Let $R$ be a commutative ring with identity. **D.G. Northcott**'s, *Finite Free Resolutions*, has:    
[![enter image description here][1]][1]
  

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


  [1]: https://i.sstatic.net/DFFoX.gif