Weak Nullstellensatz says, that for an algebraically closed field each maximal ideal
in
has form
, for
, i.e.
, the ideal of an one-element subset of
. Its proof in this generality needs quite a bit of commutative algebra. However, if we futher assume that
is uncountable (thus covering a very important case
) we can give a much quicker proof.
Theorem 1 Let
be an uncountable algebraically closed field, e.g.
, and
a proper maximal ideal in
. Then there exists
such that
.
Proof: The first step is to show that . To see this, we will show that every
is algebraic, i.e. a root a nonzero polynomial
. Note that the dimension of
as a vectorspace over
is at most countable, as
is generated by the images
of the monomials
under the ring homomorphism
, and the exponents
form a countable set. Thus for
the set
is linearly dependent, i.e. there exist such that
Thus
where and
. As
is algebraically closed, we have that
is linear, i.e.
.
Next, we observe that maps
to
, and set
, for
. As
, we see that
, for
. By maximality of
, we obtain
, as claimed.
Here one can find reduction of the general case (not assuming non-countability of ) to this one.



