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 .

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