>>14002483I don't know what your level is, but I'm gonna think out loud.
Feels like a kind of Fermat's little theorem thing.
For example, mod 3, A^3 = A for all A, but 2^2 != 2.
Or you can look mod 10, where A^5 = A for all elements, some not idempotent.
It's also obvious when A is an (n-1)th root of unity.
I'm also reminded of how if you have a ring RxS, then for any root of unity r of R, (r, 0) has this property, even though it's not a unit in RxS. The annihilators of (r,0) are precisely 0xS, so maybe it's productive to look at the annihilator of A? And then somehow use the Chinese Remainder Theorem?
The reason I'm trying to go this route is that maybe you'll find that "generalized idempotents" are just something already known, though under a different name.
So say R is the ring containing A, you look at the quotient R/Ann(A), then the image of A in this ring is a root of unity since 1 - A^(n-1) is in Ann(A), so A actually is an (n-1)th root of unity in R/Ann(A). For convenience, let B = A^(n-1).
And if we wanna use CRT, maybe we could take an ideal J containing B, making it so that Ann(A) + J = R, since (1 - B) + B = 1, so J is coprime to Ann(A). If we could prove the the intersection of J and Ann(A) is the zero ideal, then we know R is isomorphic to R/J x R/Ann(A).
Actually, let's take J = (B), and instead of taking Ann(A), we take K = (1-B) contained in Ann(A). Still J and K are coprime, and suppose we have an element e in the intersection. Then e = jB = k(1-B), so (j+k)B = k... my brain stopped working.
