Equivalency of the maximum and minimum conditions of idempotents of a ring

Question

Let $R$ be a ring with unit, and let $I$ be the set of all idempotents of $R$, that is, all $e\in R$ such that $e^2 = e$. We put a partial ordering $\leq$ on $I$ by saying $e\leq f$ if $ef=e=fe$ or equivalently if $eRe\subset fRf$. We say that $R$ satisfies the maximum condition on idempotents if every nonempty subset $A\subset I$ contains a maximum element, or equivalently, if for every chain $e_1 \leq e_2 \leq e_3 \leq \cdots$, there exists some $N$ such that for all $n\geq N$, we have $e_n=e_N$. Define the minimum condition on idempotents similarly. Note that $0\leq e\leq 1$ for all $e\in I$. We also say that a set $A$ of idempotents is orthogonal if $ef=0$ for all $e,f\in A$ where $e\neq f$.

I want to prove that the following are equivalent for a ring $R$:

1. $R$ satisfies the maximum condition on idempotents.
2. $R$ satisfies the minimum condition on idempotents.
3. $R$ satisfies the maximum condition on the left ideals $Re$ where $e$ is an idempotent (and on right ideals $eR$).
4. $R$ satisfies the minimum condition on the left ideals $Re$ where $e$ is an idempotent (and on right ideals $eR$).
5. $R$ contains no infinite orthogonal set of idempotents.

The source I am reading uses this lemma to prove the Wedderburn-Artin theorem, and unfortunately I am not very well versed in algebra and they skip over much of the proof for this lemma, proving only $(1)\implies (3)$. They say $(1)\iff (2)$, $(3)\iff(4)$, and $(3)\implies (5) \implies (1)$ are all routine, but I have struggled a lot trying to prove even $(1)\iff (2)$. I think I can prove the latter two implications, as follows.

$(5)\implies (1)$: Suppose that $e_1 \leq e_2 \leq e_3 \leq \cdots$ is a chain which does not terminate. Then define $f_1 = e_1$ and $f_{n+1} = e_{n+1}-\sum_{k=1}^{n}f_k$. Then we claim by induction this is an infinite orthogonal set. Obviously $\{ f_1 \}$ is an orthogonal set trivially. Suppose $\{ f_i \}_{i=1}^n$ is an orthogonal set of idempotents. Then

$f_{n+1}^2 = (e_{n+1}-\sum_k^{n}f_k)^2 = e_{n+1}^2 - 2e_n\sum_k^n f_k - \sum_k^n f_k^2 = e_{n+1} - \sum_k^n f_k = f_{n+1}$

where we used the fact that $\{ f_i \}_{i=1}^n$ is orthogonal and $e_{n+1}f_k=f_k$ for $k\leq n$. Similarly,

$f_{n+1}f_j = (e_{n+1} - \sum_k^{n}f_k)(f_j) = e_{n+1}f_j - \sum_k^n f_k f_j = f_j - f_j = 0$.

Hence we have constructed an infinite orthogonal set of idempotents, a contradiction. Therefore, the sequence must terminate.

$(3)\implies (5)$ can be done similarly I think, by assuming one has an infinite orthogonal set of idempotents $\{ e_n \}$ and constructing idempotents $f_n = \sum_{k=1}^n e_k$ which satisfy $f_n f_m = f_m = f_m f_n$ for $m<n$. Thus $f_n R \subset f_{n+1}R$ for all $n$ (and likewise for right ideals $Rf_n$). We can show that the subset inclusion is strict since, $f_{n+1}\notin Rf_n$. Indeed, if this were the case then $f_{n+1}=rf_n$ for some $r\in R$, but then $f_{n+1}f_n = rf_n^2 = rf_n=f_{n+1}$, but we saw already $f_{n+1}f_n = f_n$, thus $f_{n+1}=f_n$. However, by assumption each $e_i$ was distinct, so their sums must be as well, so each $f_i$ should also be distinct, a contradiction. Thus the chain $Rf_1 \subset Rf_2 \subset \cdots$ does not terminate, a contradiction, implying that there does not exist an infinite orthogonal set of idempotents.

With all this stated, my questions are as follows:

1. Is there a tidier proof of the above two implications, since they are said to be "routine"? This assumes my above proofs are correct, and if they are not, please point out where I have gone wrong.
2. Could I have some tips for proving either $(1)\iff(2)$ or $(3)\iff(4)$? I'm pretty sure if I can prove either of these two statements, the other follows by an identical argument. I have tried the usual arguments for glb property $\iff$ lub property to no avail, leading me to believe I have to use the algebraic structure somewhere.

Thanks in advance!

Answer 1

Note that $e\leq f$ iff $1-f\leq 1-e$. This allows you to turn an ascending chain into a descending chain, and hence the equivalence of 1 and 2. 3 and 4 is similar.

Proof of claim. Suppose that $e\leq f$. Hence, $ef=e=fe$ so $\left(1-e\right)\left(1-f\right)=1-e-f+ef=1-f$ and $\left(1-f\right)\left(1-e\right)=1-e-f+fe=1-f$. So, $1-f\leq 1-e$.