WEAK NULLSTELLENSATZ

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1 WEAK NULLSTELLENSATZ YIFAN WU, Abstract. We prove weak Nullstellensatz which states if a finitely generated k algebra is a field, then it is a finite algebraic field extension of k. We then use weak Nullstellensatz to prove Hilbert s Nullstellensatz. Contents. Finite Algebra versus Finite Generated Algebra 2. Weak Nullstellensatz 4 3. Hilbert s Nullstellensatz 5 References 7. Finite Algebra versus Finite Generated Algebra Let A and B be two commutative rings with. Let f : A B be a ring homomorphism that sends to. For a A and b B, define a product a b = f(a)b. This product defines a scalar multiplication in the ring B with scalars coming from A. It gives the ring B an A module structure. Definition. (A Algebra). Let A, B, f be as above. The ring B together with the A module structure as described above is said to be an A algebra. Let A, B, B 2 be commutative rings with, and let f : A B, f 2 : A B 2 be two ring homomorphisms that send s to s. B and B 2 are two A algebra as in the above definition. We now define an A algebra homomorphism between B and B 2. Definition.2 (A Algebra Homomorphism). A map h : B B 2 is said to an A algebra homomorphism if the following two conditions hold simutaneously: () h is a ring homomorphism between B and B 2. (2) h is an A module homomorphism between B and B 2. Date: First Version: May 6, 207. Last Update: May 0, 207.

2 Weak Nullstellensatz Lemma.3. Let A, B, B 2, f : A B, f 2 : A B 2 be as above. A map h : B B 2 is an A algebra homomorphism if and only if h is a ring homomorphism and h f = f 2. Proof. ( ) Assume h : B B 2 is an A algebra homomorphism, then for a A h(f (a)) f (a)=f (a) B in B h is A module homomorphism = h(f (a) B ) = h(a B ) = a h( B ) = f 2 (a)h( B ) = f 2 (a) B2 = f 2 (a). ( ) Suppose h f = f 2. We want to show that h : B B 2 is an A module homomorphism. Let a A and b B. h(a b) = h(f (a)b) h is ring homomorphism = h(f (a))h(b) = f 2 (a)h(b) = a h(b). Next we come up to two finite condition that is somehow confusing at first glance. The first one is called finite algebra. The second one is called finitely generated algebra. Definition.4 (Finite A Algebra). Let f : A B be a ring homomorphism and B be an A algebra as above. B is said to be a finite A algebra if B is finitely generated as an A module, i.e., there exists a finite set of elements b,, b n B such that B = A b + + A b n. Definition.5 (Finitely Generated A Algebra). Let f : A B be a ring homomorphism and B be an A algebra as above. B is said to be a finitely generated A algebra if B is finitely generated as a polynomial ring with coefficients from A, i.e., there exists a finite set of elements b,, b n B such that B = A[b,, b n ] = { a (e,,e n) b e b en n : a (e,,e n) A, e,, e n Z 0 }. finite sum Remark.6. Note A b + + A b n A[b,, b n ]. So if B is a finite A algebra then B is a finitely generated A algebra. The converse does not hold in general. Example.7. Let k be a field. The polynomial ring with n variables k[x,, X n ] is a finitely generately k algebra, but not a finite k algebra. Remark.8. Consider the case in which A = k is a field. Then every ring homomorphism from A = k to another ring B is always injective so we can view f : A = k B as an embedding and A = k as a subset of B. In this case, B is a finitely generated k algebra if and only if B = k[b,, b n ] with b,, b n B. We conclude this section with the following result which puts together our previous definitions on finite algebra and on finite generated algebra. 2

3 Yifan Wu, Proposition.9. Suppose A is a Noetherian commutative ring. Let A A 2 A 3 where A 2 and A 3 are also commutative rings. Suppose A 3 satisfies the following two condition simultaneously: () A 3 is a finitely generated A algebra. (2) A 3 is a fnite A 2 algebra. Then A 2 is a finitely generated A algebra. Proof. For this proof, we need some results regarding Noetherian rings and Noetherian modules. All these results can be found at [5]. Let s say c,, c m generate A 3 as A algebra, and d,, d n generate A 3 as A 2 module. More precisely, this means A 3 = A [c,, c m ] ( ) = A 2 d + + A 2 d n, c,, c m, d,, d n A 3. This relation ( ) tells us we can write ( ) c i = j e ij d j, e ij A 2, ( ) d i d j = k e ijk d k, e ijk A 2. Consider A e = A [e ij, e ijk ], i.e., the A algebra generated by all e ij, e ijk from above. We have A A e A 2. By [5, Corollary 3.3], A e is a Noetherian ring. We investigate what ( ) and ( ) tell us. The obveservation is c 2 i ( ) = ( j ( ) = j,k e ij d j ) 2 Binomial Theorem = ê jk d j d k, ê jk A e j,k ê jk e jkl d l = ê jkl d l, ê jkl A e. j,k,l l This is saying that a square c 2 i can be written in linear combination of d l s with coefficients in A e. Of course if we work hard any high power can also be written like this. So we just succesfully argue that every element in A 3 is a linear combination of d l s with coefficients from A e, so A 3 = A e d + + A e d n. By [5, Theorem 2.7], A 3 is a Noetherian A e module. Being a A e submodule, A 2 is finitely generated as A e module, A 2 = A e b + + A e b w, b,, b w A 2, A 2 = A [e ij, e ijk ] b + + A [e ij, e ijk ] b w = A [e ij, e ijk, b,, b w ] is a finitely generately A algebra. 3

4 Weak Nullstellensatz 2. Weak Nullstellensatz Theorem 2. (Weak Nullstellensatz). Let k be a field. Let B = k[b,, b n ] with b,, b n B be a finitely generated k algebra. Suppose B is a field. Then all b,, b n are algebraic over k. Hence B/k is an algebraic field extension. Before proving weak Nullstellensatz, we adapt Euclid s proof of infinitude of primes to obtain the following result: Proposition 2.2. Let k be a field, there are infinitely many irreducible polynomials in the polynomial ring with n variables k[x,, X n ]. Proof. k[x,, X n ] is a unique factorization domain. Suppose there is only finitely many irreducible polynomials f (X,, X n ),, f m (X,, X n ) k[x,, X n ]. Consider f f 2 f m + k[x,, X n ]. It cannot be factorized using f,, f m hence must be irreducible, contradicting that f,, f m are the only irreducible polynomials. Proof of 2.. Consider the field B = k[b,, b n ]. By re-indexing, we can assume b,, b s are algebraically independent over k, and b s+,, b n are algebraic over the subfield k(b,, b s ) B. Suppose on the contrary that B/k is not an algebraic extension. This means s. By hypothesis, A 3 = B is a finitely generated (A = k) algebra. Also note b s+,, b n are algebraic over A 2 so A 3 is a finite A 2 algebra. Hence we may apply.9 to A = k A 2 = k(b,, b s ) A 3 = B = k[b,, b n ], and conclude that A 2 = k(b,, b s ) is a finitely generated k algebra. By definition of finitely generated algebra, this means that This further means that k(b,, b s ) = k[c,, c t ], c,, c t k(b,, b s ). c = f (b,, b s ) g (b,, b s ),, c t = f t(b,, b s ) g t (b,, b s ), f,, f t, g,, g t k[x,, X s ]. Now by 2.2, there exists an irreducible polynomial h k[x,, X s ] which does not divide any of g,, g t k[x,, X s ]. Consider the element h(b,, b s ) k(b,, b s ) = k[c,, c t ]. 4

5 Yifan Wu, Since k[c h(b,,b s),, c t ], it must be the case that for certain C (e,,e t) k, e,, e t 0, = finite sum h(b,, b s ) = finite sum ( ) e f (b,, b s ) C (e,,e t) g (b,, b s ) C (e,,e t)c e c et t ( ) et ft (b,, b s ). g t (b,, b s ) Mutiply both side of this relation with g p g pt t for large enough p,, p t 0, in order to kill the denominators of right hand side, then it leads to a contradiction since h does not divide g,, g t. The conlusion is that it must be the case B/k is an algebraic extension, as desired. 3. Hilbert s Nullstellensatz As far as the author can see, there are a handful of verions of Hilbert s Nullstellensatz. Consider the polynomial ring of n variables k[x,, X n ] with coefficients from a field k. Consider an ideal I k[x,, X n ]. The elements in this ideal I are just polynomials f(x,, X n ). The common zeros of these polynomials form so called algebraic sets, or affine varieties. Definition 3. (Affine Varieties). The set k n is also called affine n space and denoted A n. Assign to each ideal I k[x,, X n ] the set of common zeros V(I) = {(a,, a n ) A n : f(a,, a n ) = 0 f I} A n. Subsets of A n of this form are called affine varieties. Remark 3.2. Hilbert Basis Theorem [5, Theorem 3.] states that every ideal I k[x,, X n ] is finitely generated, i.e., I(f,, f m ). So V(I) = V(f ) V(f m ). This is saying every affine variety is a finite intersection of the zero sets of single polynomial. Definition 3.3 (Ideal of a Set of Points of A n ). Conversely, given a subset X of A n. Assign to X an ideal in k[x,, X n ] I(X) = {f k[x,, X n ] : f vanishs on all of X}. Remark 3.4. It is easy to see for any ideal I k[x,, X n ], the relation I I(V(I)) always holds. What is less obvious at a first glance is that the radical I I(V(I)). The really non-trivial part is that equality holds, i.e., I = I(V(I)). 5

6 Weak Nullstellensatz Theorem 3.5 (Hilbert s Nullstellensatz). Let k be an algebraically closed field. Let I k[x,, X n ] be an ideal. Then I(V(I)) = I. Before proving this version, we might look at another version of Nullstellensatz. Theorem 3.6 (Hilbert s Nullstellensatz). Let k be an algebraically closed field. If V(I) = then I = k[x,, X n ]. Proof. We will deal with the contrapostitive, i.e., if I is a proper ideal inside k[x,, X n ], then the polynomials from I have at least one common zero. A standard result, which depends on Zorn s lemma, says the proper ideal I is contained in a maximal ideal of k[x,, X n ]. Since if the ideal becomes bigger, the set of common zeros only becomes smaller, so we might as well assume I is itself a maximal ideal. In this case, the residue ring B = k[x,, X n ]/I is a field. And it is a field extension of k. k embeds inside B = k[x,, X n ]/I as image of constant polynomials. Let X i denote the image of X i under the canonical homomorphism π : k[x,, X n ] B = k[x,, X n ]/I. Then B = k[x,, X n ] is a finitely generated k algebra. And it is a field. 2. precisely says that X,, X n are all algebraic over k. Since k is assume to be algebraically closed, it means all X,, X n k. So (X,, X n ) k n = A n. The next observation is that given a polynomial f(x,, X n ) I, plugging in X,, X n gives f(x,, X n ) = f(x,, X n ) = 0 B = k[x,, X n ]/I. So (X,, X n ) k n = A n is a common zero for all polynomials from I. We are done. Finally we come to the proof of 3.5. The proof uses so-called Rabinowitsch trick according to [4, Thereom.5, IX, ]. 6

7 Yifan Wu, Proof of 3.5. Let I k[x,, X n ] be an ideal. By 3.4, we know I I(V(I)) already. It remains to show the ( ) direction. By 3.2, assume I = (f,, f m ). Given f I(V(I)). This means f vanishes on common zeros of f,, f m. We want to show that there exists r Z >0 and g,, g m k[x,, X n ] such that ( ) f r (X,, X n ) = f (X,, X n )g (X,, X n )+ +f m (X,, X n )g m (X,, X n ). The so-called Rabinowitsch trick is to induce a new variable X n+ and then consider the following polynomials all as elements in k[x,, X n, X n+ ]: f (X,, X n ),, f m (X,, X n ), X n+ f(x,, X n ) This construction gurantees these m+ polynomials f (X,, X n ),, f m (X,, X n ), X n+ f(x,, X n ) have no common zero. Indeed if the first m of m+ polynomials vanish, then the last one evaluates to be 0. By 3.6, f (X,, X n ),, f m (X,, X n ), X n+ f(x,, X n ) generate the unit ideal k[x,, X n ]. We can find polynomials h,, h m, h k[x,, X n, X n+ ] such that = h (X,, X n, X n+ )f (X,, X n )+ +h m (X,, X n, X n+ )f m (X,, X n ) +h(x,, X n, X n+ ) ( X n+ f(x,, X n ) ) ( ). The second step of the so-called Rabinowitsch trick is to say goodbye to the new variable X n+. Replace X n+ by f(x,,x n) in ( ) throughout, we get = h (X,, X n, f(x,, X n ) )f (X,, X n ) + +h m (X,, X n, f(x,, X n ) )f m(x,, X n ) We can then multiply both side with a large enough power f(x,, X n ) r to kill denominator and conclude ( ) holds, as desired. References [] M. F. Atiyah, I. G. MacDonald, Introduction to Commutative Algebra, Westview, 206. [2] K. Hulek, Elementary Algebraic Geometry, American Mathematical Society, [3] J. Humphreys, Linear Algebraic Groups, Springer-Verlag, New York, 975. [4] S. Lang, Algebra Revised Third Edition, Springer, New York, [5] Y. Wu, Hilbert Basis Theorem, ExpositoryArticles/LinearAlgebraicGroups/Hilbert_Basis_Theorem.pdf. 3, 5 [6] O. Zariski, P. Samuel, Commutative Algebra Volume II, Van Nostrand, Princeton,