Primitive Idempotents

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

Primitive Idempotents

#1

Post by Tsakanikas Nickos »

Let \( A \) be a \( \mathbb{K} \)-algebra, where \( \mathbb{K} \) is a field, and let \( e \in A \) be a non-zero idempotent. Show that the following are equivalent:
  1. \( e \) is primitive.
  2. If \( e = e_{1}+e_{2} \), where \( e_{1} \) and \( e_{2} \) are orthogonal idempotents in \( A \), then \( e_{1}=0 \) or \( e_{2}=0 \).
  3. \( Ae \) is indecomposable.
Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Primitive Idempotents

#2

Post by Papapetros Vaggelis »

\(\displaystyle{i)\implies ii)}\) : Suppose that \(\displaystyle{e}\) is primitive.

Let \(\displaystyle{e=e_1+e_2}\), where \(\displaystyle{e_1\,,e_2}\) are orthogonal idempotents in

\(\displaystyle{A}\) . If \(\displaystyle{e_1\neq 0}\) and \(\displaystyle{e_2\neq 0}\), then

\(\displaystyle{e_1\neq 1\,\,,e_2\neq 1}\), since \(\displaystyle{e_1\,e_2=e_2\,e_1=0}\). But then,

\(\displaystyle{e_1\,,e_2}\) are non trivial idempotents in \(\displaystyle{A}\) with \(\displaystyle{e=e_1+e_2}\)

and \(\displaystyle{e}\) is primitive, a contradiction, so : \(\displaystyle{e_1=0}\) or \(\displaystyle{e_2=0}\) .

\(\displaystyle{ii)\implies iii)}\) : Let \(\displaystyle{A\,e=A_{1}\oplus A_{2}}\) . We have that

\(\displaystyle{e\in A\,e\implies \left(\exists\,e_1\in A_{1}\right)\,\left(\exists\,e_2\in A_{2}\right): e=e_1+e_2}\) .

Now, \(\displaystyle{e_2\,e_1\,\,,e_1\,e_2\in A_1\cap A_2}\), so : \(\displaystyle{e_1\,e_2=e_2\,e_1=0}\) .

Also,

\(\displaystyle{e^2=e\implies e_1^2+e_2^2=e_1+e_2\iff (e_1^2-e_1)+(e_2^2-e_2)=0}\). Since

\(\displaystyle{e_1^2-e_1\in A_{1}\,\,,e_2^2-e_2\in A_{2}}\) and the sum \(\displaystyle{A_1+A_{2}}\)

is direct, we get : \(\displaystyle{e_1^2-e_1=0=e_2^2-e_2\iff e_1^2=e_1\,\land e_2^2=e_2}\)

which means that \(\displaystyle{e_1\,,e_2}\) are orthogonal idempotents in \(\displaystyle{A}\).

According to \(\displaystyle{ii)}\), we have that \(\displaystyle{e_1=0}\) or \(\displaystyle{e_2=0}\) .

If \(\displaystyle{e_1=0}\), then \(\displaystyle{e=e_2\implies A\,e=A\,e_2=Α_1\oplus A_2}\)

and if \(\displaystyle{x\in A_{1}}\), then :

\(\displaystyle{x\in A_1+A_2\implies x\in A\,e_2\implies x\in A_2\implies x\in A_1\cap A_2=\left\{0\right\}\implies x=0}\)

so : \(\displaystyle{A_1=0}\). Similarly, if \(\displaystyle{e_2=0}\), then \(\displaystyle{A_2=0}\) .

Therefore, \(\displaystyle{A\,e}\) is indecomposable.

\(\displaystyle{iii)\implies i)}\) . Suppose that \(\displaystyle{e}\) is not primitive. Then,

\(\displaystyle{e=e_1+e_2}\), where \(\displaystyle{e_1\,,e_2}\) are non trivial orthogonal idempotents

in \(\displaystyle{A}\) .

To be continued...
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