11591. Let In be the set of all idempotent elements of Z/nZ. That is e ∈ In
if and only if e2 ≡ e (mod n). Let In1 = In , and for k ≥ 2, let Ink be the set of
all sums of the form u + v where u ∈ In , v ∈ Ink−1 , and the addition is done
modulo n. Determine, in terms of n, the least k such that Ink = Z/nZ.
Solution by Tamás Dékány, ”Fejéntaláltuka Szeged” problem solving group,
University of Szeged, Hungary. From now on we denote by Zn the ring of
modulo n remainders.
First we have to determine the idempotent elements of Zn .
αm
1 α2
Let n = pα
1 p2 · · · pm the prime factorisation of n. That is, pi is prime ∀i,
pi 6= pj ∀i 6= j and αi ≥ 1 ∀i. A well-known fact is that Zn is isomorphic to
this direct product:
Zn ∼
= Zpα1 1 × Zpα2 2 × · · · × Zpαmm .
One can easily see, that any idempotent element of Zn has idempotent components in every Zpαi . Moreover from the direct product construction of Zn , it
i
is clear that every idempotent is constructed from idempotents of Zpαi .
i
The only idempotent elements of Zpα are 0 and 1. Let us assume that x 6= 0
2
α
idempotent element of Zpα . That is x ≡ x (mod p ). We will show that x ≡ 1
(mod pα ). Let x = pr s, where 0 ≤ r < l and gcd(s, p) = 1.
x2 ≡ x
(mod pα )
p2r s2 ≡ pr s (mod pα )
pr s ≡ 1 (mod pα−r )
We got the last congruence by dividing both sides with pr and s, since gcd(s, p) =
1. The last congruence holds only if r = 0. Thus x = s ≡ 1 (mod pα ).
So we got that the set of idempotents of Zn is
In1 = (1 , . . . m ) : i ∈ {0, 1}, ∀0 ≤ i ≤ m .
From the definition of Ink , and the formulae above for In1 , one can see that
you can add zero or one to any component of an element. This means that
every element of Ink has at most k in every component:
Ink = (a1 , . . . am ) : ai ∈ {0, 1, . . . , k}, ∀0 ≤ i ≤ m .
From the above definition of Ink it is easy to see, that the minimal k which
α2
αm
1
for Ink = Zn , is k = min(pα
1 − 1, p2 − 1, . . . , pm − 1).