Associates in Ring of Polynomial Forms over Field

Theorem

Let $F \sqbrk X$ be the ring of polynomial forms over the field $F$.

Let $\map d X$ and $\map {d'} X$ be polynomial forms in $F \sqbrk X$.

Then $\map d X$ is an associate of $\map {d'} X$ if and only if $\map d X = c \cdot \map {d'} X$ for some $c \in F, c \ne 0_F$.

Hence any two polynomials in $F \sqbrk X$ have a unique monic GCD.

Proof

From the definition of associate, there exist $\map e X$ and $\map {e'} X$ \in $F \sqbrk X$ such that:

$\map d X = \map e X \cdot \map {d'} X$

$\map {d'} X = \map {e'} X \cdot \map d X$

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 $\map e X = c$ and $\map {e'} X = c'$.

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.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 28$. Highest Common Factor: Example $55$