In this section, I'll look at quotient rings of polynomial rings.

Let F be a field, and suppose . is the set of all
multiples (by polynomials) of , the * (principal) ideal
generated by* . When you form the
quotient ring , it is as if you've set multiples of equal to 0.

If , then is the * coset* of represented by .

Define ( is * congruent* to mod ) to mean that

In words, this means that and are congruent mod if they differ by a multiple of . In equation form, this says for some , or for some .

* Lemma.* Let R be a commutative ring, and
suppose
. Then if and only
if .

* Proof.* Suppose . Then for some . Hence,

Conversely, suppose . Then

Hence,

This means that .

Depending on the situation, I may write or .

* Example.* (* A quotient ring of
the rational polynomial ring*) Take in . Then two polynomials are
congruent mod if they differ by a
multiple of .

(a) Show that .

(b) Find a rational number r such that .

(c) Prove that .

(a)

(b) By the Remainder Theorem, when is divided by , the remainder is

Thus,

(c) I'll use the First Isomorphism Theorem. Define by

That is, evaluates a polynomial at . Note that

It follows that is a ring map.

I claim that . Now if and only if

That is, if and only if 2 is a root of f. By the Root Theorem, this is equivalent to , which is equivalent to .

Next, I'll show that is surjective. Let . I can think of q as a constant polynomial, and doing so, . Therefore, is surjective.

Using these results,

The first equality follows from the fact that . The isomorphism follows from the First Isomorphism Theorem. The second equality follows from the fact that is surjective.

In the last example, was a field. The next result says that this is the case exactly when is irreducible.

* Theorem.* is a field if and only if is irreducible.

* Proof.* Since is a commutative ring with identity, so is
.

Suppose is irreducible. I need to show that is a field. I need to show that nonzero elements are invertible.

Take a nonzero element of --- say , for . What does it mean for to be nonzero? It means that , so .

Now what is the greatest common divisor of and ? Well, , but is irreducible --- its only factors are units and unit multiples of .

Suppose , where and . Then , i.e. for some . But then shows that , contrary to assumption.

The only other possibility is that , where and . So I can find polynomials , , such that

Then

Hence,

This shows that is the multiplicative inverse of . Therefore, is invertible, and is a field.

Going the other way, suppose that is *not* irreducible. Then I can
find polynomials , such that , where and both have smaller degree than .

Because and have smaller degree than , they're not divisible by . In particular,

But gives

This shows that has zero divisors. Therefore, it's not an integral domain --- and since fields are integral domains, it can't be a field, either.

* Example.* (* A quotient ring
which is not an integral domain*) Prove that is not an integral domain by exhibiting a
pair of zero divisors.

and are zero divisors, because

* Example.* (* A quotient ring
which is a field*) (a) Show that is a field.

(b) Find the inverse of in .

(a) Since for all , it follows that has no rational roots. Hence, it's irreducible, and the quotient ring is a field.

(b) Apply the Extended Euclidean algorithm to and :

Therefore,

Hence,

Reducing mod , I get

Thus, is the inverse of .

* Example.* (* A field with 4
elements*) (a) Prove that is a field.

(b) Find so that

(c) Construct addition and multiplication tables for .

(a) Let . Then and . Since f has no roots in , it's irreducible. Hence, is a field.

(b) By the Division Algorithm,

This equation says that and x differ by a multiple of , so they represent the same coset mod .

Therefore,

(c) By the Division Algorithm, if , then

There are two possibilities for a and two for b, a total of 4. It follows that is a field with 4 elements. The elements are

Here are the addition and multiplication tables for :

The addition table is fairly easy to understand: For example, , because .

For the multiplication table, take as an example. ; I apply the Division Algorithm to get

So .

Alternatively, you can use the fact that in the quotient ring (omitting the coset notation), so (remember that in ).

* Remark.* In the same way, you can construct a
field of order for any prime n and
any . Just take and form the quotient ring , where is an irreducible polynomial of degree n.

* Example.* (* Computations in a
quotient ring*) (a) Show that is a field.

(b) How many elements are there in ?

(c) Compute

Express your answer in the form , where .

(d) Find .

(a) has no roots in :

Since is a cubic, it follows that it's irreducible. Hence, is a field.

(b) By the Division Algorithm, every element of can be written in the form

There are 3 choices each for a, b, and c. Therefore, has elements.

(c)

By the Division Algorithm,

Therefore,

(d) Apply the Extended Euclidean algorithm:

Therefore,

Hence,

Copyright 2018 by Bruce Ikenaga