle{R=I\oplus \left\{0\right\}}\), where \(\displaystyle{I\cap \left\{0\right\}=\left\{0\right\}}\) .

Suppose now that \(\displaystyle{I \neq \left\{0\right\}}\) and \(\displaystyle{I\neq R}\) .

According to the hypothesis, there is \(\displaystyle{e\in R-\left\{0,1\right\}}\)

such that \(\displaystyle{I=R\,e\,\,,e^2=e}\) .

Then, \(\displaystyle{e=1_{R}\cdot e\in R\,e\implies e\in I}\). Setting \(\displaystyle{f=1_{R}-e\,,J=R\,f}\), we have that

\(\displaystyle{e+f=1_{R}}\) and also that \(\displaystyle{J}\) is a left ideal( submodule) of \(\displaystyle{\left(R,+,\cdot\right)}\) .

For each \(\displaystyle{r\in R}\) we get:

\(\displaystyle{r=r\cdot 1_{R}=r\cdot \left(e+f\right)=r\cdot e+r\cdot f\in I+J}\), so: \(\displaystyle{R=I+J}\).

We''ll prove additionally that \(\displaystyle{I\cap J=\left\{0\right\}}\) . Since, \(\displaystyle{\left\{0\right\}\subseteq I\cap J}\), it is sufficient

to prove that \(\displaystyle{I\cap J\subseteq \left\{0\right\}}\). For this purpose, let \(\displaystyle{x\in I\cap J}\) . By definition,

\(\displaystyle{x\in I}\), and \(\displaystyle{x\in J}\), so: \(\displaystyle{x=r\cdot e}\) and \(\displaystyle{x=s\cdot f}\) for some \(\displaystyle{r\,,s\in R}\).

\(\displaystyle{\begin{aligned} x=x&\implies r\cdot e=s\cdot f\\&\implies r\cdot e=s-s\cdot e\\&\implies \left(r+s\right)\cdot e=s\\&\implies \left(r+s\right)\cdot e^2=s\cdot e\\&\implies \left(r+s\right)\cdot e=s\cdot e\\&\implies r\cdot e+s\cdot e=s\cdot e\\&\implies x=r\cdot e=0\in\left\{0\right\}\end{aligned}}\)

So, \(\displaystyle{R=I\oplus J}\) and \(\displaystyle{\left(R,+\cdot\right)}\) is a semisimple ring because each submodule of \(\displaystyle{_{R}\,R}\)

is a direct summand.

I can't prove the implication \(\displaystyle{1)\implies 2)}\) . What's your opinion ?

Comment

Here is another approach for the implication \(\displaystyle{1)\implies 3)}\) if we take into consideration the exercise here:

Specification Of Ideals Of A Product Ring .

The results are exactly the same for right ideals.

So, if \(\displaystyle{\left(R,+,\cdot\right)}\) is a left or right semisimple ring, then

\(\displaystyle{\left(R,+,\cdot\right)\simeq \left(\prod_{i=1}^{k}\mathbb{M}_{n_{i}}\,(D_{i}),+,\cdot\right)}\), where

\(\displaystyle{\left(D_{i},+_{i},\cdot_{i}\right)\,,1\leq i\leq k}\) are division rings.

[Tsakanikas Nickos]

We will prove the equivalence \( \displaystyle (1) \Leftrightarrow (2) \).


\( \displaystyle (1) \implies (2) : \) Let \( \displaystyle M \) be a right \( \displaystyle R \)-module and let \( \displaystyle X \) be a generating set of \( \displaystyle M \). Consider the mapping
\[ \displaystyle f : \oplus_{x\in X} R_{x} \longrightarrow M \, , \, f\left( \sum_{x \in X} r_{x} \right) = \sum_{x \in X} x \cdot r_{x} \]where \( \displaystyle R_{x}=R , \forall x \in X \). Observe that \( \displaystyle f \) is an epimorphism of right \( \displaystyle R \)-modules, since \( \displaystyle X \) is a set of generators of \( \displaystyle M \). Since, for all \( \displaystyle x \in X \), \( \displaystyle R_{x} = R_{R} \) is semisimple, \( \displaystyle R_{R} \) is a direct sum of simple right \( \displaystyle R \)-modules and, therefore, \( \displaystyle \oplus_{x \in X} R_{x} \) is a direct sum of simple right \( \displaystyle R \)-modules, which means that \( \displaystyle \oplus_{x \in X} R_{x} \) is semisimple. Due to the fact that quotients of semisimple modules are again semisimple modules, the first isomorphism theorem yields that \( \displaystyle M \) is a semisimple module.


\( \displaystyle (2) \implies (1) : \) Obviously true, as you mentioned.

[Papapetros Vaggelis]

Nickos, here is an answer about the last question you proposed.

" Conclude that a semisimple ring does not conclude any nonzero nilpotent left or right ideals "

Proof

Let \(\displaystyle{\left(R,+,\cdot\right)}\) be a semisimple ring and \(\displaystyle{I}\) be a left(right) nilpotent ideal of this ring.

According to the equivalence \(\displaystyle{(1)\iff (3)}\) and the definiton of a nilpotent ideal, there are \(\displaystyle{e\in R}\)

and \(\displaystyle{k\in\mathbb{N}}\) such that \(\displaystyle{I=R\,e\,\,(e\,R)}\) , where \(\displaystyle{e^2=e}\) and

\(\displaystyle{I^{k}=\left\{\sum_{i=1}^{n}\prod_{j=1}^{k}x_{i_{j}}\in R: x_{i_{j}}\in I\,,1\leq i\leq k\,,1\leq j\leq k\,\,,n\in\mathbb{N}\right\}=\left\{0\right\}}\) .

Note : The subset \(\displaystyle{I^{k}}\) of \(\displaystyle{R}\) is a left(right) ideal of the ring \(\displaystyle{\left(R,+,\cdot\right)}\) .

If \(\displaystyle{k=1}\), then \(\displaystyle{I=\left\{0\right\}}\). If \(\displaystyle{k\geq 2}\), then :

\(\displaystyle{e^{k}\in I^{k}\implies e\in I^{k}\implies e\in\left\{0\right\}\implies e=0\implies I=\left\{0\right\}}\) .

In conclusion, the semisimple ring \(\displaystyle{\left(R,+,\cdot\right)}\) does not contain any nonzero nilpotent left(right) ideals.