TITLE: 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?
QUESTION [23 upvotes]: In his answer to a question about simple proofs of the
Nullstellensatz
(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.

REPLY [6 votes]: The easiest way to reduce to the uncountable case may be as follows. 
Let $I$ be an ideal of $k[X_1,...,X_d]$ which does not contain $1$.
Let $P_1,\dots,P_r$ be a generating family of $I$.
Let $A=k^{\mathbf N}$ and let $m$ be a maximal ideal of $A$ which contains
the ideal $N=k^{(\mathbf N)}$ of $A$. Then $K=A/m$ is an algebraically closed field
which is has at least the power of the continuum. 
(Alternative description: let $K$ be an ultrapower of $k$, with respect to a non-principal ultrafilter.)
Lemma. For $i\in\{1,\dots,r\}$, let $a_i=(a_{i,n})\in A$. Assume that $(\bar a_1,\dots,\bar a_r)=0$ in $K^r$. Then the set of $n\in\mathbf N$ such that $(a_{1,n},\dots,a_{r,n})=0$
is infinite.
Proof. Assume otherwise. For every $n$ such that $(a_{1,n},...,a_{r,n}) \neq 0$,
choose $(b_{1,n},\dots,b_{r,n})$ such that $\sum a_{i,n}b_{i,n}=1 $,
and let $b_i=(b_{i,n})_n\in A$. Then $\sum a_i b_i - 1 $ belongs to $N^r$,
hence $\sum \bar a_i \bar b_i=1$. Contradiction.
Thanks to the lemma, one proves easily that 
the ideal $I_K$ of $K[X_1,...,X_d]$ generated by $I$
does not contain $1$. 
By the uncountable case, there exists $x=(x_1,...,x_d)\in K^d$ such 
that $P_j(x_1,...,x_d)=0$ for every $j$. 
For every $i$, let $a=(a_{n})\in A^d$ be such that $\bar a=x$.
By the lemma again the set of integers $n$ such that
$P_j(a_n) \neq 0$ for some $j$ is finite.
In particular, there exists a point $y\in k^d$ such that $P_j(y)=0$ for every $j$.