Bourbaki

Lemma 1

Tag 0CR9, permanent. Cite it as [Bourbaki, Tag 0CR9].

Commutative Algebra, chapter VIII, § 6. Dimension des anneaux gradués

a) If pp is a prime ideal of HH, pgrp^{gr} is a prime ideal.

b) Every maximal element of the set of graded ideals of HH distinct from HH is a maximal ideal of HH which contains H≥1H_{\geq 1}.

c) Every minimal prime ideal of HH is graded.

a) This follows from III, § 1, no. 4, prop. 4.

b) Let mm be a graded ideal of HH distinct from HH. Then one has

m⊂(m∩H0)+H≥1≠H. m \subset (m \cap H_0) + H_{\geq 1} \neq H.
If mm is maximal, then one has m=m0+H≥1m = m_0 + H_{\geq 1}, where m0m_0 is a maximal ideal of H0H_0, whence bb.

c) Let pp be a minimal prime ideal of HH. Since pgrp^{gr} is prime by aa and contained in pp, one has p=pgrp = p^{gr}, whence cc.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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