Proposition 19.9.10E32D710594E0

Proof. Suppose first that $d>0$. Choose sections $s_0,\ldots ,s_ d$ of positive degrees $m_0,\ldots ,m_ d$ whose normalized valuations $\nu (s_ i)/m_ i$ are affinely independent. Let $M$ be a common multiple of their degrees and put $t_ i=s_ i^{M/m_ i}\in W_ M$. The ratios $f_ i=t_ i/t_0$ belong to $K(W_\bullet )$. The flag valuation extends to quotients by subtraction, giving

These vectors are linearly independent. Distinct monomials in the $f_ i$ therefore have distinct valuations, so the term of smallest valuation in a nonzero polynomial expression cannot cancel, and the ratios are algebraically independent. Thus the transcendence degree is at least $d$; this is trivial when $d=0$.

Conversely, let $f_1,\ldots ,f_ r$, with $r\geq 1$, be algebraically independent elements of the field. Using the product of their denominators, write $f_ i=A_ i/B$ with $A_ i,B\in W_ M$ and $B\neq 0$ for a common positive degree $M$. For every integer $q\geq 1$ and every multi-index $a=(a_1,\ldots ,a_ r)\in \mathbb N^ r$ with $|a|=a_1+\cdots +a_ r\leq q$, the sections

are linearly independent, because division by $B^ q$ turns a linear relation among them into a polynomial relation among the $f_ i$. There are $\binom {q+r}{r}$ such sections. Since $W_ M\neq 0$, the degree $M$ is a multiple of $e$, and theorem 19.9.9 gives $\dim W_{Mq}=O(q^ d)$. Comparing with the binomial lower bound forces $r\leq d$, which proves equality of the transcendence degree with $d$. A body with nonempty interior has $d=n$, and this gives the stated consequence for $X$.

The field and the set of degrees with nonzero sections do not depend on the flag, so neither $d$ nor $e$ does. By theorem 19.9.9, the lattice volume is, for either flag, the limit of the same dimensions with the same normalization. It is therefore independent of the flag as well, although the bodies and their lattices may differ. $\square$