Hilbert's Nullstellensatz - Formulation

Formulation

Let k be a field (such as the rational numbers) and K be an algebraically closed field extension (such as the complex numbers), consider the polynomial ring k and let I be an ideal in this ring. The algebraic set V(I) defined by this ideal consists of all n-tuples x = (x₁,...,x_n) in Kn such that f(x) = 0 for all f in I. Hilbert's Nullstellensatz states that if p is some polynomial in k which vanishes on the algebraic set V(I), i.e. p(x) = 0 for all x in V(I), then there exists a natural number r such that pr is in I.

An immediate corollary is the "weak Nullstellensatz": The ideal I in k contains 1 if and only if the polynomials in I do not have any common zeros in Kn.

When k=K the "weak Nullstellensatz" may also be stated as follows: if I is a proper ideal in K, then V(I) cannot be empty, i.e. there exists a common zero for all the polynomials in the ideal. This is the reason for the name of the theorem, which can be proved easily from the 'weak' form using the Rabinowitsch trick. The assumption that K be algebraically closed is essential here; the elements of the proper ideal (X2 + 1) in R do not have a common zero. With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as

for every ideal J. Here, denotes the radical of J and I(U) is the ideal of all polynomials which vanish on the set U.

In this way, we obtain an order-reversing bijective correspondence between the algebraic sets in Kn and the radical ideals of K. In fact, more generally, one has a Galois connection between subsets of the space and subsets of the algebra, where "Zariski closure" and "radical of the ideal generated" are the closure operators.

As a particular example, consider a point . Then . More generally,

As another example, an algebraic subset W in Kn is irreducible (in the Zariski topology) if and only if is a prime ideal.