Question about proof of Hilbert's Nullstellensatz

Question

In the proof given here, why do we need to take $$R=\left(\frac{A}{p}\right)[f^{-1}]$$ and not just $$R=\left(\frac{A}{\sqrt{I}}\right)[f^{-1}]?$$

Answer 1

Either way works, but also either way you have to say a few more words that are not mentioned in the linked proof. The point is that you need this ring $R$ to be nonzero, so that it has a maximal ideal.

If you define $R=(A/p)[f^{-1}]$, you can prove $R$ is nonzero as follows: $p$ is a prime ideal not containing $f$, so $A/p$ is a domain and $f$ is nonzero in $A/p$, so the localization $(A/p)[f^{-1}]$ is nonzero.

If you define $R=(A/\sqrt{I})[f^{-1}]$, then you can instead say that $A/\sqrt{I}$ is a reduced ring since $\sqrt{I}$ is radical, and $f$ is nonzero in $A/\sqrt{I}$ since $f\not\in\sqrt{I}$. Since $A/\sqrt{I}$ is reduced, this means $f$ is not nilpotent in $A/\sqrt{I}$, so the localization $(A/\sqrt{I})[f^{-1}]$ is nonzero.