In his answer to a question about simple proofs of the
Nullstellensatz
([Elementary / Interesting proofs of the Nullstellensatz](https://mathoverflow.net/questions/15226/elementary-interesting-proofs-of-the-nullstellensatz)),
Qiaochu Yuan referred to a really simple proof for the case of an
uncountable algebraically closed field. 


Googling, I found this construction also in Exercise 10 of a 2008 homework
assignment from a course of J. Bernstein (see the last page of
<http://www.math.tau.ac.il/~bernstei/courses/2008%20spring/D-Modules_and_applications/pr/pr2.pdf>).
Interestingly, this exercise ends with the 
following (asterisked, hard) question: 


(\*) Reduce the case of arbitrary field $k$ to the case of
an uncountable field.


After some tries to prove it myself, I gave up and returned to googling.
I found several references to the proof provided by Qiaochu Yuan, but no
answer to exercise (\*) above. 


So, my question is: To prove the Nullstellensatz, how can the general case
of an arbitrary algebraically closed field be reduced to the
easily-proved case of an uncountable algebraically closed field? 


The exercise is from a course of Bernstein called 'D-modules and their
applications.' One possibility is that the answer arises somehow when
learning D-modules, but unfortunately I know nothing of
D-modules. Hence, proofs avoiding D-modules would be particularly
helpful. 



To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?
These logic/ZFC/model theory arguments seem out of proportion to the task at hand. Let $k$ be a field and $A$ a finitely generated $k$-algebra over a field $k$. We want to prove that there is a $k$-algebra map from $A$ to a finite extension of $k$. Pick an algebraically closed extension field $k'/k$ (e.g., algebraic closure of a massive transcendental extension, or whatever), and we want to show that if the result is known in general over $k'$ then it holds over $k$. We just need some very basic commutative algebra, as follows. 


Proof: We may replace $k$ with its algebraic closure $\overline{k}$ in $k'$ and $A$ with a quotient $\overline{A}$ of $A \otimes\_k \overline{k}$ by a maximal ideal (since if the latter equals $\overline{k}$ then $A$ maps to an algebraic extension of $k$, with the image in a finite extension of $k$ since $A$ is finitely generated over $k$). All that matters is that now $k$ is perfect and infinite. 


By the hypothesis over $k'$, there is a $k'$-algebra homomorphism 
$$A' := k' \otimes\_k A \rightarrow k',$$
or equivalently a $k$-algebra homomorphism $A \rightarrow k'$. By expressing $k'$ as a direct limit of finitely generated extension fields of $k$ such an algebra homomorphism lands in such a field (since $A$ is finitely generated over $k$). That is, there is a finitely generated extension field $k'/k$ such that the above kind of map exists. Now since $k$ is perfect, there is a separating transcendence basis $x\_1, \dots, x\_n$, so 
$k' = K[t]/(f)$ for a rational function field $K/k$ (in several variables) and a monic (separable) $f \in K[t]$ with positive degree. Considering coefficients of $f$ in $K$ as rational functions over $k$, there is a localization
$$R = k[x\_1,\dots,x\_n][1/h]$$
so that $f \in R[t]$. By expressing $k'$ as the limit of such $R$ we get such an $R$ so that there is a $k$-algebra map
$$A \rightarrow R[t]/(f).$$
But $k$ is infinite, so there are many $c \in k^n$ such that $h(c) \ne 0$. Pass to the quotient by $x\_i \mapsto c\_i$.
QED


I think the main point is twofold: (i) the principle of proving a result over a field by reduction to the case of an extension field with more properties (e.g., algebraically closed), and (ii) spreading out (descending through direct limits) and specialization are very useful for carrying out (i). 


