aa r X i v : . [ m a t h . A C ] S e p The Nullstellensatz without the Axiom of Choice
Enrique Arrondo
Abstract.
We give a short proof of the most general version of the Nullstellensatz withoutusing the Axiom of Choice.
MSC2010 classification: K , these two resultsare a consequence of the Nullstellensatz (moreover, in this case, the radical of an ideal isthe intersection of the maximal ideals containing it; hence finitely generated K -algebras arewhat is called Jacobian rings). In fact, they can be considered respectively as extensionsof the so-called weak and strong versions of the Nullstellensatz, when the ground field isan arbitrary field (i.e. not necessarily algebraically closed, which is the assumption in theclassical version of the Nullstellensatz).It is thus natural to look for proofs of the Nullstellensatz (in any possible version)avoiding any use of the Axiom of Choice. This has been approached in a more restrictiveway: finding constructive proofs (see [2] Corollary VI-1.12 or [3] Theorem V1-3.5 for the mostgeneral version we are going to deal with). However, these proofs still keep the traditionalapproach of using the notion of integral extensions of rings.The goal of this note is to find simple and short proofs of all these versions of theNullstellensatz, without any use of the Axiom of Choice. The main idea is taken from [1], inwhich we gave an elementary proof of the (classical) Weak Nullstellensatz for algebraicallyclosed fields. We state that main idea separately in Lemma 2, and it will be the keystonefor our main proofs. We want to stress that –forgetting about using or not the Axiom ofChoice– we do not know of any previous simple proof of the most general version of theNullstellensatz.In the case of an algebraically closed field K , the Nullstellensatz implies that maxi-mal ideals of a finitely generated K -algebra are in bijection with the closed points of thecorresponding affine variety. When K is arbitrary, the generalized Nullstellensatz impliesthat maximal ideals correspond now to sets of conjugate points in finite extensions of K (this is the spirit of Theorem 4). Hence, a generalized Weak Nullstellensatz should state theexistence of maximal ideals, while a generalized Strong Nullstellensatz should state that theradical of an ideal is the intersection of the maximal ideals containing the ideal. We willprove all these results without using the Axiom of Choice. Although some of the proofs arewell-known, we will include them here for the sake of completeness.1e want to thank Henri Lombardi for indicating us the references [2] and [3]. Thisresearch was developed in the framework of the project MTM2015-65968-P funded by theSpanish Government.We start with the generalized Noether Normalization Lemma. We reproduce here thestandard proof when K is arbitrary (the case when K is infinite is simple and it is alsostandard, and can be found in [1]). Lemma 1. If K be any field and f is a nonconstant polynomial in K [ x , . . . , x n ] with n ≥ ,then there is an automorphism of K [ x , . . . , x n ] transforming f into a monic polynomial inthe varible x n . Proof.
Let m be larger than any exponent i j of any monomial x i . . . x i n n appearing in f .An automorphism of K [ x , . . . , x n ] leaving invariant x n and mapping each other x i to x i + x m n − − i n sends any monomial x i . . . x i n n to a polynomial whose monomial of highest degreeis x i m n − + ... + i n − m + i n n . Regarding this degree as a number written in base m , it followsthat i m n − + . . . + i n − m + i n ≤ i ′ m n − + . . . + i ′ n − m + i ′ n if and only if ( i , . . . , i n − , i n ) ≤ ( i ′ , . . . , i ′ n − , i ′ n ) in the lexicographical order. Therefore, such automorphism maps f to amonic polynomial in x n of degree i m n − + . . . + i n − m + i n , where ( i , . . . , i n − , i n ) is themaximum of the set of exponents of f when ordered lexicographically.The main technical lemma we will need is the following (which keeps the main idea of[1], and we essentially copy the proof there adapted to our more general context). Lemma 2.
Let I ⊂ K [ x , . . . , x n ] be a proper ideal containing a polynomial g that is monicin the variable x n . If M ′ ⊂ K [ x , . . . , x n − ] is a proper ideal containing I ∩ K [ x , . . . , x n − ] ,then the ideal of K [ x , . . . , x n ] generated by I and M ′ is proper. Proof.
Suppose for contradiction that there exist f in I and f ′ generated by M ′ such1 = f + f ′ . Thus we can write f = f + f x n + . . . + f d x dn , with all the f i in K [ x , . . . , x n − ],and such that f − , f , . . . , f d ∈ M ′ . On the other hand, we can express the monicpolynomial g in the form g = g + g x n + . . . + g e − x e − n + x en with g j in K [ x , . . . , x n − ] for j = 0 , . . . , e − R be the resultant of f and g with respect to the variable x n . In other words, R isthe polynomial in K [ x , . . . , x n − ] given by the determinant R = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f f . . . f d . . . f . . . f d − f d . . .
0. . .0 . . . f f . . . f d − f d g g . . . g e − . . . g . . . g e − g e − . . .
0. . . . . .0 . . . g g . . . g e − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e rows d rows .
2e recall the well known result that R belongs to I : in the above determinant defining R ,add to the first column the second one multiplied by x n , then the third column multiplied by x n , and so on until one adds the last column multiplied by x d + e − n ; developing the resultingdeterminant by the first column shows that R is a linear combination of f and g , so it is in I .Therefore R is a member of I ∩ K [ x , . . . , x n − ], and hence of M ′ . But a direct inspection ofthe determinant defining the resultant shows that, when quotienting modulo M ′ , it reducesto the determinant of a lower-triangular matrix whose entries on its main diagonal are all1s. Hence R is not in M ′ , which is a contradiction.From this, we can obtain the main versions of the Nullstellenstaz: Theorem 3.
Any proper ideal I ⊂ K [ x , . . . , x n ] is contained in a maximal ideal. Proof.
Let us assume I = 0, since otherwise the result is trivial. We prove the theorem byinduction on n . The case n = 1 is immediate, because any nonzero proper ideal I of K [ x ]is generated by a nonconstant polynomial. The ideal generated by any irreducible factor ofsuch polynomial is a maximal ideal containing I .We assume now n >
1. Lemma 1 allows us to suppose that I contains a polynomial g monic in the variable x n . Fixing such a polynomial g , we consider the ideal I ′ := I ∩ K [ x , . . . , x n − ]. Since 1 is not in I , it follows that I ′ is a proper ideal. Therefore, by theinduction hypothesis there is a maximal ideal M ′ of K [ x , . . . , x n − ] containing I ′ .If we consider the field K ′ = K [ x , . . . , x n − ] /M ′ , there exists a natural surjective map K ′ [ x n ] → K [ x , . . . , x n ] / (cid:0) I + ( M ′ ) (cid:1) . By Lemma 2, its kernel J is a proper ideal of K ′ [ x n ],hence K ′ [ x n ] /J , and thus also K [ x , . . . , x n ] / (cid:0) I + ( M ′ ) (cid:1) possesses a maximal ideal. Thepullback to K [ x , . . . , x n ] of such maximal ideal yields a maximal ideal containing I , whichcompletes the proof. Theorem 4.
A prime ideal I ⊂ K [ x , . . . , x n ] is maximal if and only if the quotient K [ x , . . . , x n ] /I is a finite extension of K . Proof.
We first prove that the finite dimension of K [ x , . . . , x n ] /I implies that I is maximal.Equivalently, we need to prove that the quotient K [ x , . . . , x n ] /I is a field. To that purpose,we fix a non-zero element of the quotient, i.e. the class of a polynomial f not in I . Sincethe classes of 1 , f, f , ... are not K -linearly independent modulo I , there is a nontrivialcombination λ + λ f + . . . + λ d f d (with λ , . . . , λ d ∈ K ) that is in I . Since f is not in I and I is prime, we can assume λ = 0. Hence the class of λ − ( − λ − . . . − λ d f d − ) is an inverseof the class of f . This proves that K [ x , . . . , x n ] /I is a field, and hence I is a maximal ideal.We prove now the converse by induction on n , the case n = 1 being trivial, since thequotient of K [ x ] by the ideal generated by an irreducible polynomial has finite dimension.So assume n ≥ I is a maximal ideal. By Lemma 1, we can suppose I to contain a polynomial g that is monic in x n . Since I ∩ K [ x , . . . , x n − ] is proper, becauseit does not contain 1, it is contained in a maximal ideal M ′ ⊂ K [ x , . . . , x n − ] (see Theorem3). By Lemma 2, the ideal generated by I and M ′ is proper. Since I is maximal, this impliesthat M ′ is contained in I . In other words, I ∩ K [ x , . . . , x n − ] = M ′ .By induction hypothesis, the extension K ⊂ K [ x , . . . , x n − ] /M ′ is finite. On the otherhand, the extension K [ x , . . . , x n − ] /M ′ ⊂ K [ x , . . . , x n ] /I is also finite, because of theexistence of a monic polynomial g in I . This proves that K [ x , . . . , x n ] /I is finite over K ,as wanted. Remark 5.
When K is algebraically closed, any finite extension of it is isomorphic to K .Hence any maximal ideal is the kernel of a morphism of K -algebras K [ x , . . . , x n ] → K . Thismorphisms is determined by the images a , . . . , a n of x , . . . , x n , hence it is the evaluationat the point ( a , . . . , a n ) ∈ A nK . In other words, the maximal ideals of K [ x , . . . , x n ] are, inthis case, the ideals of points. In this context, Theorem 3 is saying that, if I ⊂ K [ x , . . . , x n ]is a proper ideal, there is a point in A nK such that all polynomials of I vanish at it. This isthe classical statement of the Weak Nullstellensatz.When K is an arbitrary field, what we get is that maximal ideals are now kernels of amorphism of K -algebras K [ x , . . . , x n ] → K ′ , where K ′ is a finite extension of K , i.e. theycorrespond to points whose coordinates are in a finite extension of K . Of course, any K -automorphism of K ′ (i.e. any element of the Galois group of the extension) provides anothermorphism with the same kernel. Therefore a maximal ideal of K [ x , . . . , x n ] can be regardedas the ideal of a set of conjugate points in a finite extension of K . Theorem 6. If I ⊂ K [ x , . . . , x n ] is a proper ideal, its radical is the intersection of allmaximal ideals containing I . Proof.
It is clear that the radical of I is contained in any maximal ideal containing I , sothat we only need to check that a polynomial f in all maximal ideals containing I is inthe radical of I . Assume, for contradiction, that f is not in √ I . This is equivalent to saythat the K -algebra (cid:0) K [ x , . . . , x n ] /I (cid:1) ¯ f is not zero (where ¯ f stands for the class of f modulo I ). Since this K -algebra is isomorphic to the quotient of K [ x , . . . , x n , x n +1 ] by the idealgenerated by I and x n +1 f −
1, Theorem 3 implies that it possesses a maximal ideal ˜ M .Consider now the kernel P of the natural map K [ x , . . . , x n ] → (cid:0) K [ x , . . . , x n ] /I (cid:1) ¯ f / ˜ M ,which is a prime ideal containing I . Since the image of f is the class of a unit, P doesnot contain f . Hence, we will find the wanted contradiction if we prove that P is in fact amaximal ideal. And this holds by Theorem 4, because K [ x , . . . , x n ] /P can be regarded asa subspace of (cid:0) K [ x , . . . , x n ] /I (cid:1) ¯ f / ˜ M , which is finite dimensional. Remark 7.
According to Remark 5, Theorem 6 can be read, when K is algebraically closed,as saying that the radical of an ideal I ⊂ K [ x , . . . , x n ] is the ideal formed by all polynomialsvanishing at V ( I ) (where V ( I ) is the set of points of A nK that kill all polynomials of I ). Thisis the classical statement of the Strong Nullstellensatz. The proof of Theorem 6 (which isthe standard one) can be considered as the generalization of the classical Rabinovich trickthat allows to prove the strong version of the Nullstellensatz from the weak one.4 eferences. [1] E. Arrondo, Another Elementary Proof of the Nullstellensatz , Amer. Math. Monthly113 (2006), no. 2, 169-171.[2] H. Lombardi, C. Quitt´e,
Commutative algebra: Constructive methods. Finite projectivemodules , arXiv: 1605.04832.[3] R. Mines, F. Richman, W. Ruitenburg,