Maps between commutative rings are just as important as these between abelian groups. As you should anticipate, our maps should preserve ring operations.

A map between commutative rings and is called a

homomorphismif is a group homomorphism and for any

Here, as usual, we use and to denote addition and multiplications in respective rings. You should notice that the second property is essentially the same as for the addition in the group homomorphism definition. Analogously, we define the kernel and the image of a homomorphism: As well, a ring *isomorphism* is a homomorphism that is a bijection. Again, we would be interested in the properties of rings that are not changing under an isomorphism.

The following is why the ideals of are so important.

The kernel of a homomorphism is an ideal in

To see this, recall first that the kernel of the group homomorphism is a subgroup of Let We have to show that for any By the homomorphism property,

Given an ideal we can define a multiplication on the abelian group by setting To see that this is well-defined, we take and compute i.e. the product does not depend upon a particular choice of coset representatives of in We should also check the distributivity, certainly.

Unlike for quotient groups, the equation does not mean an equality of *sets*. Indeed, it can already happen that while (Such an example can already be obtained by taking )

It is a good point to discuss this phenomenon a bit more.

**Nilpotent elements and ideals**

A ring element is called *nilpotent* if there exists a natural number such that and the minimal such is called the *degree of nilpotency*. For instance is nilpotent, and has nilpotency degree 2. More generally, we can talk about a subset satisfying Such subsets, and in particular, ideals, are also called nilpotent.

**Exercise.** Prove that the sum of two nilpotent elements of a commutative ring is nilpotent. Derive that the set of all nilpotent elements of is an ideal.

Here I wrote more about various other operations on ideals.

Just as for abelian groups, we have an isomorphism theorem:

Let be a ring homomorphism, then s.t. is a ring isomorphism.

By the analogous theorem for groups, we already know that is a bijection. To prove the rest, we only need to check that for any Let and Then, as is a homomorphism, and so

**Examples.**

- Let be given by for a fixed prime This is a ring homomorphism, and thus
- Let and be given by Then this is a ring homomorphism, and thus
- Let and Then

**Multiplicative inverses.** (A bit of number theory.)

Let be a commutative ring with identity. How far is it from a field? In a field all the non-zero elements form an abelian (or better said, a commutative) group (w.r.t. to the multiplication), denoted

For instance, for one has isomorphic to the cyclic group of order This follows from a more general result, that we state here.

**Exercise** Any finite subgroup of is cyclic. Prove this using the observation that the polynomial equation has at most solutions in

More generally, for one has the number of integers between 1 and that are relatively prime to Thus in particular we have *Euler Theorem* for any such that

The isomorphism of the abelian groups is in fact also a ring isomorphism: In particular, this implies

**Excercise**. Show that where ‘s are the distinct prime divisors of

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