Vtyjkyuk

Discrete Valuation
of a field $K$ is a function $nu: K^timesto mathbbZ$ that
satisfies the following properties.

(i) $nu$ is surjective.

(ii) $nu(xy) = nu(x) + nu(y)$.

(iii) $nu(x + y) ≥$ min $nu(x), nu(y)$,

(iv) $nu(1)=0$.

How can I prove that in a Discrete Valuation Ring, $$v(x)=infty ~textif and only if~x=0.$$ Using those conditions?

As stated, the question does not make sense: $v$ is not defined on $0,$ and $infty$ is not an element of the codomain of $v.$ To make sense of it, we must first set some notations. Let $Gamma = Bbb Zcupinfty$ be the ordered monoid under addition defined by addition on $Bbb Z$ with the additional rules that $inftygeq a$ for all $ainGamma$ and $infty + a = infty$ for any $ainGamma.$

Suppose $v : K^timestoBbb Z$ is a discrete valuation in the sense of the question. Suppose that $v' : KtoGamma$ is a function such that:

1. $left.v'right|_K^times = v,$


2. $v'(xy) = v'(x) + v'(y),$ and


3. $v'(x + y)geqminv'(x),v'(y).$

I claim that a unique such $v'$ exists, and that $v'(0) = infty.$

Defining a $v'$ satisfying these conditions is easy: we already know what $v'$ must do on $K^times,$ so we simply set $v'(0)=infty.$ Then we only need to check conditions 2 and 3. To check 2, note that if $x$ and $y$ are both nonzero, then the relationship holds because $v'(xy) = v(xy).$ Suppose then that $x = 0.$ Then
$$
v'(0cdot y) = v'(0) = infty = infty + v'(y) = v'(0) + v'(y),
$$
so that 2 holds. To check 3, again note that if $xneq -y$ and $xneq 0,$ we already know this to be true. If $x = 0,$ we have
$$
v'(0 + y) = v'(y)geqmininfty, v'(y) = v'(y)quadcheckmark
$$
If $y = -x,$ we have
$$
infty = v'(0) = v'(x + (-x))geq minv'(x),v'(-x)quadcheckmark
$$
by the properties of $infty$ in the ordered monoid $Gamma.$

To check that $v'$ is unique, suppose $v'$ is some function satisfying the given conditions, with $v'(0) = gammainGamma.$ Then we must have
$$
gamma = v'(0) = v'(0cdot x) = v'(0) + v'(x) = gamma + v'(x)
$$
for all $xin K.$ Taking $x$ to be any element of $K^times$ such that $v'(x) = 1$ implies that we must have $gamma = gamma + 1,$ and the only element of $Gamma$ satisfying this property is $infty,$ so that $v'(0) = infty,$ as desired.

By the definition and uniqueness of $v',$ it then follows that $v'(x) = infty$ if and only if $x = 0,$ because $v'(x)inBbb Z$ for any $xin K^times,$ and we have already seen that any $v'$ as above must have $v'(0) = infty.$