Corollary

If a finite sequence of elements $(x_i)_{1 \leq i \leq n}$ of $C$ (for $n \geq 2$) is such that $\sum_{i=1}^n x_i = 0$, there exist at least two distinct indices $j, k$ such that

If there were only a single index $k$ such that $v(x_k) = \inf_{1 \leq i \leq n} v(x_i)$, Proposition 1 would show that $v(x_k) = v(0) = +\infty$, whence $v(x_i) = +\infty$ for all $i$, contrary to the relation $n \geq 2$ and the hypothesis made on $k$.

Remarks (1) If $v : C \to \Gamma_\infty$ is a valuation on $C$ and $u : B \to C$ a homomorphism of a ring $B$ to $C$, it is immediate that the composite mapping $B \xrightarrow{u} C \xrightarrow{v} \Gamma_\infty$ is a valuation on $B$ with values in $\Gamma$.

(2) Conditions (VL$_I$) and (VL$_II$) show immediately that the set $\bar{v}^{-1}(+\infty)$ is a two-sided ideal $\mathfrak{p}$ in $C$ distinct from $C$ by virtue of (VL$_III$); moreover, if $x, y$ are two elements of $C$ such that $v(xy) = +\infty$, it follows from (VL,) that necessarily $v(x) = +\infty$ or $v(y) = +\infty$; in other words, the quotient ring $C/\mathfrak{p}$ has no divisor of 0 other than 0; it is immediately verified that the mapping $\bar{v} : C/\mathfrak{p} \to \Gamma_\infty$ derived from $v$ by passing to the quotient is a valuation on $C/\mathfrak{p}$, the inverse image of $+\infty$ under this valuation reducing to 0.