Orthogonally finite ring and primitive idempotent

Question

If a ring is given to be an orthogonally finite (that is it does not contain an infinite set of mutually orthogonal idempotents), how it ensures the existence of a primitive idempotent (that is a nonzero idempotent which can not be written as a sum of two mutually orthogonal idempotents)? Please explain the existence in details.

Answer 1

To show orthogonally finite implies existence of primitive idempotent, we may argue the contrapositive: no primitive idempotents implies orthogonally infinite.

The takeaway from a lack of primitive idempotents is that every idempotent may be decomposed into a sum of two orthogonal idempotents, which can be repeated indefinitely.

As an illustration idempotents, consider any direct sum $\mathbb{F}\oplus\cdots\oplus\mathbb{F}=\mathbb{F}^n$ of isomorphic copies of a field $\mathbb{F}$. The idempotents correspond to subsets $S$ of $\{1,\cdots,n\}$, describing which components are unity $1$ versus which are zero $0$. Decomposing such an idempotent into a sum of orthogonal ones corresponds to partitioning a set into disjoint subsets. This implies refining a decomposition retains orthogonality. That is, if $e_0=f_1+e_1$ with $f_1\perp e_1$ and we also decompose $e_1=f_2+e_2$ also with $f_2\perp e_2$, then $f_1,f_2,e_3$ are all orthogonal. We can investigate if this holds generally.

Suppose $R$ is a ring with no primitive idempotents, and decompose

$$ e_0=f_1+e_1 \\ e_1=f_2+e_2 $$

which implies $e_0=f_1+f_2+e_2$. We want to show $f_1,f_2,e_2$ are all orthogonal. We know $f_1\perp e_1$ and $e_1=f_2+e_2$ so we may write $f_1(f_2+e_2)=0$, then right-multiply by $f_2$ or $e_2$ to get the desired orthogonality relations $f_1f_2=0$ and $f_1e_2=2$ (using $f_2\perp e_2$), QED.

This can be extended indefinitely:

$$ e_0=f_1+e_1 \\ e_1=f_2+e_2 \\ e_2=f_3+e_3 \\ \vdots $$

and subsequently, $e_0=f_1+\cdots+f_n+e_n$ for all $n$. We may show by induction that the finite set $\{f_1,\cdots,f_n,e_n\}$ is orthogonal for all $n$. Given the truth of this claim for a given $n$, we may show it is true for $n+1$ by setting $f=f_1+\cdots+f_n$ so $e_0=f+f_{n+1}+e_{n+1}$, then right-multiply $f(f_{n+1}+e_{n+1})=0$ by $f_{n+1}$ and $e_{n+1}$ to get $ff_{n+1}=0$ and $fe_{n+1}=0$, then left-multiply by $f_i$ (where $1\le i\le n$) to get $f_if_{n+1}=0$ and $f_ie_{n+1}=0$, QED.

We conclude $\{f_1,f_2,f_3,\cdots\}$ is an infinite orthogonal set. Recall the assumption we made in the beginning was that there are no primitive idempotents. Therefore, we conclude the contrapositive: orthogonally finite implies the existence of a primitive idempotent.