I have been thinking about the question "What is the best -- i.e., some combination of shortest, most natural, easiest -- proof of the Nullstellensatz?" recently on the eve of a commutative algebra course.

In my notes up to this point I had been following Kaplansky's treatment of Goldman domains and Hilbert-Jacobson rings. This places the problem in a more general context and allows for an attractively thorough analysis. At the end one comes out with the following results:

(1) The polynomial ring $k[t_1,\ldots,t_n]$ is a Jacobson ring -- i.e., every radical ideal is the intersection of the *maximal ideals* containing it.

(2) (Zariski's Lemma): If $\mathfrak{m}$ is a maximal ideal of $k[t_1,\ldots,t_n]$, then $k[t_1,\ldots,t_n]/\mathfrak{m}$ is a finite degree field extension of $k$.

(That Hilbert's Nullstellensatz follows from (1) and (2) is an easy, standard argument that I won't discuss here.)

But it is well-known that to prove the Nullstellensatz one needs only (2), because then (1) follows by a short and easy argument that everyone seems to like: **Rabinowitsch's Trick**. So perhaps this is a sign that developing the theory of (Hilbert-)Jacobson rings to prove the Nullstellensatz is overkill.

So the question seems to be: what is the best proof of Zariski's Lemma?

After looking around at various proofs, here is what I think the answer is now: it is an easy consequence of the following result.

Theorem (Artin-Tate Lemma): Let $R \subset T \subset S$ be a tower of rings such that

(i) $R$ is Noetherian,

(ii) $S$ is finitely generated as an $R$-algebra, and

(iii) $S$ is finitely generated as a $T$-module.

Then $T$ is finitely generated as an $R$-algebra.

Proof: Let $x_1,\ldots,x_n$ be a set of generators for $S$
as an $R$-algebra, and let $\omega_1,\ldots,\omega_m$ be a set of
generators for $S$ as a $T$-module. For all $1 \leq i \leq n$, we may
write
\begin{equation}
\label{ARTINTATEEQ1}
x_i = \sum_j a_{ij} \omega_j, \ a_{ij} \in T.
\end{equation}
Similarly, for all $1 \leq i,j \leq m$, we may write
\begin{equation}
\label{ARTINTATEEQ2}
\omega_i \omega_j = \sum_{i,j,k} b_{ijk} \omega_k, \ b_{ijk} \in T.
\end{equation}
Let $T_0$ be the $R$-subalgebra of $T$ generated by the $a_{ij}$ and $b_{ijk}$. Since $T_0$ is a finitely generated algebra over the Noetherian ring $R$, it
is itself a Noetherian ring by the Hilbert Basis Theoerem. \ \indent
Now each element of $S$ may be expressed as a polynomial in the $x_i$'s
with $R$-coefficients. Making substitutions using the two equations above shows that $S$ is a finitely generated $T_0$-module. Since
$T_0$ is Noetherian, the submodule $T$ is also finitely generated as a $T_0$-module. This immediately implies that $T$ is finitely
generated as a $T_0$-algebra and then in turn that $T$ is finitely generated
as an $R$-algebra, qed!

Proof that Artin-Tate implies Zariski's Lemma:

It suffices to prove the following: let $K/k$ be a field extension which is finitely generated as a $k$-algebra. Then $K/k$ is algebraic. Indeed, suppose otherwise: let $x_1,\ldots,x_n$ be a transcendence basis for
$K/k$ (where $n \geq 1$ since $K/k$ is transcendental), put $k(x) = k(x_1,\ldots,x_n)$ and consider the tower of rings

$k \subset k(x) \subset K$.

To be sure, we recall the definition of a transcendence basis: the elements $x_i$ are algebraically equivalent over $k$ and $K/k(x)$ is algebraic. But since $K$ is a finitely generated $k$-algebra, it is certainly a finitely generated $k(x)$-algebra and thus $K/k(x)$ is a finite degree field extension. Thus the Artin-Tate Lemma applies to our tower: we conclude that $k(x)/k$ is a finitely generated $k$-algebra. But this is absurd. It implies
the much weaker statement that $k(x) = k(x_1,\ldots,x_{n-1})(x_n)$ is finitely
generated as a $k(x_1,\ldots,x_{n-1})[x_n]$-algebra, or weaker yet, that
there exists some field $F$ such that $F(t)$ is finitely generated as
an $F[t]$-algebra: i.e., there exist finitely many rational functions
$\{r_i(t) = \frac{p_i(t)}{q_i(t)} \}_{i=1}^N$ such that every rational function
is a polynomial in the $r_i$'s with $k$-coefficients. But $F[t]$ is a PID with infinitely many nonassociate nonzero prime elements $q$ (e.g. adapt Euclid's argument of the infinitude of the primes), so we may choose a nonzero prime element $q$ which does not divide $q_i(t)$ for any $i$. It is then clear that $\frac{1}{q}$ cannot be a polynomial in the $r_i(t)$'s: for instance,
evaluation at a root of $q$ in $\overline{F}$ leads to a contradiction.

Note that this is almost all exactly as in Artin-Tate's paper, except for the endgame above, which has been made a little more explicit and simplified: their conclusion seems to depend upon unique factorization in $k[t_1,\ldots,t_n]$, which does not come until later on in my notes.

Further comments:

(i) The proof is essentially a reduction to Noether's normalization in the case of field extensions, which becomes the familiar result about existence of transcendence bases. Thus it is not so far away from the most traditional proof of the Nullstellensatz. But I think the Artin-Tate Lemma is easier than Noether Normalization.

(ii) Speaking of Noether: the proof of the Artin-Tate Lemma is embedded in the standard textbook proof that if $R$ is a finitely generated $k$-algebra and $G$ is a finite group acting on $R$ by
ring automorphisms, then $R^G$ is a finitely generated $k$-algebra. In fact I had already typed this proof up elsewhere in my notes. Realizing that the Artin-Tate Lemma is something I was implicitly proving in the course of another result anyway was part of what convinced me that this was an efficient route to the Nullstellensatz. Note that the paper of Artin and Tate doesn't make any connection with Noether's theorem and conversely the textbooks on invariant theory that prove Noether's theorem don't seem to mention Artin-Tate. (However, googling -- Artin-Tate Lemma, Noether -- finds several research papers which allude to the connection in a way which suggests it is common knowledge among the cognoscenti.)

**Added**: It turns out this is the proof of Zariski's Lemma given in Chapter 7 of Atiyah-Macdonald. I had missed this because (i) they give another (nice) proof using valuation rings in Chapter 5 and (ii) they do not attribute the Artin-Tate Lemma to Artin and Tate, although their treatment of it is even closer to the Artin-Tate paper than mine is above. (In the introduction of their book, they state cheerfully that they have not attributed any results. I think this is a drawback of their otherwise excellent text.)