# Help with proof of showing idempotents in set of Integers Modulo a prime power are $0$ and $1$

I was reading online for a project I’m currently doing and came across the following claim and proof. The statement would be useful to me, and although I’ve spent a long time looking at it there’s still a few steps that I just don’t follow and I was hoping someone could help me.

Claim: If $p$ is a prime number and $k$ a positive integer then the only idempotent elements in $\mathbb{Z}$/$p^k\mathbb{Z}$ are $0$ and $1$.

Proof:

An element, $x$, of $\mathbb{Z}$/$n\mathbb{Z}$ is idempotent if it is a solution to the congruence $x^2\equiv x \pmod {p^k}$. Clearly $x=0$ and $x=1$ are solutions to this congruence.

Say that $x\neq0$. Write $x=p^rs$ where $0\leq r<k$ and $\gcd(p,s)=1$.Then $s(p^rs-1)\equiv 0\pmod {p^{k-r}}$ which implies that $p^rs \equiv 1\pmod {p^{k-r}}$.

Hence $r$ must equal $0$ and $x=s\equiv 1\pmod {p^k}$.∎

Choosing $x\neq0$ and showing it to be equal to $1$ makes sense. However I don’t follow the next few congruences, why are they true? Since $\gcd(p,s)=1$, doesn’t this tell us that ${p^{k-r}}$ can’t divide $s$? In turn implying from the first congruence that ${p^{k-r}}$ divides $p^{r}s-1$. I don’t see why this is true? As for the following steps, I can see why $r$ must equal $0$, but unfortunately not the rest.

Sorry for the long post, any help is greatly appreciated.

#### Solutions Collecting From Web of "Help with proof of showing idempotents in set of Integers Modulo a prime power are $0$ and $1$"

The ring $\mathbb{Z}/p^k \mathbb{Z}$ is a local ring: it has a unique maximal ideal, the set of all multiples of $p$. (The ideals in the quotient ring $\mathbb{Z}/p^k \mathbb{Z}$ are precisely the ideals of $\mathbb{Z}$ containing $p^k$. Since $\mathbb{Z}$ is a PID, the result comes down to the fact that the only prime dividing $p^k$ is $p$.)

An element $e$ of a commutative ring is an idempotent if $e^2 = e$.

Here are some very easy facts about idempotents:

1) If $e$ is an idempotent, so is $1-e$: $(1-e)(1-e) = 1 – 2e + e^2 = 1-2e + e = 1-e$.
2) If an idempotent $e$ is a unit, then $e = 1$: multiply $e^2 = e$ by $e^{-1}$.
3) One says that an idempotent $e$ is nontrivial if neither $e$ nor $1-e$ is $0$. It follows from 2) that neither $e$ nor $1-e$ is a unit.

Proposition: A local ring has no nontrivial idempotents.

Proof: Let $\mathfrak{m}$ be the unique maximal ideal of $R$. If $e$ is a nontrivial idempotent, then by 3) above $e,1-e \in \mathfrak{m}$, hence $e + (1-e) = 1 \in \mathfrak{m}$, contradiction.

Theorem. Let $b,c,d\in \mathbb{Z}$ with $d\neq 0$. Then
$$ad\equiv bd \pmod{c}$$
$$a \equiv b \pmod{\frac{c}{\gcd(d,c)}}$$
Proof. Let $\alpha=\gcd(d,c)$, and set $c=c’\alpha$, with $\gcd(\alpha,c’)=1$. Similarly, set $d=d’\alpha$, with $\gcd(\alpha,d’)=1$. Also, $\gcd(c’,d’)=1$, by maximality of gcd. Now, suppose $ad\equiv bd \pmod{c}$. Then $c|(ad-bd)$. Translating, we get $c’\alpha | \alpha ad'(a-b)$. Cancel the $\alpha$ and we get $c’|d'(a-b)$. Since $\gcd(c’,d’)=1$, in fact $c’|(a-b)$. The other direction is simpler: suppose $c’|(a-b)$. Then $c’\alpha | \alpha ad'(a-b)$.