Proof that K[x] is a Euclidean domain

Proof

Let $K$ be a field.

1. The field $K$ is an integral domain: $1 \neq 0$, and if $a b = 0$ with $a \neq 0$, then $b = a^{-1} a b = 0$.
2. By the theorem on polynomial rings, $K[x]$ is an integral domain.
3. For nonzero $h \in K[x]$, the degree $\deg h$ is an integer at least $0$. So $\delta(h) = \deg h$ defines a function $\delta \colon K[x] \setminus \set{0} \to \N$.
4. Let $f, g \in K[x]$ with $g \neq 0$. Let $c$ be the leading coefficient of $g$, so $c \neq 0$ and $c$ has an inverse $c^{-1}$ in $K$.
5. Put $g_1 = c^{-1} g$. Its coefficients are those of $g$ times $c^{-1}$. Since $c^{-1} a = 0$ only for $a = 0$, the polynomial $g_1$ has degree $\deg g$ and leading coefficient $c^{-1} c = 1$.
6. So $g_1$ is monic. By the division algorithm in $K[x]$, there are $q_1, r \in K[x]$ with $f = q_1 g_1 + r$ and $\deg r < \deg g_1 = \deg g$.
7. Put $q = c^{-1} q_1$. Then $q g = q_1 c^{-1} g = q_1 g_1$, so $f = q g + r$.
8. If $r \neq 0$, then step 6 gives $\delta(r) = \deg r < \deg g = \delta(g)$. So $r = 0$ or $\delta(r) < \delta(g)$.
9. By steps 2, 3 and 8, $K[x]$ is a Euclidean domain with the degree as Euclidean function.