# Quotient Rings of Polynomial Rings

• Let F be a field. is a field if and only if is irreducible.

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 . For example,

If I take an arbitrary rational polynomial and apply the Division Algorithm to divide it by , I'll get a rational number as a remainder. For example,

The equation shows that and differ by a multiple of . So

Thus, mod every rational polynomial is congruent to a rational number.

This makes the following isomorphism reasonable:

To prove this, I'll use the First Isomorphism Theorem. Define by

That is, evaluates a polynomial at . Since

it follows that is a ring map.

I claim that .

First, let , where . Then

Therefore, , so .

Conversely, suppose that . Then , or . This means that is a root of , so is a factor of , by the Root Theorem. So for some , and .

This proves that , and hence .

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) In , and are zero divisors, because

This is a replay of the last part of the proof, and demonstrates in this case that is not a domain, hence not a field.

Example. ( A quotient ring which is a field) Consider . is irreducible in . Hence, is a field.

Now if this is really true, I ought to be able to take a nonzero element and find a multiplicative inverse. For example, I'll find the inverse of .

Apply the Extended Euclidean algorithm to and :

Therefore,

Hence,

Going mod , I get

Thus, is the reciprocal of .

Example. ( A field with 4 elements) Consider in . You can check that it has no roots, so it's irreducible. Therefore, is a field.

By the Division Algorithm, I can reduce every polynomial in to a polynomial mod . For example, take in . By the Division Algorithm,

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

Therefore,

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 .

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.

has no roots in :

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

(b) How many elements are there in ?

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

Therefore, has elements.

(c) Express

in the form , where .

By the Division Algorithm,

Therefore,

(d) Find .

Apply the Extended Euclidean algorithm:

Therefore,

Hence,

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Copyright 2007 by Bruce Ikenaga