= Polynomial ring over a field is Euclidean
{title2=$\nu(f)=\deg f\quad(f\ne0)$}
For a <field> $F$, <polynomial division> makes the one-variable <polynomial ring> $F[X]$ a <Euclidean domain> with <Euclidean function> equal to the <polynomial degree>. For a divisor $g\ne0$, subtract a suitable scalar multiple of $X^{\deg f-\deg g}g$ to cancel the <leading coefficient> of the current remainder. Its degree strictly decreases. Iteration terminates at $f=qg+r$, where $r=0$ or $\deg r<\deg g$. Division by the leading coefficient is valid because it is nonzero in a <field>.
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