Polynomial Forms over Field is Euclidean Domain

Theorem

Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental in $F$.

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

Then $F \sqbrk X$ is a Euclidean domain.

Proof

From Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors we have that:

$\forall a, b \in F \sqbrk X, a \ne 0_F, b \ne 0_F: \map \deg {a b} \ge \map \deg a$

where $\map \deg a$ denotes the degree of $a$.

From Division Theorem for Polynomial Forms over Field:

$\forall a, b \in F \sqbrk X, b \ne 0_F: \exists q, r \in F \sqbrk X: a = q b + r$

where $\map \deg r < \map \deg b$ (or $r = 0$), and $q$ and $r$ are unique.

So $\deg$ is a Euclidean valuation on $F \sqbrk X$.

Hence the result.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Example $53$