For an algebraically closed field and an ideal , the strong Hilbert Nullstellensatz is
Its weak form says that an ideal with empty affine zero set is the unit ideal. Equivalently, if finitely many polynomials have no common zero, a polynomial combination of them equals one.
For an algebraically closed field , every maximal ideal of has the form
for a unique point . Equivalently, every proper ideal has a common zero.
If a field is finitely generated as an algebra over a subfield , then is a finite algebraic extension of . Applying this to a residue field of a polynomial ring proves the Weak Hilbert Nullstellensatz.
For an ideal over an algebraically closed field,
Let . If , then
for all but finitely many primes . The Nullstellensatz puts a power of each generator of in ; clearing denominators leaves only finitely many exceptional characteristics.
To prove the Strong Hilbert Nullstellensatz from its weak form, adjoin a variable and the equation . If vanishes on , the ideal has no common zero and is therefore the unit ideal. Substituting and clearing denominators yields for some .
A complex affine algebraic set is either finite or has the cardinality of . One proof takes a positive-dimensional irreducible component and applies Noether normalization: the resulting finite dominant map to a positive-dimensional affine space has a point above every closed point.
Let be a homogeneous ideal over an algebraically closed field, and let
be the irrelevant ideal. Then exactly when either or . For a proper homogeneous ideal this is also equivalent to for some .
In the standard graded homogeneous coordinate ring of , the irrelevant ideal is
It vanishes only at the origin of the corresponding affine cone, which does not represent a projective point.
Every open cover of an affine variety has a finite subcover. Refine it by distinguished opens . Their complements have no common point, so the Nullstellensatz writes as a finite combination of the ; the corresponding finite family of distinguished opens covers.

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