Polynomial Rings over Fields

GCD for Polynomials

Let

R

be a domain, and let

f (x), g (x) \in R [x]

. The greatest common divisor (gcd) of

f (x)

and

g (x)

is a polynomial

d (x) \in R [x]

such that:

$d (x)$ is a common divisor of $f (x)$ and $g (x)$ ; that is, $d ∣ f$ and $d ∣ g$ ;
if $c (x)$ is any common divisor of $f (x)$ and $g (x)$ , then $c (x) ∣ d (x)$ ;
$d (x)$ is monic.

GCD for Polynomials Is Unique

As per title.

Note that gcd (denoted by $d$ ) of $f$ and $g$ , if it exists, is unique. If we had $d^{'}$ which is another gcd, then since it also divides both polynomials we know that $d^{'} ∣ d$ .

Similarly, $d ∣ d^{'}$ if one regards $d$ merely as a common divisor. By Exercise 4, $d^{'} = u d$ for some unit $u \in F [x]$ ; that is, $d^{'} = u d$ for some nonzero constant $u$ (prove why). Since $d$ and $d^{'}$ are both monic, however, $u = 1$ and $d^{'} = d$ .

GCD as a Linear Combination

Let

F

be a field and let

f (x), g (x) \in F [x]

with

g (x) \neq 0

. Then the

\gcd (f (x), g (x)) = d (x)

exists, and it is a linear combination of

f (x)

and

g (x)

; that is, there are polynomials

a (x)

and

b (x)

with

d (x) = a (x) f (x) + b (x) g (x)

Every Polynomial Ideal is Principal

F

is a field, then every ideal in

F [x]

is a principal ideal

Principal Ideal Domain

A crone

R

is called a principal ideal domain if it is a domain such that every ideal is a principal ideal.

Show that

Z

is a Principal Ideal Domain

As per title.

F

is a Field then

F [x]

is a Principal Ideal Domain

As per title.

Let $I$ be an ideal in $F [x]$ , if $I = {0}$ then $I = {(0)}_{⋄}$ , which shows that it is a principal ideal. Now suppose that $I \neq {0}$ , and let $m (x) \in I$ of smallest degree, we'll show that $I = {(m (x))}_{⋄}$ .

We can see that ${(m (x))}_{⋄} \subseteq I$ because for any element in an ideal, all of it's multiples are a part of it. Looking at ${(m (x))}_{⋄} \supseteq I$ , then given some $f (x) \in I$ then by the division algorithm we have $q (x), r (x)$ such that $f (x) = q (x) m (x) + r (x)$ where either $r (x) = 0$ or $\deg (r) < \deg (m)$ but then $r (x) = f (x) - q (x) m (x) \in I$ . Now if $\deg (r) < \deg (m)$ then we've found a polynomial with a smaller degree than $m (x)$ in $I$ (namely $r (x)$ ), so therefore we must have that $r (x) = 0$ which means that $f (x) = q (x) m (x) \in {(m (x))}_{⋄}$ .