Question About Rings

Groups, Rings, Domains, Modules, etc, Galois theory
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Tsakanikas Nickos
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Question About Rings

#1

Post by Tsakanikas Nickos »

Let \( \displaystyle R \) be an associative ring with unity \( \displaystyle 1_{R} \). If \( \displaystyle \left\{ I_{k} \right\}_{k \in K} \) is a family of non zero right (left) ideals of \( \displaystyle R \) and if \( \displaystyle R \) is the direct sum of that family (\( \displaystyle R = \oplus_{k \in K} I_{k} \)), show that \( \displaystyle K \) is a finite index set ( \( \displaystyle \big| K \big| < \infty \) ).

When can \( \displaystyle R \) be written as a finite direct sum of right (left) ideals?
Papapetros Vaggelis
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Joined: Mon Nov 09, 2015 1:52 pm

Re: Question About Rings

#2

Post by Papapetros Vaggelis »

Suppose that \(\displaystyle{R=\oplus_{k\in K}\,I_{k}}\) is a family of non zero right ideals of \(\displaystyle{\left(R,+,\cdot\right)}\) .

\(\displaystyle{\oplus_{k\in K}\,I_{k}=\left\{\sum_{i=1}^{n}x_{i}\in R: x_{i}\in I_{i}\,,n\in\mathbb{N}\right\}}\) .

\(\displaystyle{1\in R\implies \exists\,n\in\mathbb{N}: 1=e_{k_{1}}+...+e_{k_{n}}\,\,,e_{k_{i}}\in I_{k_{i}}\,,k_{i}\in K\,,1\leq i\leq n}\) .

If \(\displaystyle{r\in R}\), then:

\(\displaystyle{r=1\cdot r=\left(e_{k_{1}}+...+e_{k_{n}}\right)\cdot r=e_{k_{1}}\cdot r+...+e_{k_{n}}\cdot r}\),

where \(\displaystyle{e_{i}\,r\in I_{i}\,,1\leq i\leq n}\)

because \(\displaystyle{e_{i}\in I_{i}\,,r\in R}\) and \(\displaystyle{I_{i}}\) is right ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) , so :

\(\displaystyle{r\in \oplus_{i=1}^{n}I_{i}}\). Therefore, \(\displaystyle{R=\oplus_{i=1}^{n}I_{i}}\).

If \(\displaystyle{k\in K-\left\{k_1,...,k_{n}\right\}}\), then consider the non-zero right ideal \(\displaystyle{I_{k}}\) . Then, there is \(\displaystyle{x\in R}\)

such that \(\displaystyle{x\in I_{k}}\) and \(\displaystyle{x\neq 0}\). So,

\(\displaystyle{x=1\cdot x=e_{k_{1}}\cdot x+...+e_{k_{n}}\cdot x\iff e_{k_{1}}\cdot x+...+(-x)+...+e_{k_{n}}\cdot x=0}\) and since the sum

\(\displaystyle{\sum_{k\in K}I_{k}}\) is direct, we get : \(\displaystyle{e_{k_{i}}\,x=0\,,\forall\,i\in\left\{1,...,n\right\}\,\,,x=0}\) , so :

\(\displaystyle{I_{k}=\left\{0\right\}}\), a contradiction.

Therefore,

\(\displaystyle{K-\left\{k_1\,...,k_{n}\right\}=\varnothing\implies K=\left\{k_1,...,k_n\right\}\implies \left|K\right|<\infty}\) .
Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Question About Rings

#3

Post by Tsakanikas Nickos »

As far as the second part is concerned I had in mind the following result:

\( \displaystyle R \) contains a partition of unity \( \displaystyle \{ e_{i} \}_{i=1}^{n} \) if and only if \( \displaystyle R \) can be written as a direct sum of left or right ideals (generated by the elements \( \displaystyle \{ e_{i} \}_{i=1}^{n} \) respectively).
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Question About Rings

#4

Post by Papapetros Vaggelis »

2nd part

Suppose that \(\displaystyle{R=\oplus_{k=1}^{n}I_{k}}\), where \(\displaystyle{I_{k}\,,1\leq k\leq n}\) are non-zero right ideals of \(\displaystyle{\left(R,+,\cdot\right)}\).

\(\displaystyle{1\in R\implies 1=e_{1}+...+e_{n}\,,e_{k}\in I_{k}\,,1\leq k\leq n}\) .

Obviously, \(\displaystyle{e_{k}\,R\subseteq I_{k}\,,1\leq k\leq n}\) .

Let \(\displaystyle{x\in I_{k}}\) .

Then, \(\displaystyle{x=1\cdot x=e_{1}\cdot x+...+e_{k}\cdot x+....+e_{n}\cdot x\implies}\)

\(\displaystyle{\implies e_{1}\cdot x+...+\left(e_{k}\,x-x\right)+...+e{n}\cdot x=0}\)

and since the sum \(\displaystyle{\sum_{k=1}^{n}I_{k}}\) is direct, we have that:

\(\displaystyle{e_{i}\cdot x=0\,,\forall\,i\in\left\{1,...,n\right\}-\left\{k\right\}\,\,,e_{k}\cdot x-x=0\iff x=e_{k}\cdot x\in e_{k}\,R}\) .

In conclusion, \(\displaystyle{e_{k}\,I=I_{k}\,,1\leq k\leq n}\). Also, for every \(\displaystyle{k\in\left\{1,...,n\right\}}\), :

\(\displaystyle{0=e_{1}\cdot e_{k}+...+\left(e_{k}^2-e_{k}\right)+...+e_{n}\cdot e_{k}\implies e_{i}\cdot e_{k}=0\,\forall\,i\in\left\{1,...,n\right\}-\left\{k\right\}\,\,,e_{k}^2=e_{k}}\) .

Therefore, the set \(\displaystyle{\left\{e_1,...,e_{n}\right\}}\) is a partition of unity of \(\displaystyle{\left(R,+,\cdot\right)}\) .

On the other hand, suppose that \(\displaystyle{\left\{e_1,...,e_n\right\}}\) is a partition of unity of \(\displaystyle{\left(R,+,\cdot\right)}\).

Consider the right ideals \(\displaystyle{I_{k}=e_{k}\,R\,,1\leq k\leq n}\) . We have that \(\displaystyle{1=e_1+...+e_n}\) .

Let \(\displaystyle{r\in R}\) . Then,

\(\displaystyle{r=1\cdot r=e_1\cdot r+...+e_n\cdot r}\), where \(\displaystyle{e_{k}\cdot r\in I_{k}}\), so :

\(\displaystyle{R=\sum_{k=1}^{n}I_{k}}\) . We''ll prove that this sum is direct. For this purpose, suppose that

\(\displaystyle{x_1+...+x_n=0}\), where \(\displaystyle{x_{k}\in I_{k}\,,1\leq k\leq n}\) . Then,

\(\displaystyle{\left(\forall\,k\in\left\{1,...,n\right\}\right)\,\left(\exists\,r_{k}\in R\right): x_{k}=e_{k}\cdot r_{k}}\) and thus

\(\displaystyle{e_{1}\cdot r_{1}+...+e_{n}\cdot r_{n}=0}\). Let \(\displaystyle{k\in\left\{1,...,n\right\}}\).

Since \(\displaystyle{\left\{e_1,...,e_n\right\}}\) is a partition of unity, we have that

\(\displaystyle{e_{k}\cdot e_{m}=0=e_{m}\cdot e_{k}\,,\forall\,m\in\left\{1,...,n\right\}-\left\{k\right\}\,\,,e_{k}^2=e_{k}}\) .

Therefore,

\(\displaystyle{\begin{aligned} \sum_{i=1}^{n}e_{k}\cdot r_{k}=0&\implies e_{k}\cdot \left(e_{1}\cdot r_{1}+...+e_{n}\cdot r_{n}\right)=e_{k}\cdot 0\\&\implies \sum_{i=1}^{n}\left(e_{k}\cdot e_{i}\right)\cdot r_{i}=0\\&\implies e_{k}^2\cdot r_{k}=0\\&\implies e_{k}\cdot r_{k}=0\\&\implies x_{k}=0\end{aligned}}\)

So, \(\displaystyle{x_{k}=0\,,\forall\,k\in\left\{1,...,n\right\}}\) and then the sum \(\displaystyle{\sum_{k=1}^{n}I_{k}}\) is direct.

In conclusion, as Nickos said :

The ring \(\displaystyle{\left(R,+,\cdot\right)}\) contains a partition of unity \(\displaystyle{\left\{e_1,...,e_n\right\}}\) if, and only if, \(\displaystyle{R}\)

can be written as a direct sum of right(left) ideals.
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