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.

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