Lemma 2

Let $p$ and $q$ be prime ideals of $H$ such that $q \subset p$ and $q \neq p$. If $q^{gr} = p^{gr}$, then $q$ is graded, $p$ is not, and $ht(p/q) = 1$.

\* Remark 1. --- Let us take up again the notation of example 1 of No. 1. Lemma 2 implies that, if two irreducible subvarieties $Y$ and $Z$ of $\mathbf{C}^{n+1}$ have the same projecting cone and if $Z \subset Y$ and $Z \neq Y$, then $Y$ is the projecting cone of $Z$, and $Z$ is of codimension 1 in $Y$. \*

Replacing $H$ by $H/q^{gr}$, one is reduced to the case where $q^{gr} = \{0\}$. Then $H$ is an integral domain (lemma 1, $a$), $p^{gr} = 0$, and it is a matter of proving that $ht(p) \leq 1$: this will indeed imply that $ht(q) = 0$, hence that $q = \{0\}$. Since $p^{gr} = \{0\}$, one has $p \cap H_n = \{0\}$ for every $n$, and $p$ is disjoint from the multiplicative subset $T = \bigcup (H_n - \{0\})$. The ring $H_p$ is therefore isomorphic to a ring of fractions of $T^{-1}H$, and one has therefore

(§ 1, No. 3, prop. 6 and 7). But, by lemma 4 of V, § 1, No. 8, $T^{-1}H$ is a field or is isomorphic to a ring $K[X, X^{-1}]$, where $K$ is a field; hence one has $\dim(T^{-1}H) \leq 1$ and $ht(p) \leq 1$, which was to be proved.