Associates in Ring of Polynomial Forms over Field

Theorem
Let $F \left[{X}\right]$ be the ring of polynomial forms over the field $F$.

Let $d \left({X}\right)$ and $d\,' \left({X}\right)$ be polynomial forms in $F \left[{X}\right]$.

Then $d \left({X}\right)$ is an associate of $d\,' \left({X}\right)$ iff $d \left({X}\right) = c \cdot d\,'\left({X}\right)$ for some $c \in F, c \ne 0_F$.

Hence any two polynomials in $F \left[{X}\right]$ have a unique monic GCD.

Proof
From the definition of associate, there exist $e \left({X}\right)$ and $e\,' \left({X}\right)$ \in $F \left[{X}\right]$ such that:


 * $d \left({X}\right) = e \left({X}\right) \cdot d\,' \left({X}\right)$
 * $d\,' \left({X}\right) = e\,' \left({X}\right) \cdot d \left({X}\right)$

From Field is Integral Domain, $F$ is an integral domain.

From Degree of Product of Polynomials over Integral Domain it follows that necessarily $\deg e = \deg e\,' = 0$, as $F$ has no proper zero divisors.

Thus for some $c, c\,' \in F$, it must be that $e \left({X}\right) = c$ and $c\,' \left({X}\right) = c\,'$.

From the two equations above it follows that $c \cdot c\,' = 1_F$, where $1_F$ is the unity of $F$.

Hence, it follows that $c \ne 0_F$.

The result follows.