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

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, \(d \mid f\) and \(d \mid g\);
- if \(c(x)\) is any common divisor of \(f(x)\) and \(g(x)\), then \(c(x) \mid d(x)\);
- \(d(x)\) 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 \(d^{\prime}\) which is another gcd, then since it also divides both polynomials we know that \(d^{\prime} \mid d\).

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

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 \(\operatorname{gcd}(f(x), g(x))=d(x)\) 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

If \( F \) is a field, then every ideal in \( F \left[ x \right] \) 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 \( \mathbb{ Z } \) is a Principal Ideal Domain

As per title.

If \( F \) is a Field then \( F \left[ x \right] \) is a Principal Ideal Domain

As per title.

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

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