Proposition 2

a) We have $\dim_{\mathrm{gr}}(H) \leq \dim(S^{-1}H) \leq \dim(H)$. b) For every graded prime ideal $p$ of $H$, we have $\operatorname{ht}_{\mathrm{gr}}(p) \leq \operatorname{ht}(S^{-1}p) = \operatorname{ht}(p)$.

For every ideal $a$ of $H$, let us denote by $a^{\mathrm{gr}}$ the greatest graded ideal contained in $a$; we have $a^{\mathrm{gr}} = \sum_n (a \cap H_n)$.