Formal power series ring, norm. Let $k$ be a field. Let $R$ be the formal power series ring $k[[x]]$. Define $N$ on $R \setminus \{0\}$ as follows: $N(f)$ is the smallest $n$ of which the coefficient of $x^n$ in $f$ is nonzero.
(a) Prove that $R$ is a Euclidean domain with Euclidean norm $N$.
(b) For $a$, $b$, $a+b$ nonzero elements of $R$, prove that $N(a+b)$ cannot be bounded as a function of $N(a)$ and $N(b)$.
(c) Prove that if $a$ and $b$ are two power series such that $b$ does not divide $a$ (and $b \neq 0$), there are infinitely many pairs $(q, r)$ for which $a = bq + r$ and $N(r) < N(b)$.
 A: (a) $N$ being a Euclidean norm follows directly from (c), at least according to the Wikipedia definition of Euclidean domain, that only requires one pair $(q,r)$ from (c) to exist for every $a,b$ with $b$ nonzero. If you have another definition, please let us know.
(b) To show that $N(a+b)$ cannot be bounded by any function $f(N(a),N(b))$, consider the following: Given any $n\in\mathbb{N}$, set $a\in k[[x]]$ by $a_i=1$ for all $i\in\mathbb{N}$ and define $b\in k[[x]]$ by setting $b_i=-a_i=-1$ for $0\le i<n$ and $b_i=a_i=1$ for $i\ge n$. Then obviously $N(a)=N(b)=0$ and $N(a+b)=n$.
(c) Because $k$ is a field, $b\in k[[x]]$ divides $a\in k[[x]]$ iff $N(b)\le N(a)$. To see direction we need, the "if" part, assume that $b$ doesn't divide $a$ and $N(b)\le N(a)$. Then divide both by $x^{N(b)}$ and apply the formula for dividing power series where the denominator has invertible $0$-coefficient:
$$ c_n = \frac{1}{b_0}\left(a_n - \sum_{k=1}^n b_k c_{n-k}\right). $$
Now that we know that $N(b)>N(a)$, we can multiply $b$ by any power series $q\in k[[x]]$ we want, because still $N(bq)\ge N(b) > N(a)$ and fix it afterwards by setting $r=a-bq$. Note that $N(r)=N(a)<N(b)$ because the $N(a)$-th coefficient of $a$ is not modified by the substraction, because $N(b)>N(a)$.
So there are infinitely many pairs $(q,r)$, and in fact, given an arbitrary $q\in k[[x]]$ we can choose an appropriate $r\in k[[x]]$ such that the condition is fulfilled.
