All rings in this post are assumed to have the multiplicative identity

In this post, we showed that the ring has idempotents, where is the number of prime divisors of Clearly is also the number of prime (= maximal) ideals of (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let be the Euler’s totient function. The number of units of is If is the prime factorization of then If is odd, then is even for all and hence divides So if is odd, then the number of idempotents of divides the number of units of As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring is called **semilocal** if the number of maximal ideals of is finite and it is called **local** if it has only one maximal ideal.

**Example 1**. If is a prime and is an integer, then is a local ring with the unique maximal ideal If is an integer, then is a semilocal ring (what are its maximal ideals?).

**Example 2**. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let be an Artinian ring and let be the set of all finite intersections of maximal ideals of Since is Artinian, has a minimal element Let be any maximal ideal of Since and is a minimal element of we must have So for some and hence Therefore are all the maximal ideals of

**Problem 1**. Show that if is a local ring, then are the only idempotents of

**Solution**. Let be the maximal ideal of and let be an idempotent of Then Thus either or If then because otherwise which is false. So is a unit because there’s no other maximal ideal to contain So for some and thus Similarly, if then and thus is a unit. So for some implying that and hence

**Problem 2 **Show that the number of idempotents of a commutative Arinian ring is where is the number of maximal ideals of

**Solution**. Since is Artinian, it has only finitely many maximal ideals, say (see Example 2). The Jacobson radical of is nilpotent, hence there exists an integer such that

Thus, by the Chinese remainder theorem for commutative rings, Since each is a local ring, with the unique maximal ideal it has only two idempotents, by Problem 1, and so has exactly idempotents.

**Problem 3**. Let be a finite commutative ring. Show that if the number of elements of is odd, then the number of idempotents of divides the number of units of

**Solution**. Since is finite, it is Artinian. Let be the set of maximal ideals of By problem 2, the umber of idempotents of is and for some integer

Since is odd, each is odd too because is a subgroup of and so divides Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each is which is an even number because both and are odd. So the number of units of which is the product of the number of units of is divisible by which is the number of idempotents of

**Remark 1**. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, has one unit and two idempotents. However, the result is true in which has units and two idempotents.

Can we find all even integers for which the result given in Problem 3 is true in ? Probably not because this question is equivalent to finding all integers such that where is the number of prime divisors of and that is not an easy thing to do.

**Remark 2**. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider the ring of matrices with entries from the field of order three. Then has units (see Problem 3 in this post!) but, according to my calculations, has idempotents and does not divide