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Let $R$ be a commutative ring with identity. D.G. Northcott's, Finite Free Resolutions, has:
enter image description here

and in Theorem 16 of Chapter 5 proves that: $p.grade(I,M) = p.grade(P,M)$ for some prime ideal $P$ containing $I$ .

Question. Let $I$ be a finitefinitely generated ideal. Can one haveclaim that there is a finitefinitely generated prime ideal $P$ containing $I$ such that $p.grade(I,M) = p.grade(P,M)$  ?

Thank you

Let $R$ be a commutative ring with identity. D.G. Northcott's, Finite Free Resolutions, has:
enter image description here

and in Theorem 16 of Chapter 5 proves that: $p.grade(I,M) = p.grade(P,M)$ for some prime ideal $P$ containing $I$ .

Question. Let $I$ be finite. Can one have a finite prime ideal $P$ containing $I$ such that $p.grade(I,M) = p.grade(P,M)$  ?

Thank you

Let $R$ be a commutative ring with identity. D.G. Northcott's, Finite Free Resolutions, has:
enter image description here

and in Theorem 16 of Chapter 5 proves that: $p.grade(I,M) = p.grade(P,M)$ for some prime ideal $P$ containing $I$ .

Question. Let $I$ be a finitely generated ideal. Can one claim that there is a finitely generated prime ideal $P$ containing $I$ such that $p.grade(I,M) = p.grade(P,M)$?

Thank you

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Theorem 16 , Chapter 5 of Northcott 's, Finite Free Resolutions: p.grade

Let $R$ be a commutative ring with identity. D.G. Northcott's, Finite Free Resolutions, has:
enter image description here

and in Theorem 16 of Chapter 5 proves that: $p.grade(I,M) = p.grade(P,M)$ for some prime ideal $P$ containing $I$ .

Question. Let $I$ be finite. Can one have a finite prime ideal $P$ containing $I$ such that $p.grade(I,M) = p.grade(P,M)$ ?

Thank you