Proposition 3

Let $p$ be a prime ideal of $H$. If $p \neq p^{gr}$, one has $ht(p^{gr}) = ht(p) - 1$.

By Lemma 1, $a$, the ideal $p^{gr}$ is prime and contained in $p$, hence $ht(p^{gr}) \leq ht(p) - 1$. Since the proposition is trivial when $ht(p^{gr}) = + \infty$, one may suppose that $ht(p^{gr}) < + \infty$. Let us prove the inequality $ht(p) \leq ht(p^{gr}) + 1$ by induction on $ht(p^{gr})$. It is enough to prove that, for every prime ideal $q$ contained in $p$ and distinct from $p$, one has $ht(q) \leq ht(p^{gr})$. Let us distinguish two cases according as $q^{gr} \neq p^{gr}$ or $q^{gr} = p^{gr}$. If $q^{gr} \neq p^{gr}$, then one has $ht(q^{gr}) < ht(p^{gr})$; one has

by the induction hypothesis if $q \neq q^{gr}$ and trivially if $q = q^{gr}$; consequently, one has $ht(q) \leq ht(q^{gr}) + 1 \leq ht(p^{gr})$, which was to be proved. If $q^{gr} = p^{gr}$, then one has $q = q^{gr}$ by Lemma 2, hence $ht(q) = ht(q^{gr}) \leq ht(p^{gr})$, whence again the desired conclusion.