Polynomial Forms over Field is Euclidean Domain

Theorem

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

Let $X$ be transcendental in $F$.

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

Then $F \left[{X}\right]$ is a Euclidean domain.

Proof

From Degree of Product of Polynomials over Integral Domain we have that:

$\forall a, b \in F \left[{X}\right], a \ne 0_F, b \ne 0_F: \deg \left({a b}\right) \ge \deg \left({a}\right)$

where $\deg \left({a}\right)$ denotes the degree of $a$.

From Division Theorem for Polynomial Forms over a Field, we have that:

$\forall a, b \in F \left[{X}\right], b \ne 0_F: \exists q, r \in F \left[{X}\right]: a = q b + r$

where $\deg \left({r}\right) < \deg \left({b}\right)$ (or $r = 0$), and $q$ and $r$ are unique.

So $\deg$ is a Euclidean valuation on $F \left[{X}\right]$.

Hence the result.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.27$: Example $53$