Show that the ring of formal power series is a Euclidean domain.

Question

Here is the question I want to solve part $(c)$ of it:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$ In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

$(a)$ Let $(R, \mathfrak{m})$ be a local ring. Show that $R^* = R\setminus \mathfrak{m}.$

$(b)$ Show that, for a field $K,$ $R = K[[x]]$ is a local ring.

$(c)$ Show that $R = K[[x]]$ is a Euclidean domain.

I got hint to use that $(R, \mathfrak{m})$ be a local ring but I still do not know how to do that, could anyone help me in this proof please?

Euclidean Domain definition:

A euclidean domain $R$ is an integral domain together with a function $\sigma : R \rightarrow \mathbb N$ where $\sigma (r) = 0 \iff r= 0 $ (well ordered monoid) and such that, for each $a,b \in R,$ there exist $q,r \in R$ where $a = bq + r$ and $\sigma(r) < \sigma(b).$

Answer 1

Hint:

Use as a Euclidean function $\sigma$ the order of a formal power series: $$\sigma(a_0+a_1X+a_2X^2+\dots)=\min\{i\in\mathbf N\mid a_i\ne 0\}$$ and prove the existence of $q$ and $r$ by comparing the orders of the dividend $f$ and divisor $g$.

Some details:

Let $f,g\in K[[X]]$ and $\sigma(g)=r$.

1. If $\sigma(f)<\sigma(g)$, the solution is obvious: $q=0$, $\:r=f$.
2. If $\sigma(f)\ge\sigma(g)$, I'll give an example: suppose $\:f=X^3(X^2+2X^3+3X^4+4X^5+\dots)\:$ and $g=X^3+X^4+X^5=X^3(1+X+X^2)$ (a polynomial is also a formal power series). Note that $u=1+X+X^2$ is a unit in $K[[X]]$, which is a discrete valuation ring. We deduce that \begin{align} f&=(X^2+2X^3+3X^4+4x^5+\dots)X^3\\ &=\underbrace{(X^2+2X^3+3X^4+4x^5+\dots)u^{-1}\!}_{\textstyle q}\;g, \end{align}