Exercise A0.V.2.14

Proposition

Let $k$ be a field. Then $k[[x]]$ is an Euclidean domain.

Proof

We define $v : k[[x]] \setminus \{0\} \to \mathbb{Z}_{\geq 0}$ by setting $v(f)$ to be the degree of the dominant term of $f$ (which we will refer to as the order of $f$).

Suppose $a, b \in k[[x]]$ with $b$ non-zero. If $b$ is a unit, then we can divide by $b$ without remainder, which suffices for Euclidean valuation. Suppose otherwise.

Let $r \in k[[x]]$ be the polynomial containing all monomials of $a$ whose degree in $x$ is less than the order of $b.$ Note that, by construction, $v(r) < v(b).$

We want to find $q \in k[[x]]$ such that $a = bq + r.$ To do this, we will define a series of approximations that converge to the desired $q.$

Let $q_0 = 0,$ and let $a_0 = a - r.$ Observe that the invariant $a_0 = a - r - q_0b$ holds at this step.

Suppose $q_{n - 1}$ and $a_{n - 1}$ are defined, and the required invariant holds for them. Then we define their successors as follows. If $a_{n - 1} = 0$ then $a_n = 0$ and $q_n = q_{n - 1}.$ Otherwise, let $sx^m$ be the dominant monomial of $a_{n - 1}.$ Since $k$ is a field and $m \geq v(b),$ there exists a monomial $s'x^{m'}$ such that the dominant term of $s'x^{m'}b$ is $sx^m.$ Let $q_n = q_{n - 1} + s'x^{m'}$ and let $a_n = a_{n - 1} - s'x^{m'}b.$

Since $a_{n - 1} = a - r - q_{n - 1}b,$ we observe that

$$a_n = a - r - q_{n - 1}b - s'x^{m'}b = a - r - (q_{n - 1} + s'x^{m'})b,$$

which validates our invariant. Also note that, as long as both $a_{n -1}$ and $a_n$ are non-zero, $v(a_{n - 1}) < v(a_n),$ as subtracting $s'x^{m'}b$ removes the dominant monomial of $a_{n - 1}.$

Define a metric $d(f,g)$ on $k[[x]]$ by $d(f,g) = 2^{-v(f - g)}$ if $f \neq g$ and $d(f,g) = 0$ if $f = g.$ By construction, $d(a - r, bq_n) \to 0$ as $n \to 0.$ By completeness of $k[[x]],$ this limit exists.

Taking $q$ to be the limit of $q_0, q_1, \ldots,$ we now can write $a = bq + r.$ Since $r$ is either $0$ or has valuation $v(r)$ less than $b,$ $v$ is indeed a Euclidean valuation on $k[[x]].$ Thus, $k[[x]]$ is a Euclidean domain.

$\blacksquare$