If R is a ring, the * ring of polynomials in x with
coefficients in R* is denoted . It consists of all
formal sums

Here for all but finitely many values of i.

If the idea of "formal sums" worries you, replace a formal sum with the infinite vector whose components are the coefficients of the sum:

All of the operations which I'll define using formal sums can be defined using vectors. But it's traditional to represent polynomials as formal sums, so this is what I'll do.

A nonzero polynomial has * degree n* if and , and n is the largest integer
with this property. The zero polynomial is defined by convention to
have degree . (This is necessary in order to make the
degree formulas work out.) Alternatively, you can say that the degree
of the zero polynomial is undefined; in that case, you will need to
make minor changes to some of the results below.

Polynomials are added componentwise, and multiplied using the "convolution" formula:

These formulas say that you compute sums and products as usual.

* Example.* (* Polynomial
arithmetic*) (a) Compute

(b) Compute

(a)

(b)

Let R be an integral domain. Then If , write to denote the degree of f. It's easy to show that the degree function satisfies the following properties:

The verifications amount to writing out the formal sums, with a
little attention paid to the case of the zero polynomial. These
formulas *do* work if either f or g is equal to the zero
polynomial, provided that is understood to
behave in the obvious ways (e.g. for
any ).

* Example.* (* Degrees of
polynomials)* (a) Give examples of polynomials such that

(b) Give examples of polynomials such that .

(a)

This shows that equality might not hold in .

(b)

* Proposition.* Let F be a field, and let be the polynomial ring in one variable over F. The
units in are exactly the nonzero elements of F.

* Proof.* It's clear that the nonzero elements of
F are invertible in , since they're already invertible
in F. Conversely, suppose that is
invertible, so for some . Then , which
is impossible unless f and g both have degree 0. In particular, f is
a nonzero constant, i.e. an element of F.

* Theorem.* (* Division
Algorithm*) Let F be a field, and let . Suppose that . There are unique
polynomials such that

* Proof.* The idea is to imitate the proof of the
Division Algorithm for .

Let

The set is a subset of the nonnegative integers, and therefore must contain a smallest element by well-ordering. Let be an element in S of smallest degree, and write

I need to show that .

If , then since , I have .

Suppose then that . Assume toward a contradiction that . Write

Assume , and .

Consider the polynomial

Its degree is less than n, since the n-th degree terms cancel out.

However,

The latter is an element of S.

I've found an element of S of smaller degree than , which is a contradiction. It follows that .

Finally, to prove uniqueness, suppose

Rearranging the equation, I get

Then

But . The equation can only hold if

This means

Hence, and .

* Example.* (* Polynomial
division*) Divide by in .

Remember as you follow the division that , , and --- I'm doing arithmetic mod 5.

If you prefer, you can do long division without writing the powers of x --- i.e. just writing down the coefficients. Here's how it looks:

Either way, the quotient is and the remainder is :

* Definition.* Let R be a commutative ring and
let . An element is a * root* of if .

Note that polynomials are actually formal sums, not functions. However, it is obvious how to plug a number into a polynomial. Specifically, let

For , define

Observe that a polynomial can be * nonzero as a
polynomial* even if it equals 0 for every input! For example,
take is a nonzero
polynomial. However, plugging in the two elements of the coefficient
ring gives

* Theorem.* Let F be a field, and let , where .

(a) (* The Root Theorem*) c is a root of in F if and only if .

(b) has at most n roots in F.

* Proof.* (a) Suppose . Write

Then or .

In the first case, r is a nonzero constant. However, this implies that

This contradiction shows that , and .

Conversely, if is a factor of , then for some . Hence,

Hence, c is a root of f.

(b) If are the distinct roots of f in F, then

Taking degrees on both sides gives .

* Example.* (* Applying the Root
Theorem*) In , show:

(a) is a factor of .

(b) is a factor of for any .

(a) If , then . Hence, 1 is a root of , and by the Root Theorem is a factor of .

(b) If , then . Hence, 1 is a root of , so is a factor of by the Root Theorem.

* Example.* (* Applying the Root
Theorem*) Prove that is
divisible by in .

Plugging in into gives

Since is a root, is a factor by the Root Theorem.

* Remark.* If the ground ring isn't a field, it's
possible for a polynomial to have more roots than its degree. For
example, the quadratic polynomial has roots , , , . The previous result does not apply, because is not a field.

* Corollary.* (* The Remainder
Theorem*) Let F be a field, , and let . When is divided by , the remainder is .

* Proof.* Divide by :

Since , it follows that is a constant. But

Therefore, the constant value of is .

* Example.* (* Applying the
Remainder Theorem*) Suppose leaves a
remainder of 5 when divided by and a remainder of
-1 when divided by . What is the remainder when is divided by ?

By the Remainder Theorem,

Now divide by . The remainder has degree less than , so for some :

Then

Solving the two equations for a and b, I get and . Thus, the remainder is .

* Definition.* Let R be an integral domain.

(a) If , then x * divides* y
if for some . Write to mean that x divides y.

(b) x and y are * associates* if , where u is a unit.

(Recall that a * unit* in a ring is an element
with a multiplicative inverse.)

(c) An element is *
irreducible* if , x is not a unit, and if implies either y is a unit or z is a unit.

(d) An element is * prime* if
, x is not a unit, and implies or .

* Proposition.* A nonzero nonconstant polynomial
is irreducible if and only if implies that either g or h is a constant.

* Proof.* Suppose is irreducible and . Then
one of , is a unit. But we
showed earlier that the units in are the constant
polynomials.

Suppose that is a nonzero nonconstant polynomial, and implies that either g or h is a constant.

Since f is nonconstant, it's not a unit. Note that if , then , since .

Therefore, the condition that implies that either g or h is a constant means that implies that either or is a unit --- again, since the nonzero constant polynomials are the units in . This is what it means for f to be irreducible.

* Example.* Show that is irreducible in but not in .

has no real roots, so by the Root Theorem it has no linear factors. Hence, it's irreducibile in .

However, in .

* Corollary.* Let F be a field. A polynomial of
degree 2 or 3 in is irreducible if and only if it
has no roots in F.

* Proof.* Suppose has degree 2 or 3.

If f is not irreducible, then , where neither g nor h is constant. Now and , and

This is only possible if at least one of g or h has degree 1. This means that at least one of g or h is a linear factor , and must therefore have a root in F. Since , it follows that f has a root in F as well.

Conversely, if f has a root c in F, then is a factor of f by the Root Theorem. Since f has degree 2 or 3, is a proper factor, and f is not irreducible.

* Remark.* The result is false for polynomials of
degree 4 or higher. For example, has no roots
in , but it is not irreducible over .

* Example.* (* Checking for
irreducibility of a quadratic or cubic*) Show that is irreducible.

Since this is a cubic polynomial, I only need to see whether it has any roots.

Since has no roots in , it's irreducible.

* Proposition.* In an integral domain, primes are
irreducible.

* Proof.* Let x be prime. I must show x is
irreducible. Suppose . I must show either y or z is
a unit.

, so obviously . Thus, or . Without loss of generality, suppose .

Write . Then , and since (primes are nonzero) and we're in a domain, . Therefore, z is a unit, and x is irreducible.

* Definition.* Let R be an integral domain, and
let . is a * greatest common divisor* of x and y if:

(a) and .

(b) If and , then .

The definition says "a" greatest common divisor, rather than "the" greatest common divisor, because greatest common divisors are only unique up to multiplication by units.

The definition above is the right one if you're dealing with an
arbitrary integral domain. However, if your ring is a polynomial
ring, it's nice to single out a "special" greatest common
divisor and call it *the* greatest common divisor.

* Definition.* A * monic
polynomial* is a polynomial whose leading coefficient is 1.

For example, here are some monic polynomials over :

* Definition.* Let F be a field, let be the ring of polynomials with coefficients in F,
and let , where f and g are not both zero.
*The* * greatest common divisor* of f and
g is the monic polynomial which is a greatest common divisor of f and
g (in the integral domain sense).

* Example.* (* Polynomial
greatest common divisors*) Find the greatest common divisor of
and in .

is a greatest common divisor of and :

Notice that any nonzero constant multiple of is also a greatest common divisor of and (in the integral domain
sense): For example,
works. This makes sense, because the units in are the nonzero elements of . But by convention, I'll refer to --- the monic greatest common divisor --- as
*the* greatest common divisor of and .

The preceding definition assumes there *is* a greatest common
divisor for two polynomials in . In fact, the
greatest common divisor of two polynomials exists --- provided that
both polynomials aren't 0 --- and the proof is essentially the same
as the proof for greatest common divisors of integers.

In both cases, the idea is to use the Division Algorithm repeatedly
until you obtain a remainder of 0. This must happen in the polynomial
case, because the Division Algorithm for polynomials specifies that
the remainder has strictly smaller *degree* than the divisor.

Just as in the case of the integers, each use of the Division Algorithm does not change the greatest common divisor. So the last pair has the same greatest common divisor as the first pair --- but the last pair consists of 0 and the last nonzero remainder, so the last nonzero remainder is the greatest common divisor.

This process is called the * Euclidean
algorithm*, just as in the case of the integers.

Let h and be two greatest common divisors of f and g.
By definition, and . From this, it follows that h and have the same degree, and are constant multiples of
one another. If h and are both *monic* --- i.e.
both have leading coefficient 1 --- this is only possible if they're
equal. So there is a *unique* monic greatest common divisor
for any two polynomials.

Finally, the same proofs that I gave for the integers show that you
can write the greatest common divisor of two polynomials as a linear
combination of the two polynomials. You can use the *
Extended Euclidean Algorithm* that you learned for integers to
find a linear combination. To summarize:

* Theorem.* Let F be a field, , f and g not both 0.

(a) f and g have a unique (monic) greatest common divisor.

(b) There exist polynomials such that

* Example.* (* Applying the
Extended Euclidean Algorithm*) Find the greatest common divisor
of and in and express the
greatest common divisor as a linear combination of and with coefficients in .

The greatest common divisor is . The greatest common divisor is only determined up to multiplying by a unit, so multiplying by gives the monic greatest common divisor .

You can check that

* Example.* (* Applying the
Extended Euclidean Algorithm*) Find the greatest common divisor
of and in and express the greatest common divisor as
a linear combination of and with coefficients in .

The greatest common divisor is , and

The greatest common divisor is only determined up to multiplying by a unit. So, for example, I can multiply the last equation by 2 to get

Copyright 2018 by Bruce Ikenaga