AATA Polynomial Rings

Section 17.1 Polynomial Rings

Throughout this chapter we shall assume that

R

is a commutative ring with identity. Any expression of the form

f (x) = \sum_{i = 0}^{n} a_{i} x^{i} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n},

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where

a_{i} \in R

and

a_{n} \neq 0,

is called a polynomial over

R

with indeterminate

x .

The elements

a_{0}, a_{1}, \dots, a_{n}

are called the coefficients of

f .

The coefficient

a_{n}

is called the leading coefficient. A polynomial is called monic if the leading coefficient is 1. If

n

is the largest nonnegative number for which

a_{n} \neq 0,

we say that the degree of

f

n

and write

\deg f (x) = n .

If no such

n

exists—that is, if

f = 0

is the zero polynomial—then the degree of

f

is defined to be

- \infty .

We will denote the set of all polynomials with coefficients in a ring

R

R [x] .

Two polynomials are equal exactly when their corresponding coefficients are equal; that is, if we let

\begin{aligned} p (x) & = a_{0} + a_{1} x + \dots + a_{n} x^{n} \\ q (x) & = b_{0} + b_{1} x + \dots + b_{m} x^{m}, \end{aligned}

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then

p (x) = q (x)

if and only if

a_{i} = b_{i}

for all

i \geq 0 .

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To show that the set of all polynomials forms a ring, we must first define addition and multiplication. We define the sum of two polynomials as follows. Let

\begin{aligned} p (x) & = a_{0} + a_{1} x + \dots + a_{n} x^{n} \\ q (x) & = b_{0} + b_{1} x + \dots + b_{m} x^{m} . \end{aligned}

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Then the sum of

p (x)

and

q (x)

p (x) + q (x) = c_{0} + c_{1} x + \dots + c_{k} x^{k},

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where

c_{i} = a_{i} + b_{i}

for each

i .

We define the product of

p (x)

and

q (x)

to be

p (x) q (x) = c_{0} + c_{1} x + \dots + c_{m + n} x^{m + n},

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where

c_{i} = \sum_{k = 0}^{i} a_{k} b_{i - k} = a_{0} b_{i} + a_{1} b_{i - 1} + \dots + a_{i - 1} b_{1} + a_{i} b_{0}

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for each

i .

Notice that in each case some of the coefficients may be zero.

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Example 17.1.

Suppose that

p (x) = 3 + 0 x + 0 x^{2} + 2 x^{3} + 0 x^{4}

and

q (x) = 2 + 0 x - x^{2} + 0 x^{3} + 4 x^{4}

are polynomials in $Z [x] .$ If the coefficient of some term in a polynomial is zero, then we usually just omit that term. In this case we would write $p (x) = 3 + 2 x^{3}$ and $q (x) = 2 - x^{2} + 4 x^{4} .$ The sum of these two polynomials is

p (x) + q (x) = 5 - x^{2} + 2 x^{3} + 4 x^{4} .

The product,

p (x) q (x) = (3 + 2 x^{3}) (2 - x^{2} + 4 x^{4}) = 6 - 3 x^{2} + 4 x^{3} + 12 x^{4} - 2 x^{5} + 8 x^{7},

can be calculated either by determining the $c_{i}$ s in the definition or by simply multiplying polynomials in the same way as we have always done.

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Example 17.2.

Let

p (x) = 3 + 3 x^{3} and q (x) = 4 + 4 x^{2} + 4 x^{4}

be polynomials in $Z_{12} [x] .$ The sum of $p (x)$ and $q (x)$ is $7 + 4 x^{2} + 3 x^{3} + 4 x^{4} .$ The product of the two polynomials is the zero polynomial. This example tells us that we can not expect $R [x]$ to be an integral domain if $R$ is not an integral domain.

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Theorem 17.3.

Let $R$ be a commutative ring with identity. Then $R [x]$ is a commutative ring with identity.

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Proof.

Our first task is to show that $R [x]$ is an abelian group under polynomial addition. The zero polynomial, $f (x) = 0,$ is the additive identity. Given a polynomial $p (x) = \sum_{i = 0}^{n} a_{i} x^{i},$ the inverse of $p (x)$ is easily verified to be $- p (x) = \sum_{i = 0}^{n} (- a_{i}) x^{i} = - \sum_{i = 0}^{n} a_{i} x^{i} .$ Commutativity and associativity follow immediately from the definition of polynomial addition and from the fact that addition in $R$ is both commutative and associative.

To show that polynomial multiplication is associative, let

\begin{aligned} p (x) & = \sum_{i = 0}^{m} a_{i} x^{i}, \\ q (x) & = \sum_{i = 0}^{n} b_{i} x^{i}, \\ r (x) & = \sum_{i = 0}^{p} c_{i} x^{i} . \end{aligned}

Then

\begin{aligned} [p (x) q (x)] r (x) & = [(\sum_{i = 0}^{m} a_{i} x^{i}) (\sum_{i = 0}^{n} b_{i} x^{i})] (\sum_{i = 0}^{p} c_{i} x^{i}) \\ = [\sum_{i = 0}^{m + n} (\sum_{j = 0}^{i} a_{j} b_{i - j}) x^{i}] (\sum_{i = 0}^{p} c_{i} x^{i}) \\ = \sum_{i = 0}^{m + n + p} [\sum_{j = 0}^{i} (\sum_{k = 0}^{j} a_{k} b_{j - k}) c_{i - j}] x^{i} \\ = \sum_{i = 0}^{m + n + p} (\sum_{j + k + l = i} a_{j} b_{k} c_{l}) x^{i} \\ = \sum_{i = 0}^{m + n + p} [\sum_{j = 0}^{i} a_{j} (\sum_{k = 0}^{i - j} b_{k} c_{i - j - k})] x^{i} \\ = (\sum_{i = 0}^{m} a_{i} x^{i}) [\sum_{i = 0}^{n + p} (\sum_{j = 0}^{i} b_{j} c_{i - j}) x^{i}] \\ = (\sum_{i = 0}^{m} a_{i} x^{i}) [(\sum_{i = 0}^{n} b_{i} x^{i}) (\sum_{i = 0}^{p} c_{i} x^{i})] \\ = p (x) [q (x) r (x)] \end{aligned}

The commutativity and distribution properties of polynomial multiplication are proved in a similar manner. We shall leave the proofs of these properties as an exercise.

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Proposition 17.4.

Let $p (x)$ and $q (x)$ be polynomials in $R [x],$ where $R$ is an integral domain. Then $\deg p (x) + \deg q (x) = \deg (p (x) q (x)) .$ Furthermore, $R [x]$ is an integral domain.

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Proof.

Suppose that we have two nonzero polynomials

p (x) = a_{m} x^{m} + \dots + a_{1} x + a_{0}

and

q (x) = b_{n} x^{n} + \dots + b_{1} x + b_{0}

with $a_{m} \neq 0$ and $b_{n} \neq 0 .$ The degrees of $p (x)$ and $q (x)$ are $m$ and $n,$ respectively. The leading term of $p (x) q (x)$ is $a_{m} b_{n} x^{m + n},$ which cannot be zero since $R$ is an integral domain; hence, the degree of $p (x) q (x)$ is $m + n,$ and $p (x) q (x) \neq 0 .$ Since $p (x) \neq 0$ and $q (x) \neq 0$ imply that $p (x) q (x) \neq 0,$ we know that $R [x]$ must also be an integral domain.

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We also want to consider polynomials in two or more variables, such as

x^{2} - 3 x y + 2 y^{3} .

Let

R

be a ring and suppose that we are given two indeterminates

x

and

y .

Certainly we can form the ring

(R [x]) [y] .

It is straightforward but perhaps tedious to show that

(R [x]) [y] ≅ R ([y]) [x] .

We shall identify these two rings by this isomorphism and simply write

R [x, y] .

The ring

R [x, y]

is called the ring of polynomials in two indeterminates

x

and

y

with coefficients in

R .

We can define the ring of polynomials in

n

indeterminates with coefficients in

R

similarly. We shall denote this ring by

R [x_{1}, x_{2}, \dots, x_{n}] .

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Theorem 17.5.

Let $R$ be a commutative ring with identity and $α \in R .$ Then we have a ring homomorphism $ϕ_{α} : R [x] \to R$ defined by

ϕ_{α} (p (x)) = p (α) = a_{n} α^{n} + \dots + a_{1} α + a_{0},

where $p (x) = a_{n} x^{n} + \dots + a_{1} x + a_{0} .$

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Proof.

Let $p (x) = \sum_{i = 0}^{n} a_{i} x^{i}$ and $q (x) = \sum_{i = 0}^{m} b_{i} x^{i} .$ It is easy to show that $ϕ_{α} (p (x) + q (x)) = ϕ_{α} (p (x)) + ϕ_{α} (q (x)) .$ To show that multiplication is preserved under the map $ϕ_{α},$ observe that

\begin{aligned} ϕ_{α} (p (x)) ϕ_{α} (q (x)) & = p (α) q (α) \\ = (\sum_{i = 0}^{n} a_{i} α^{i}) (\sum_{i = 0}^{m} b_{i} α^{i}) \\ = \sum_{i = 0}^{m + n} (\sum_{k = 0}^{i} a_{k} b_{i - k}) α^{i} \\ = ϕ_{α} (p (x) q (x)) . \end{aligned}

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The map

ϕ_{α} : R [x] \to R

is called the evaluation homomorphism at

α .

Abstract Algebra: Theory and Applications

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Section 17.1 Polynomial Rings

Example 17.1.

Example 17.2.

Theorem 17.3.

Proof.

Proposition 17.4.

Proof.

Theorem 17.5.

Proof.