Let 𝐾 =Frac⁡𝑅 be the field of fractions.

First we consider the case 𝐼 =𝑅, i.e. we must show 𝔭−1 ≠𝑅, whereby it is sufficient to find 𝑥 ∈𝔭−1 ∖𝑅. By definition, 𝑥 ∈𝔭−1 iff 𝑥𝔭 ⊆𝑅.

We will try to find 𝑎,𝑏 ∈𝑅 so that 𝑏𝔭 ⊆⟨𝑎⟩, but 𝑏 ∉⟨𝑎⟩, so that 𝑎−1𝑏 is the appropriate 𝑥. To this end, let 0 ≠𝑎 ∈𝔭. By ^P1, we have 𝔭1⋯𝔭𝑟 ⊆⟨𝑎⟩ for some nonzero prime ideals, where we are free to assume that 𝑟 is minimal. Since ⟨𝑎⟩ ⊆𝔭 it follows by ^D2 𝔭𝑖 ⊆𝔭. say 𝑖 =1. But since dim⁡𝑅 =1, 𝔭𝑖 is maximal, so 𝔭 =𝔭1.

If 𝑟 =1, then again by maximality 𝔭 =⟨𝑎⟩, whence 𝔭−1 =𝑅𝑎−1 cannot be equal to 𝑅 or else 𝑎 is a unit and ⟨𝑎⟩ =⟨1⟩ which is not prime.

Now consider 𝑟 ≥2, whence 𝔭2⋯𝔭𝑟 ⊊⟨𝑎⟩ by minimality of 𝑟. Hence there exists 𝑏 ∈𝔭2⋯𝔭𝑟 such that 𝑏 ∉⟨𝑎⟩. By construction 𝑏𝔭 ⊆⟨𝑎⟩, and it follows that 𝑥 =𝑎−1𝑏 ∈𝔭−1 ∖𝑅, proving the case 𝐼 =𝑅.

More generally, use the Noetherian nature of 𝑅 to write 𝐼 =⟨𝛼1,…,𝛼𝑛⟩. Suppose towards contradiction 𝔭−1𝐼 =𝐼. Then for every 𝑥 ∈𝔭−1, we may write

for some 𝐴 =(𝑎𝑖𝑗) ∈M𝑛⁡(𝑅). Let 𝑇 =𝑥1𝑛 −𝐴, so that

whence det𝑇 =0. But det𝑇 is a monic polynomial in 𝑥 with coëfficients in 𝑅, whence 𝑥 is integral over 𝑅. But 𝑅 is integrally closed, hence 𝔭−1 =𝑅, contradicting the above special case.

For invertibility, note 𝑥𝔭 ⊆𝑅 for all 𝑥 ∈𝔭−1, and 𝑅 ⊆𝔭−1, so 𝔭 ⊆𝔭−1𝔭 ⊆𝑅. Since 𝔭−1𝔭 ≠𝔭 but is an ideal, and 𝔭 is maximal, it follows 𝔭−1𝔭 =(1).