# Details in Eisenbud's proof of the general Nullstellensatz

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?

As you have pointed out, to prove the general nullstellensatz, it suffices to prove the following statement.

Let $$R$$ be a ring, and let $$S$$ be a finitely generated $$R$$-algebra. Fix a prime ideal $$P\subseteq S$$. If $$R$$ is Jacobson and $$S/P$$ contains a nonzero $$b$$ such that $$(S/P)[b^{-1}]$$ is a field, then $$S/P$$ is a field.

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.

Let $$R$$ be a ring, and let $$S$$ be a finitely generated $$R$$-algebra which is an integral domain. If $$R$$ is Jacobson and $$S$$ has a nonzero $$b$$ such that $$S[b^{-1}]$$ is a field, then $$S$$ is a field.

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 $$S$$.

• Ah, this clearifies things alot. Just to be sure I’m with you, in the last part the point is that $S$ is also an $R’$-algebra and since $R’$ is also Jacobson, the statement is equivalent to the second formulation you present with the added assumption that $R$ is also an integral domain? Sep 21, 2022 at 19:35
• I think it should be equivalent (both statements will be true, so I guess they are equivalent). But I was originally just thinking in one direction: suppose that $R$ is Jacobson and $S$ satisfies all the properties. Then $R'$ is also Jacobson, and is an integral domain, and $S$ a f.g. $R'$-algebra with $R'\hookrightarrow S$. Now we use these things to prove that $S$ has a nonzero $b$ such that...
– Dave
Sep 21, 2022 at 19:44
• Okay, thanks for the answer. I think I understand now. It doesn’t matter if we consider $S$ as an $R$-algebra or an $R’$-algebra since both are Jacobson and the «base ring» doesn’t affect the «Jacobsonness» of $S$ since that is a property of its ring structure rather than its algebra structure. Sep 21, 2022 at 19:49
• Yeah exactly. $S$ being Jacobson doesn't depend on the base ring $R$.
– Dave
Sep 21, 2022 at 20:58