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?


1 Answer 1


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 $S$.

  • $\begingroup$ 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? $\endgroup$ Sep 21, 2022 at 19:35
  • $\begingroup$ 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... $\endgroup$
    – Dave
    Sep 21, 2022 at 19:44
  • $\begingroup$ 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. $\endgroup$ Sep 21, 2022 at 19:49
  • $\begingroup$ Yeah exactly. $S$ being Jacobson doesn't depend on the base ring $R$. $\endgroup$
    – Dave
    Sep 21, 2022 at 20:58

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