Let R be a ring, and let I be a (two-sided) ideal. Considering just
the operation of addition, R is a group and I is a subgroup. In fact,
since R is an *abelian* group under addition, I is a
*normal* subgroup, and the quotient group is defined. Addition of cosets is defined by
*adding* coset representatives:

The zero coset is , and the additive inverse of a coset is given by .

However, R also comes with a multiplication, and it's natural to ask
whether you can turn into a ring
by *multiplying* coset representatives:

I need to check that that this operation is well-defined, and that
the ring axioms are satisfied. In fact, everything works, and you'll
see in the proof that it depends on the fact that I is an
*ideal*. Specifically, it depends on the fact that I is closed
under multiplication by elements of R.

By the way, I'll sometimes write " " and sometimes " "; they mean the same thing.

* Theorem.* If I is a two-sided ideal in a ring
R, then has the structure of a ring under coset addition and
multiplication.

* Proof.* Suppose that I is a two-sided ideal in
R. Let .

Coset addition is well-defined, because R is an abelian group and I a normal subgroup under addition. I proved that coset addition was well-defined when I constructed quotient groups.

I need to show that coset multiplication is well-defined:

As before, suppose that

Then

The next-to-last equality is derived as follows: , because I is an ideal; hence . Note that this uses the
multiplication axiom for an ideal; in a sense, it explains why the
multiplication axiom requires that an ideal be closed under
multiplication *by ring elements on the left and right*.

Thus, coset multiplication is well-defined.

Verification of the ring axioms is easy but tedious: It reduces to the axioms for R.

For instance, suppose I want to verify associativity of multiplication. Take . Then

(Notice how I used associativity of multiplication in R in the middle of the proof.) The proofs of the other axioms are similar.

* Definition.* If R is a ring and I is a
two-sided ideal, the * quotient ring* of R mod I
is the group of cosets with the
operations of coset addition and coset multiplication.

* Proposition.* Let R be a ring, and let I be an
ideal

(a) If R is a commutative ring, so is .

(b) If R has a multiplicative identity 1, then is a multiplicative identity for . In this case, if is a unit, then so is , and .

* Proof.* (a) Let .
Since R is commutative,

Therefore, is commutative.

(b) Suppose R has a multiplicative identity 1. Let . Then

Therefore, is the identity of .

If is a unit, then

Therefore, .

* Example.* (* A quotient ring of
the integers*) The set of even integers is an ideal in . Form the quotient ring .

Construct the addition and multiplication tables for the quotient ring.

Here are some cosets:

But two cosets and are the same exactly when a and b differ by an even integer. Every even integer differs from 0 by an even integer. Every odd integer differs from 1 by an even integer. So there are really only two cosets (up to renaming): and .

Here are the addition and multiplication tables:

You can see that is isomorphic to .

In general, is isomorphic to . I've been using " " informally to mean the set with addition and multiplication mod n, and taking for granted that the usual ring axioms hold. This example gives a formal contruction of as the quotient ring .

* Example.* is the ring
of polynomials with coefficients in . Consider the
ideal .

(a) How many elements are in the quotient ring ?

(b) Reduce the following product in to the form :

(c) Find in .

The ring is analogous to . In the case of , you do computations mod n: To "simplify", you divide the result of a computation by the modulus n and take the remainder. In , the polynomial acts like the "modulus". To do computations in , you divide the result of a computation by and take the remainder.

(a) By the Division Algorithm, any can be written as

This means that , where . Then

Since there are 3 choices for a and 3 choices for b, there are 9 cosets.

(b) First, multiply the coset representatives:

Dividing by , I get

Then

(c) To find multiplicative inverses in , you use the Extended Euclidean Algorithm. The same idea works in quotient rings of polynomial rings.

Thus,

* Example.* (a) List the elements of the cosets
of in the ring .

(b) Is the quotient ring an integral domain?

(a) If x is an element of a ring R, the ideal consists of all multiples of x by elements of R. It is not necessarily the same as the additive subgroup generated by x, which is

In this example, the additive subgroup generated by is

As usual, I get it by starting with the zero element and the generator , then adding until I get back to .

This set is *contained* in the ideal ; I need to check whether it is *the
same* as the ideal.

If , then

Thus, an element of the ideal consists of a pair , where each component is even. There are two even elements in (namely 0 and 2) and 3 even elements in (namely 0, 2, and 4), so there are such pairs. Thus, the ideal has a maximum of 6 elements. Since the additive subgroup above already has 6 elements, it must be the same as the ideal.

I can list the elements of the cosets of the ideal as I would for subgroups.

(b) Note that

Hence, is not an integral domain.

* Example.* In the ring , consider the principal ideal
.

(a) List the elements of .

(b) List the elements of the cosets of .

(c) Is the quotient ring a field?

(a) Note that the additive subgroup generated by has only two elements. It's not the same as the ideal generated by , so I can't find the elements of the ideal by taking additive multiples of . I'll find the elements of the ideal by multiplying by the elements of , then throwing out duplicates. The computation is routine, if a bit tedious.

Removing duplicates, I have

(b) Since the ideal has 4 elements and the ring has 20, there must be 5 cosets.

(c) Note that is the identity.

Since every nonzero coset has a multiplicative inverse, the quotient ring is a field.

Copyright 2018 by Bruce Ikenaga