The ring is more interesting than it is often given credit for. This post will tell you the isomorphism type of its group of units.

Why care about such a simple thing? Well first, the rings are universal, in the following sense. If is any ring with unity, there exists a unique and injective homomorphism . here is called the characteristic of . It’s easy to find this homomorphism: there is a natural map , and we factor out its kernel (which is of the form , being a principal ideal domain). So we should expect to come up anywhere that rings do (though only applies to a single ring, and is disproportionately common).

Theorem: If , where the product is over odd primes, the group of units can be written as a product of cyclic groups when and when . In particular, the group of units is cyclic exactly when is a power of an odd prime, twice a power of an odd prime, or .

The rest of this post will prove this theorem. First, the Chinese remainder theorem tells us that if factors into primes as , then . So we only need to address the question when is a power of a prime.

What are the invertible elements in ? If , then , so cannot be invertible. Those who are quick on the draw with the number theory will be able to quickly show that if does not divide , then is invertible in . (the standard technique here is to note that and are relatively prime, so we can choose integers and such that )

Since there are elements divisible by , it follows that has units. (notice that this, along with the Chinese remainder theorem, gives us a nice formula for Euler’s phi function)

Proposition 1: When is odd, the group of units of is a cyclic group of order .

Proof: The case is actually the most difficult. It is a special case of a theorem which says that a finite subgroup of the multiplicative group of any field is cyclic. For a proof of this case, see e.g. here. (I will certainly discuss this topic in detail in a future post)

Suppose that is a generator for the group of units of . Then the order of is divisible by , so it is of the form . Then has order . We claim that has order , from which it follows that has order .

What we really need to show here is that is divisible by but not . But this is straightforward induction: if with not divisible by , then taking -th powers we see that .

Proposition 2: If , the group of units of is the product of two cyclic groups, one of order , and one of order . If , the group of units of is cyclic.

Proof In fact, we will show that generates a cyclic group of order that does not contain , an element of order . (there is nothing special about here—any choice that is will work) We first show, by induction on , that , the case being clear.

For , the inductive hypothesis now tells us that , so there is some such that . Squaring, we get .

Since is an element of order , this shows that generates a cyclic group of order . So we just need to make sure that . But a cyclic group can contain at most one element of order , and we have already shown .

The second statement in the proposition is easily verified.

Exercise: Prove the second sentence of the theorem! (hint: for one direction, use the Chinese remainder theorem for cyclic groups, and for the other, count elements of order )

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