Associates in Ring of Polynomial Forms over Field

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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 $d' \left({X}\right)$ if and only if $d \left({X}\right) = c \cdot d' \left({X}\right)$ 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.

$\blacksquare$


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