A question on Euclidean Domains Let $D$ be a Euclidean domain whose function δ satisfies: 
$$
(1) \qquad δ(ab)=δ(a)δ(b)$$
$$
(2)\qquad δ(a+b)≤ \max(δ(a),δ(b)).
$$
Show that either $D$ is a field or $D=F[x]$, $F$ a field, $x$ an indeterminate.
Thanks for reading!
 A: First, recall that the elements which take on the minimum value of $\delta$ are the units [$u$ a unit iff $\delta(u) \leq \delta(x)$ for all non-zero $x$ in $D$.] 
Note that $\delta(1)=0$ implies that $\delta(a)=\delta(a)\delta(1)=0$. Thus if $\delta$ takes on the value 0, everything takes on the value $0$ and so $D$ is a field. Let's suppose $D$ is not a field. [Thus $\delta(1)>0$] Since $0<\delta(1)=\delta(1\cdot 1)=\delta(1)\delta(1)$ we must have $\delta(1)=1$.
Let $\mathbb{F}$ be the units of $D$ along with $0$. This set is closed under multiplication.
Suppose $a,b \in \mathbb{F}-\{0\}$. Either $a-b=0 \in \mathbb{F}$ or $a-b \not=0$ and so $\delta(a-b) \leq \mathrm{max}\{\delta(a),\delta(-b)\}=\delta(a)=\delta(-b)=\delta(1)=1$ [since $a$, $-b$, and $1$ are units which share the min value of $\delta$]. Thus $\mathbb{F}$ is closed under subtraction. Thus it is a field (all its non-zero elements are units).
Let $0 \not= t \in D$ be a non-unit element of minimal $\delta$-value (the well ordering principal says such an $t$ exists). Now $t$ is not a unit so $0<1=\delta(1)<\delta(t)$ and so $\delta(t)>1$. This implies that $\delta(t)<\delta(t)\delta(t)=\delta(t^2)<\delta(t)\delta(t)\delta(t)=\delta(t^3)<\cdots$. Now if $t$ were algebraic over $\mathbb{F}$, it would be the root of some monic polynomial $g(x)\in\mathbb{F}[x]$, say $g(x)=x^m+b_{m-1}x^{m-1}+\cdots+b_0=x^m+h(x)$. $\delta(-h(t))$ is at most the maximum of all $\delta(-b_kt^k)=\delta(-b_k)\delta(t^k)=\delta(t)^k$ (recall $-b_k$ is a unit so $\delta(-b_k)=1$) so this is at most $\delta(t)^{m-1}$. Thus we have found $t^m=-h(t)$ and $\delta(t^m)=\delta(t)^m \leq \delta(t)^{m-1}$ (contradiction). Therefore, $t$ is transcendental over $\mathbb{F}$. Therefore, $\mathbb{F}[t]$ is a ring of polynomials over $\mathbb{F}$.
Finally, suppose $0 \not= a \in D$. Dividing by $t$ gives $a=qt+r_0$ where either $r_0=0$ or $\delta(r_0)<\delta(t)$. But $\delta(t)$ had minimal $\delta$-value among non-units. Thus either $r_0=0$ or $r_0$ is a unit, so $r_0 \in \mathbb{F}$. Next, note that $\delta(q) < \delta(q)\delta(t)=\delta(qt)=\delta(a-r_0)\leq \delta(a)$ (by property (2)). Now divide $q$ by $t$ and continue inductively. Eventually $\delta$ bottoms out and this process stops. We have just shown that $a=(\cdots (r_\ell t+r_{\ell-1})t\cdots)t+r_0 \in \mathbb{F}[t]$. Therefore, $D=\mathbb{F}[t]$.
