This refers to David Eisenbud's "Commutative Algebra With a View Toward Algebraic Geometry". He proves the following version of the Nullstellensatz
General Nullstellensatz [Thm 4.9 in Eisenbud] Let $R$ be a Jacobson ring. If $S$ is a finitely generated $R$-algebra, $S$ is
Jacobson as well.
Before proving the theorem, he proves the following version of the Rabinowitch-trick:
Rabinowitch-trick [Lemma 4.20 in Eisenbud] Let R be a ring. The following are equivalent.
a. $R$ is Jacobson.
b. If $P$ is a prime ideal in $R$ and $S = R/P$ contains an element $b$
$\neq 0$ such that $S[b^{-1}]$ is a field, then $S$ is a field.
Hence, if $R$ is Jacobson and we want to show that so is $S$ it suffices to prove that if $P$ is a prime ideal of $S$ and $S' = S/P$ contains a non-zero $b$ such that $S'[b^{-1}]$ is a field, then $S'$ is a field. Now comes the part that confuses me. Eisenbud writes
Replacing $S$ by $S'$, and factoring out the preimage of $P$ from $R$,
we may assume that $R$ is a domain contained in $S$, and that $b \in$
$S$ is such that $S[b^{-1}]$, and we must show that $S$ is a field.
I understand that $R / R\cap P$ is an integral domain since $R \cap P$ is prime in $R$ when $P$ is prime in $S$. However, I don't understand how passing to this quotient covers the initial case. Why does it suffice to consider the quotient, how does one go back to the initial case knowing that $S/P$ is a field for every prime $P \subseteq S$. Or have I misunderstood what Eisenbud means with "replacing" in this context?
Best Answer
As you have pointed out, to prove the general nullstellensatz, it suffices to prove the following statement.
Now let's just think about proving this statement. Notice that $S':=S/P$ is also a finitely generated $R$-algebra. So the above statement is equivalent to the following.
Now we just need to be able to replace $R$ with its quotient by the restriction of the zero ideal $\langle 0\rangle_S$ in $S$ (since we're now working with $S$ an integral domain). But this just follows from the fact that a quotient of a Jacobson ring by a prime ideal is also a Jacobson ring (e.g. see Quotient of Jacobson ring is Jacobson as in Eisenbud). Thus, if $R$ is Jacobson, then $R':=R/R\cap\langle 0\rangle_S$ is Jacobson, and this latter ring $R'$ is an integral domain which is embedded in $S$.