The first step in a proof of the general form of the Nullstellensatz

Question

I'm reading an article about the proof of the general form of nullstellensatz : http://www.math.uwaterloo.ca/~jpbell/nullstellensatz.pdf.

Recall that a ring $R$ is called Jacobson if the Jacobson radical $J(R/P) =0 $ for every prime ideal $P$ of $R$.

In the article, he states a general form of nullstellensatz as follow.

Theorem 1. Let $R$ be a Jacobson ring and $S$ be a finitely generated $R$-algebra. Then $S$ is a Jacobson ring and if $M$ is a maximal ideal of $S$ then $N:=M \cap R$ is a maximal ideal of $R$ and $S/M$ is a finite field extension of $R/N$.

In the proof of the theorem, he uses next Theorem, called the Rabinowitch Trick :

Theorem 2. Let $R$ be a ring. Then $R$ is a Jacobson ring if and only if whenever $P$ is a prime ideal of $R$ and $T:=R/P$ has the property that $T_{b}$ (localization) is a field for some nonzero $b\in T$ then $T$ is a field.

To prove the theorem 1, he approaches through three main steps:

(1) To show that the Nullstellensatz holds when $S=R[x]$ (generated by 1 element)

(2) Use induction to show that it holds for $S=R[x_1, \dots ,x_n]$

(3) Show that the Nullstellensatz holds for homomorphic images of $R[x_1, \dots ,x_n]$.

Every finitely generated $R$-algebras is isomorphic to an algebra of the form $R[x_1, \dots ,x_n]/I$ for some ideal $I$ and so we get the result from (1)-(3)

Following next image is about Step (1):

Why the underlined statements true? For the first statement, I'm now constructing explicit ring homomorphism as follows (we denote $0 \neq b := f(x) \in R'[x]$ ) :

$$\alpha: K[x]_b \to R'[x]_b$$

$${({a_n \over b_n} x^n + \cdots + {a_0 \over b_0}) \over (\bar{f}(x))^m} \mapsto {1 \over (f(x))^m} \big[\psi({a_n \over b_n})({x^n \over 1}) + \cdots + \psi({ a_0 \over b_0})({1 \over 1}) \big],$$ where $\bar{f}$ is the image of $f(x) \in R'[x]$ to $K[x]$ and $\psi : K \to R'[x]_b$ be the induced homomorphism from $\varphi : R' \to R'[x]_b$ (the existence is gauranteed by the universal property of the field of fractions). Explicitly, $\psi({a \over c}) := \varphi{(a)} \varphi{(c)}^{-1}= ({a \over 1})({c \over 1})^{-1}$.

And through trial, It seems to be true that the above map is well-defined ring isomorphism.

But for the second underlined statement, similar strategy (constructing explicit homomorphism mimicking as above) seems does not work by the well-definedness problem. Anyway this strategy also works?

Is there another approach to show the underlined statement, by simply noting some point?

Answer 1

Whenever I go over this proof, rather than finding an explicit homomorphism between these two rings, I've always found it more illuminating to quickly prove statement (1) without actually using the underlined statements, and then it will be easy to see why the underlined statements are true.

In light of the Rabinowitch trick, the statement we're trying to prove is really a fact about the ring $S/P$ for some prime ideal $P$. As such, rather than using the above notation $R'$, we can factor out a prime ideal $P$ from $S$ (and the corresponding ideal $R\cap P$ from $R$) as preprocessing and assume from the beginning that $S$ and $R$ are integral domains. In fact, what we will end up showing is that if such an element $b$ exists, then both $S$ and $R$ are both fields. If $S$ is already a field, then such a $b$ exists (trivially), so we get the second part of the Nullstellensatz for free by factoring out $M$ and $N$ via the same proof.

As such, the statement (1) can be reformulated as:

---

Let $R$ be a ring and $S$ an $R$-algebra generated by one element. Suppose $R$ is Jacobson. If $R$ and $S$ are both integral domains and there exists $b\in S$ such that $S[b^{-1}]$ is a field, then $S$ is a field. Furthermore, if $S$ is a field, then $R$ is a field such that $S$ is a finite field extension of $R$.

Proof: Since $S$ is generated over $R$ by a single element, $S = R[x]/Q$ for some prime ideal $Q$. Suppose $Q= 0$ and we have $b\in R[x]$ such that $R[x][b^{-1}]$ is a field. Write $K$ as the field of fractions of $R$. Since $R[x][b^{-1}]$ is a field, it must contain all inverses of $R$, so it must contain $K$, meaning $K[x][b^{-1}]$ is a field as well. It is easy to show that $K[x]$ is Jacobson, but $K[x]$ is not a field, contradicting the Rabinowitch trick. Therefore, $Q\ne 0$.

Suppose $p\in Q$ such that $p(x) = p_mx^m+\ldots+p_0 = 0$ in $S$ with coefficients in $R$. Then, multiplying the polynomial by $p_m^{-1}$, $$x^m+\ldots+p_m^{-1} p_0 = 0,$$ so $S[p_m^{-1}]$ is integral over $R[p_m^{-1}]$. Since $b\in S[p_m^{-1}]$, $b$ also satisfies a monic polynomial with coefficients in $R[p_m^{-1}]$, say $$b^m+\ldots+ r_0 = 0.$$ Clearing denominators and relabeling if necessary, we see that $b$ satisfies $$r_mb^m+\ldots+r_0 = 0$$ with coefficients in $R$. Without loss of generality, since $S$ is an integral domain, we can assume that $r_0\ne 0$. Setting $\beta = b^{-1}$ and dividing by $r_0$, we get a monic relation for $\beta$: $$\beta^m+\ldots+r_0^{-1} r_m = 0.$$ Here would be a good time to mention the following fact: If $R\subseteq S$ is an integral extension of rings, then $R$ is a field iff $S$ is a field. (A proof can be found here.)

As such, since $R[(ab)^{-1}] = R[a^{-1}][b^{-1}]$ and $S[\beta]$ is already a field so $S[p_m^{-1}\beta] = S[\beta]$, we see that $S[\beta]$ is integral over $R[(p_mr_0)^{-1}]$. But $S[\beta]$ is a field, so $R[(p_mr_0)^{-1}]$ is a field. Since $R$ is Jacobson, $R$ must also be a field. In particular, $R[p_m^{-1}] = R$, so $S$ is actually integral over $R$. Again, this means $S$ is also a field, as desired.

---

Indeed, we have just proven that if such an element $b$ exists, $R$ is a field, so $R=K$. Since $S$ is also a field in this case, $$S=S[b^{-1}] = K[x]/QK[x] = (K[x]/QK[x])[b^{-1}],$$ and we see that both underlined statements are true, and $S$ is a finite field extension of $R=K$.