I was recently asked to prove a nonexhaustive set of cases for the weak Nullstellensatz, and the proof was so deceptively simple I decided to share it. The main theorem goes along the lines of this.
Nullstellensatz: Let be an algebraically closed field and an ideal of . Then is maximal if and only if .
It is not hard to see that is a maximal ideal. We can compute an isomorphism by looking at the surjective ring map . It remains to show that all maximal ideals are of this form. We will prove a subset of cases for as follows.
Cheap Nullstellensatz: Let be an algebraically closed field of uncountable cardinality and an ideal of . If is maximal then .
Proof: We do it in a few parts.

Every field extension has an embedding , where is the set of rational polynomials of .
If is a nontrivial extension of (i.e. ), then must have a transcendental element , and for all nonzero . Thus, we can define the map and we claim this is an embedding of into . It is clearly welldefined because if , then does not divide by zero. It is also easy to check that this is a ring map and because , we can see that it is injective because its kernel is .

is a vector space and has uncountable dimension. It is not too hard to see that is a vector space, and to show that its dimension is uncountable, we show that the set of functions of the form , where , is a linearly independent set. Thus, we wish to find coefficients such that
We clear denominators and call the resulting function , or
Now consider the ring homomorphisms sending to and for all , compute
The product is a product of nonzero elements and thus invertible, so we can calculate for all , and thus the set is linearly independent of uncountable cardinality, so has uncountable dimension over .

Let be a finitely generated algebra. Then has countable dimension over as a vector space. To see this, let be the generators of . We can write as
for some so is some subalgebra of the free algebra . The free algebra has countable basis so must also have countable basis.
We now have all the building blocks to finish the proof. We notice that is a finitely generated algebra, and its quotient field must also be a finitely generated field extension. If is a transcendental extension, then it has uncountable dimension over because it contains as a subfield, but this dimension is also countable because is finitely generated. Thus cannot be transcendental so it must be an algebraic extension, but is algebraically closed so .