Want to show that $\mathbb{F}[x]$, $\Bbb{F}$ a field, is a Euclidean ring Let $d$ be a mapping such that $d(f) = 1$ for all $f \in \mathbb{F}[x]$
Take $f(x),g(x)  \in \mathbb{F}[x]$
Then $d(f) = 1 \le d(fg) = 1$
Now, we want to find $q, r$ such that 
(*) $f(x) = q(x) \cdot g(x) + r(x)$ with $r(x) = 0$ or $d(r) < d(g)$
Let $f(x) = q(x) \cdot g(x) + r(x)$
If $r(x) = 0$ then we have $f(x) = q(x) \cdot g(x)$
and we can choose $q(x)$ to be an associate of $f(x)$ and (*) holds.
Now suppose $d(r) \ge d(g)$ and consider
$r(x) = f(x) - q(x) \cdot g(x)$
I think I'm on the right track to come up with a contradiction but I can't see where to go from here?
 A: Indeed, postulating $d(\cdot)=1$ identically forces $d(r)=0$ in every case. However given $f(x)$ and $g(x)$ you cannot in general choose an associate $q(x)$ of $f(x)$ for which $$f(x)=q(x)g(x)+0$$
holds. For example if $f(x)=x$ and $g(x)=x^2$ you're saying $x=\lambda x^2$ for some scalar $\lambda\in\Bbb F$, because the units of $\Bbb F[x]$ are the scalars from $\Bbb F$, but this is impossible. Indeed no $q(x)$ fits the bill here, associate to $f(x)$ or otherwise.
One thing to note about your choice of Euclidean function and way of proof is that it uses absolutely no properties of $\Bbb F[x]$, so if it were valid then every domain would be Euclidean (false).
You need a Euclidean function that works. One thing to note about the form $q(x)g(x)+r(x)$ is that it is exactly the form in the division algorithm. If you long-divide $f(x)$ by $g(x)$, in what sense is the remainder $r(x)$ "smaller" than $g(x)$? There is a specific number related to $r(x)$ which will be smaller than the number associated to $g(x)$. What kinds of numbers do polynomials have associated to them? Their degrees.
