Recall the statements of the the weak and strong forms Hilberts's Nullstellensatz and of Noether's Nomalisation Lemma.

Theorem 1.1. (Hilbert's Nullstellensatz (weak form) Let K be an algebraically closed field and M be a maximal ideal of K[X1,…,Xn], then M=<X1−a1,…,Xn−an> for some ai∈K.

Theorem 1.2. (Hilbert's Nullstellensatz (full form) Let K be an algebraically closed field and f,g1,…,gm∈K[X1,…,Xn], then there exists a natural number k such that fk∈<g1,…,gm> if and only if for each (a1,…,an)∈Kn with gi(a1,…,an)=0 for 1≤i≤n, implies that f(a1,…,an)=0.

Theorem 1.3. (Noether's Normalization Lemma) Let K be an arbitrary field and T=K[α1,…,αn] be an integral domain, then there exist β1,…,βm in T such that β1,…,βm are algebraically independent over K and T is a finitely generated as a module over the subring S:=K[β1,…,βm]. In particular, T is integral over S and m is the transcendence degree of K(α1,…,αn) over K.

Usually, the weak form of Hilbert's Nullstellensatz is proved by some technical and classical facts such as Noether's Normalization Lemma, Zariski's Lemma, Artin-Tate Lemma and the concept of G-domains and Hilbert rings. Good references for this are [5, § 31], and the set of references it gives on p. 433; see also [1]. Proofs of the full form of Hilbert's Nullstellensatz are usually depend on adding a new variable, or using Rabinowiteh's trick, or by the concept of Hilbert rings, see [7], [4] and [6], respectively. For the general version of Noether's Normalization see [2, Lemma 10.10.1] or [7, Theorem 14.14].

In this short note we prove that Noether's Normalization Lemma also implies the full form of Hilbert's Nullstellensatz Theorem.

Let us recall some standard definitions and facts from commutative ring theory which will be used in this note. For a ring R, the Jacobson radical of R is denoted by J(R). It is clear that if R⊆T is an integral extension of rings, then J(R)=R∩J(T). If B⊆A is an algebraic extension of integral domains, and I be a nonzero ideal of A, then one can easily see that I∩B≠0. In particular, if R⊆T is an integral extension of integral domains with J(T)≠0, then J(R)≠0. Finally note that it is manifest that if K is an arbitrary field and R=K[X1,…,Xn] is the polynomial ring of algebraically independent variables X1,…,Xn over K, then J(R)=0, for a more general result see [3].

Proof. Hilbert's Nullstellensatz (full form). Note that it is clear that if there exists a natural number k such that fk∈<g1,…,gm>, then for each (a1,…,an)∈Kn with gi(a1,…,an)=0 for 1≤i≤m, we have f(a1,…,an)=0. Conversely, assume that for each natural number k, we have fk∉I:=<g1,…,gm>. Therefore I≠R:=K[X1,…,Xn] and so if S:={1,f,f2,…,fk,…}, then we infer that I∩S=∅. Hence there exists a prime ideal P of R such that I⊆P, P∩S=∅ and P is maximal with respect to these properties. We claim that P is a maximal ideal of R. Otherwise, note that if P is not a maximal ideal of R, then for each maximal ideal Q of R which contains P, we infer that f∈ Q, by the maximality of P. Hence, 0≠f+P∈J(RP)≠0. Let A:=RP=K[α1,…,αn], where αi=Xi+P. Thus by our assumption A is an integral domain which is not a field and therefore A is not algebraic over K. Now by Noether's Normalization Lemma, there exist β1,…,βr in A such that β1,…,βr are algebraically independent over K and B is a finitely generated as a module over the subring B:=K[β1,…,βr]. Therefore B⊆A is an integral extension of integral domains and since A is not algebraic over K we infer that r≥ 1. Now since J(A)≠0, we immediately conclude that J(B)≠0 which is absurd. Thus P is a maximal ideal of R. Therefore by the weak form of Hilbert Nullstellensatz, we infer that there exist a1,…,an in K such that P=<X1−a1,⋯,Xn−an>. Since, I⊆P, we deduce that for each i, gi(a1,…,an)=0 and since f∉P, we infer that f(a1,…,an)≠0 which is absurd. Thus there exists a natural k, such that fk∈I and hence we are done.

This work was supported by Research Council of Shahid Chamran University of Ahvaz (Ahvaz-Iran) Grant Number: SCU.MM1400.721.