The material in this chapter is fairly informal. Unlike earlier chapters, no

attempt is made to rigorously prove the results contained here

**definition 1.** Let R be a commutative ring, and x a variable. Then R[x]

is the set of polynomial
with coefficients

Example. Observe that
is in Z[x]. It is also in Q[x], in R[x],

and in C[x] since
Observe that
is in

Q[x] but not in Z[x]. Observe that
is in R[T] but not

in Q[T]. Observe that Z - i is in C[Z] but not in C[S].

If
is a polynomial with coefficients
,
we adopt

the convention that a_{i} = 0 for all values of i not occurring in the expression

For example, when writing
in the form

we consider
and

Two polynomials

are defined to be equal if and only if
for all

Example. Observe

Among the polynomials in R[x] are the constant polynomials a_{0}. In other

words,
can be thought of as both an element of R and as a constant

polynomial in R[x]. Thus

We define multiplication and addition of polynomials in the usual way.

For example, in
the product of
with
can be

computed as follows

Exercise 1. Multiply

The set R[x] is closed under addition and multiplication. So + and *

give two binary operations

**Lemma 1.** If R is a commutative ring, then addition and multiplication
are

associative and commutative on R[x]. In addition, the associative law holds.

**Lemma 2. **If R is a commutative ring, then the constant polynomial 0 is
an

additive identity for R[x], and the constant polynomial 1 is a multiplicative

identity for R[x].

**Theorem 3.** If R is a commutative ring, then R[x] is also a commutative

ring.

**definition 2.** If
then f(a) denotes what we get when we substitute

a for x in f. It is defined whenever the substitution makes sense (typically

when a is in R, or when a is in a ring containing R).

Example. If in then

Example. If
in
then
.
(Did

you see what happened to the linear term?).

Example. If
,
and y is another variable, then f(y) is in R[y] and has

the same coefficients. However, if x and y are di erent variables, then f(x)

is not considered to be equal to f(y) unless f is a constant polynomial.

Example. Let
Observe that f(x) is just f itself since when we

replace x with x we get what we started with. So f(x) is another way of

writing f. So we can write f as f(x) when we want to emphasize that f is

a polynomial in x.

Example. Here is an amusing example. Let
Then

So

polynomials cannot be treated as functions when R is nite: two distinct

polynomials, for example f and
as above, can have identical values. This

cannot happen for functions.

**definition 3** (Root of a polynomial). Let
and
If f(a) = 0

then a is called a root of

The above example (preceding the definition) shows that every element

of
is a root of

Exercise 2. Find the roots of in Find the roots of

Let F be a field. The ring of polynomials F[x] has a quotient-remainder

theorem. To state this theorem we need to discuss a notion of size for F[x]

traditionally called the degree:

definition 4. Let
where R is a commutative ring. If f has form

then the degree of f is defined to be n

and the leading coefficient is defined to be

If f = 0 then the degree of f is said to be undefined (some authors give

it degree
).

Be careful when using this definition in modular arithmetic. For example,

6x^{3} + 2x^{2} - x + 1 in F_{3}[x] has only degree 2, and 6x^{3} + 2x^{2}
- x + 1 in F2[x]

has degree 1. However, 6x^{3} + 2x^{2} - x + 1 in F_{5}[x] has degree 3

You would hope that the degree of fg would be the sum of the degrees of

f and g individually. However, examples such as (2x^{2} + 3x + 1)(3x^{2} + 2) =

3x^{3} +x^{2} +2: in Z_{6}[x] spoil our optimism. However, if the
coefficients are in

a field F then it works.

**Theorem 4**. If f,
are non-zero polynomials where F is a field,

then

deg(fg) = deg f + deg g:

Informal Exercise 3. Justify the above theorem. Explain why the proof does

not work if the coefficients are in Z_{m} where m is composite. Hint: focus on

the leading coefficients.

As mentioned above, the degree of a polynomial is a measure of size.

When we divide we want the size of the remainder to be smaller than the

size of the quotient. This leads to the following:

Theorem 5 (Quotient-Remainder Theorem). Let f,
be polynomials

where F is a field. Assume g is not zero. Then there are unique polynomials

q(x) and r(x) such that (i) f(x) = q(x)g(x) + r(x), and (ii) the polynomial

r(x) is either the zero polynomial or has degree strictly smaller than g(x).

Remark 1. The polynomial q(x) in the above is called the quotient and the

polynomial r(x) is called the remainder.

Remark 2. This theorem extends to polynomials in R[x] where R is a com-

mutative ring that is not a field, as long as we add the extra assumption

that the leading coefficient of g is a unit in R.

Remark 3. We could use this theorem as a basis to prove theorems about

GCD's and unique factorization in F[x].

As an important special case of the above theorem, consider g(x) = x-a

where a ∈,R. Then the remainder r(x) must be zero, or have
degree zero.

So r = r(x) is a constant polynomial. What is this constant? To find out,

write f(x) = q(x)(x - a) + r. When we substitute x = a we get

f(a) = q(a)(a - a) + r = 0 + r = r:

In other words, r = f(a). This gives the following:

Corollary 6. Let a ∈,F where F is a field, and let f ∈,F[x]. Then there is

a unique polynomial q ∈,F[x] such that

f(x) = (x - a)q(x) + f(a):

Remark 4. This actually works for commutative rings as well as for
fields F

since the leading coefficient of g(x) = x - a is 1 which is always a unit.

The following is a special case of the above corollary (where f(a) = 0).

Corollary 7. Let a ∈,F where F is a field, and let f ∈,F[x]. Then a is a

root of f if and only if (x - a) divides f.

Theorem 8. Let f ∈,F[x] be a nonzero polynomial with
coefficients in a

field F. Then f has at most n = deg f roots in F.

Proof. This is proved by induction. Let S be the set of natural numbers n

such that every polynomial f that has degree n has at most n roots in F.

Our goal is to show that S = N.

Showing 0 ∈S is easy. If f is a non-zero constant
polynomial of degree 0,

then it has 0 roots since it is a nonzero constant polynomial.

Suppose that k ∈S. We want to show k + 1
∈,S. To do so, let f be a

polynomial of degree k+1. If f has no roots, then the statement is trivially

true. Suppose that f does have a root a ∈,F. Then, by
Corollary 7,

f(x) = q(x)(x - a):

By Proposition 4, deg f = 1 + deg q. In other words, deg q = k. By the

inductive hypothesis k 2 S, the polynomial q has at most k roots.

We will now show that the only possible root of f that is not a root of

q is a (but a could also be a root of q). Suppose that f has a root b
≠, a.

Then 0 = f(b) = q(b)(b-a). Since b-a ≠, 0, we can multiply both sides by

the inverse: 0(b-a)-1 = q(b)(b-a)(b-a)^{-1}. Thus 0 = q(b). So every root

of f not equal to a must be a root of q(x). Since q(x) has at most k roots,

it follows that f(x) must have at most k + 1 roots. So k + 1 2 S.

By the principle of mathematical induction, N = S. The result follows.

Remark 5. Observe how this can fail if m is not a prime. The polynomial

x^{2} - 1 2 Z_{8} has degree 2, yet it has four roots! (Can you find them?)

Informal Exercise 4. Show that if f, g ∈,F[x]
are non-zero polynomials

where F is a field, then the set of roots of fg is the union of the set of roots

of f with the set of roots of g.

Exercise 5. Show that the result of the above exercise does not hold Z_{8}[x]

by looking at

Informal Exercise 6. Although the result of Exercise 4 does not hold if F is

replaced by a general commutative ring (such as Z_{m} where m is composite),

one of the two inclusions does hold. Which one and why?

One can prove unique factorization into irreducible polynomials for F[x].

A polynomial f ∈,F[x] is said to be
irreducible if it is not a constant and

if it has no divisors g with 0 <, deg g <, deg f. These polynomials play the

role of prime numbers in polynomial rings. One can use the methods of

Chapter 4 to prove that every polynomial is the product of a constant times

irreducible polynomials.

Finally, even if F is nite, one can prove that there are an in nite number

of irreducible polynomials in F[x] using a similar argument to that used in

showing that there are an in nite number of primes.

**Theorem 9 **(Fundamental Theorem of Algebra, Part 1). Every nonconstant

polynomial in C[X] has a root in C.

**Corollary 10.** Every non-constant polynomial with real or complex coe -

cients has a root in C.

**Corollary 11** (Fundamental Theorem of Algebra, Part 2). Every non-

constant polynomial in C[x] is the product of linear polynomials in C[x].

For real roots we get the following weaker results:

**Theorem 12.** Every polynomial of odd degree in R[x] has a root in R.

Real polynomials do not always factor into linear real polynomials. The

following weaker result is true:

**Theorem 13.** Every non-constant polynomial in R[x] factors into a product

of linear and quadratic real polynomials in R[x]

In other words, we have to allow for the possibility of quadratic factors

that have no real roots. The irreducible polynomials of R[x] are the lin-

ear polynomials and the quadratic polynomials with negative discriminate.

Contrast this with C[x] where the irreducible polynomials are just the linear

polynomials.

In Q[x] the situation is even worse. We can find polynomials of any degree

that have no roots in Q, and we can find polynomials of any degree that are

irreducible, and do not factor into smaller degree factors.

Exercise 7. Factor x^{4}-1 into irreducible polynomials in C[x]. Factor x^{4}-1

into irreducible polynomials in R[x].