stylistic:tone, quality:correction, structural:scholar_edition
Formulation

Let k {\displaystyle k} be a field (such as the rational numbers) and K {\displaystyle K} be an algebraically closed field extension of k {\displaystyle k} (such as the complex numbers). Consider the polynomial ring k {\displaystyle k} and let I {\displaystyle I} be an ideal in this ring. The algebraic set V (I) {\displaystyle \mathrm {V} (I)} defined by this ideal consists of all n {\displaystyle n} -tuples x = (x 1, …, x n) {\displaystyle \mathbf {x} =(x_{1},\dots,x_{n})} in K n {\displaystyle K^{n}} such that f (x) = 0 {\displaystyle f(\mathbf {x})=0} for all f {\displaystyle f} in I {\displaystyle I}. Hilbert's Nullstellensatz states that if p is some polynomial in k {\displaystyle k} that vanishes on the algebraic set V (I) {\displaystyle \mathrm {V} (I)}, i.e. p (x) = 0 {\displaystyle p(\mathbf {x})=0} for all x {\displaystyle \mathbf {x} } in V (I) {\displaystyle \mathrm {V} (I)}, then there exists a natural number r {\displaystyle r} such that p r {\displaystyle p^{r}} is in I {\displaystyle I}.

An immediate corollary is the weak Nullstellensatz: The ideal I ⊆ k {\displaystyle I\subseteq k} contains 1 if and only if the polynomials in I do not have any common zeros in K. The weak Nullstellensatz may also be formulated as follows: if I is a proper ideal in k, {\displaystyle k,} then V(I) cannot be empty, i.e. there exists a common zero for all the polynomials in the ideal in every algebraically closed extension of k. This is the reason for the name of the theorem, the full version of which can be proved easily from the 'weak' form using the Rabinowitsch trick. The assumption of considering common zeros in an algebraically closed field is essential here; for example, the elements of the proper ideal (X + 1) in R {\displaystyle \mathbb {R} } do not have a common zero in R. {\displaystyle \mathbb {R}.}

With the notation common in algebraic geometry, the Nullstellensatz can also be formulated as

for every ideal J. Here, J {\displaystyle {\sqrt {J}}} denotes the radical of J and I(U) is the ideal of all polynomials that vanish on the set U.

In this way, taking k = K {\displaystyle k=K} we obtain an order-reversing bijective correspondence between the algebraic sets in K and the radical ideals of K. {\displaystyle 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 P = (a 1, …, a n) ∈ K n {\displaystyle P=(a_{1},\dots,a_{n})\in K^{n}}. Then I (P) = (X 1 − a 1, …, X n − a n) {\displaystyle I(P)=(X_{1}-a_{1},\ldots,X_{n}-a_{n})}. More generally,

Conversely, every maximal ideal of the polynomial ring K {\displaystyle K} (note that K {\displaystyle K} is algebraically closed) is of the form (X 1 − a 1, …, X n − a n) {\displaystyle (X_{1}-a_{1},\ldots,X_{n}-a_{n})} for some a 1, …, a n ∈ K {\displaystyle a_{1},\ldots,a_{n}\in K}.

As another example, an algebraic subset W in K is irreducible (in the Zariski topology) if and only if I (W) {\displaystyle I(W)} is a prime ideal.