Problem. Let n > 1 be an integer and consider the ring R = \mathbb{Z}/n \mathbb{Z}. Show that the number of idempotents of R is 2^m, where m is the number of distinct prime divisors of n.

Solution. Let n = \prod_{i=1}^m p_i^{n_i} be the prime factorization of n and let R_i = \mathbb{Z}/p_i^{n_i} \mathbb{Z}. By the Chinese remainder theorem we have

R \cong R_1 \times \cdots \times R_m. \ \ \ \ \ \ \ \ \ (*)

Claim. If p is a prime and \ell > 0 is an integer, then the only idempotent elements of \mathbb{Z}/p^{\ell} \mathbb{Z} are 0 and 1.

Proof of the cliam. So we want to show that, modulo p^{\ell}, the equation x^2 \equiv x \mod p^{\ell} has only two trivial solutions x =0, 1. Suppose that x \neq 0 is a solution of x^2 \equiv x \mod p^{\ell}. Let x=p^r s, where 0 \leq r < \ell and \gcd(s,p)=1. Then s(p^rs-1) \equiv 0 \mod p^{\ell - r} which gives us p^rs \equiv 1 \mod p^{\ell - r}. Thus r = 0 and hence x=s \equiv 1 \mod p^{\ell}

It is clear now from (*) and the claim that the number of idempotents of the ring R is 2^m. \ \Box

Example. Find all idempotents of R = \mathbb{Z}/60\mathbb{Z}.

Solution. By the above problem, we know that R has 8 idempotents, two of them being 0, 1 \mod 60. Let R_1= \mathbb{Z}/4\mathbb{Z}, \ R_2 = \mathbb{Z}/3\mathbb{Z} and R_3 = \mathbb{Z}/5\mathbb{Z}. Then R \cong R_1 \times R_2 \times R_3. All idempotents of R_1 \times R_2 \times R_3 are (0,0,0), (1,0,0), (1,1,0), (0,1,0), (0,1,1), (0,0,1), (1,0,1) and (1,1,1). So we just need to find the preimage of each idempotent in R. Obviously the preimages of (0,0,0), (1,1,1) are 0, 1 \mod 60 respectively.
Now, let’s find the preimage of, say, b=(0,1,1). Let a = m + 60\mathbb{Z} be the preimage. Then the image of a is (m + 4\mathbb{Z}, m + 3\mathbb{Z}, m + 5\mathbb{Z})=b = (4\mathbb{Z},1 + 3\mathbb{Z},1 + 5\mathbb{Z}). So m is divisible by 4 and is equivalent to 1 modulo 3 or 5. It follows that a = 16 + 60\mathbb{Z} . \Box

  1. Nour says:

    How to find all idempotent elements in the ring z modulo n

  2. Seth Troisi says:

    This really helped solidify some math I was working on thanks!

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