Let $C$ be a smooth projective algebraic curve over an algebraically closed field $k$ of characteristic zero. Let $K$ be the function field of $C$. We take rational functions $f_0,...,f_n \in K$, not all zeroes, and set $$e_P := -\min\{v_P(f_0),...,.v_P(f_n)\},$$ where $v_P(f)$ is the order of $f$ at a point $P$ of $C$.
We define a descending sequence of linear subspaces $\mathbb{P}^n = V_0 \supseteq V_1 \supseteq V_2 \supseteq...$ such that, for each $m \geq 0$, $$V_m(k) = \{[a_0:...:a_n] \in \mathbb{P}^n(k):v_P(\sum_{i=0}^na_if_i)+e_P \geq m\}.$$ It is obvious by definition of $V_m$ that if $m_1 \leq m_2$, then $V_{m_1} \supseteq V_{m_2}$. What baffles me is the condition $v_P(\sum_{i=0}^na_if_i)+e_P \geq m$. Here don't we always have $$v_P(\sum_{i=0}^na_if_i) = \min\{v_P(f_0),...,v_P(f_n)\}?$$ Thus wouldn't $v_P(\sum_{i=0}^na_if_i)+e_P$ just be zero everytime?
Another question is, whenever we have a strict inclusion $V_m \supsetneq V_{m+1}$, how do we see that $V_{m+1}$ is precisely of codimension one in $V_m$?
EDIT. I realised for the first question it should be $$v_P(\sum_{i=0}^na_if_i) = \min\{v_P(a_0f_0),...,v_P(a_nf_n)\}.$$ Thus when appropriate $a_i$'s are zero, we get varying values for the sum with $e_P$. So there is no problem here.