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 $$f \left( x \right) = q \left( x \right) m \left( x \right) \in \left( m \left( x \right) \right) _ \diamond$$.